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Chapter 3, , Graphing and, Maximum Minimum, -, , Differential calculus provides tests for locating the key features of graphs., Now that we know how to differentiate, we can use this information to assist, us in plotting graphs. The signs of the derivative and the second derivative of, a function will tell us which way the graph of the function is "leaning" and, "bending.", Using the derivative to predict the behavior of graphs helps us to find the, points where a function takes on its maximum and minimum values. Many, interesting word problems requiring the "best" choice of some variable involve, searching for such points., In Section 3.1, we study the geometric aspects of continuity. This will, provide a useful introduction to graphing. In Section 3.6, we use the ideas of, maxima and minima to derive an important theoretical result-the mean, value theorem. One consequence of this theorem is a fact which we used in, connection with antiderivatives: a function whose derivative is zero must be, constant., , 3.1 Continuity and the, Intermediate Value Theorem, If a continuous function on a closed interval has opposite signs at the endpoints, it, must be zero at some interior point., In Section 1.2, we defined continuity as follows: "A function f ( x ) is said to be, continuous at x,, if lin,,,o f ( x ) = f(x,)." A function is said to be continuous on, a given interval if it is continuous at every point on that interval. If a function f, is continuous on the whole real line, we just say that "f is continuous." An, imprecise but useful guide is that a function is continuous when its graph can, be drawn "without removing pencil from paper." In Figure 3.1 .l, the curve on, the left is continuous at x,, while that on the right is not., , Copyright 1985 Springer-Verlag. All rights reserved.

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"140, , Chapter 3 Graphing and Maximum-Minimum Problems, , a discontinuous curve, (right)., , Example 1, , I, , 0, , I, , +, , ,"0, , Y, , X0, , I, , Decide where each of the functions whose graphs appear in Fig. 3.1.2 is, continuous. Explain your answers., , Ftwe 3.1.2. Where are, , these functions continuous?, , Solution (a) This function jumps in value at each of the points x , = 0, x , = 2 1 ,, x0 = f2, . . . , so limx,xo f ( x ) does not exist at these points and thus f is, not continuous there; however, f is continuous on each of the intervals, between the jump points., ( b ) This function jumps in value at x, = - 1 and x , = 1 , and so, lim,,,,, f ( x ) and limx,-, f ( x ) do not exist. Thus f is not continuous at, x , = +- 1; it is continuous on each of the intervals (- m, - l), (- 1, l), and, , +, , (1900)., , (c) Even though this function has sharp corners on its graph, it is continuous;, lim,,xo f ( x ) = f(x,) at each point x,., (d) Here lim,,, f ( x ) = 1, so the limit exists. However, the limit does not equal, f(1) = 2. Thus f is not continuous at x , = 1. It is continuous on the, intervals ( - m , 1) and ( 1 , m ) . A, , In Section 1.2, we used various limit theorems to establish the continuity of, functions that are basic to calculus. For example, the rational function rule for, limits says that a rational function is continuous at points where its denominator, does not vanish., Example 2, Solution, , Example 3, , Show that the function f ( x ) = ( x - 1)/3x2 is continuous at x ,, , = 4., , This is a rational function whose denominator does not vanish at x , = 4, so it, is continuous by the rational function rule. A, Let g ( x ) be the step function defined by, , Show that g is not continuous at x, = 0. Sketch., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.1 Continuity and the Intermediate Value Theorem, , 141, , Figure 3.1.3. This step, , function is discontinuous at, x,, = 0., Solution, , The graph of g is shown in Fig. 3.1.3. Since g approaches (in fact, equals) 0 as, x approaches 0 from the left, but approaches 1 as x approaches 0 from the, right, lim,,,g(x), does not exist. Therefore, g is not continuous at x, = 0. A, , Example 4, , Using the laws of limits, show that iff and g are continuous at x,, so is fg., , Solution, , We must show that limXjxo( fg)(x) = (fg)(x,). By the product rule for limits,, limxjxo[f(x)g(x)l = Ilim,,,of(x>l[limx jxog(x)l.= f(xo)g(xo), since f and g are, continuous at x,. But f(x,)g(x,) = (fg)(x,), and so lim,,,o< fg>(x) = (fg)(xo),, as required. A, In Section 1.3, we proved the following theorem: f is differentiable at x,,, then f is continuous at x,. Using our knowledge of differential calculus, we can, use this relationship to establish the continuity of additional functions or to, confirm the continuity of functions originally determined using the laws of, limits., , Example 5, , (a) Show that f(x), continuous?, , = 3x2/(x3-, , (b) Show thaiflx) =, Solution, , 2) is continuous at x,, is continuous at x, , =, , I . Where else is it, , = 0., , (a) By our rules for differentiation, we see that this function is differentiable, at x, = 1; indeed, x3 - 2 does not vanish at x, = 1. Thus f is also, continuous at x, = 1. Similarly, f is continuous at each x, such that, x i - 2.5: 0, i.e., at each x, # 3$., (b) This function is the composition of the square root function h(u) = & and, the function g(x) = x 2 + 2x + 1; f(x) = h(g(x)). Note that g(0) = 1 > 0., Since g is differentiable at any x (being a polynomial), and h is differentiable at u = l , f is differentiable at x = 0 by the chain rule. Thus f is, continuous at x = 0. A, According to our previous discussion, a continuous function is one whose, graph never "jumps." The definition of continuity is local since continuity at, each point involves values of the function only near that point. There is a, corresponding global statement, called the intermediate value theorem, which, involves the behavior of a function over an entire interval [a, b]., , Let f be continuous1 on [a, b] and suppose that, for some number c,, f ( a ) < c < f (b) or f ( a ) > c >f (b). Then there is some point x , in (a, b), such that f (x,) = c., , ' Our definition of continuity on [ a ,b ] assumes that f is defined near each point 2 of [ a ,b ] ,, including the endpoints, and that lim,, f ( x ) =f(SZ). Actually, at the endpoints, it is enough to, assume that the one-sided limits (from inside the interval) exist, rather than the two-sided ones., , Copyright 1985 Springer-Verlag. All rights reserved.

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142, , Chapter 3 Graphing and Maximum-Minimum Problems, , In geometric terms, this theorem says that for the graph of a continuous, function to pass from one side of a horizontal line to the other, the graph must, meet the line somewhere (see Fig. 3.1.4). The proof of the theorem depends on, a careful study of properties of the real numbers and will be omitted. (See the, references listed in the Preface.) However, by drawing additional graphs like, those in Fig. 3.1.4, you should convince yourself that the theorem is reasonable., , Figure 3.1.4. The graph off, must pierce the horizontal, liney = c if it is to get, across., , Example 6, , Solution, , Show that there is a number x, such that x i - x,, , = 3., , Let f(x) = x5 - x. Then f(0) = 0 and f(2) = 30. Since 0 < 3 < 30, the intermediate value theorem guarantees that there is a number x, in (0,2) such that, f(x,) = 3. (The function f is continuous on [O, 21 because it is a polynomial.) A, Notice that the intermediate value theorem does not tell us how to find the, number x, but merely that it exists. (A look at Fig. 3.1.4 should convince you, that there may be more than one possible choice for x,.) Nevertheless, by, repeatedly dividing an interval into two or more parts and evaluating f(x) at, the dividing points, we can solve the equation f(xo) = c as accurately as we, wish. This method of bisection is illustrated in the next example., , @ Example 7, , (The method of bisection) Find a solution of the equation x5 - x = 3 in (0,2), to within an accuracy of 0.1 by repeatedly dividing intervals in half and, testing each half for a root., , Solution, , In Example 6 we saw that the equation has a solution in the interval (0,2). To, locate the solution more precisely, we evaluate f(l) = l 5 - 1 = 0. Thus f(1) < 3, < f(2), so there is a root in (1,2). Now we bisect [1,2] into [ l , 1.51 and [1.5,2], and repeat: f(1.5) x 6.09 > 3. so there is a root in (1, 1.5); f(1.25) = 1.80 < 3,, so there is a root in (1.25, 1.5); f(1.375)% 3.54 > 3, so there is a root in (1.25,, 1.375); thus x, = 1.3 is within 0.1 of a root. Further accuracy can be obtained, by means of further bisections. (Related techniques for root finding are, suggested in the exercises for this section. Other methods are presented in, Section 1 1.4.) A, There is another useful way of stating the intermediate value theorem (the, contrapositive statement)., , (Second Version), Let f be a function which is continuous on [a, b] and suppose that, f(x) # c for all x in [a, b]. If f(a) < c, then f(b) < c as well. (See Fig., 3.1.5.) Similarly, if f(a) > c, then f(b) > c as well., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.1 Continuity and the Intermediate Value Theorem, , 143, , y=c, , Figure 3.1.5. The graph, starts below the line y = c, and never pierces the line,, so it stays below the line., , In geometric language, this second version of the theorem says: "The graph of, a continuous function which never meets a horizontal line must remain on one, side of it." The first version says: "'If the graph of a continuous function, passes from one side of a horizontal line to the other, the graph must meet the, line somewhere." You should convince yourself that these two statements, really mean the same thing., In practice, the second version of the intermediate value theorem can be, useful for determining the sign of a function on intervals where it has no roots,, as in the following example., Example 8, , Suppose that f is continuous on [0,3],that f has no roots on the interval, and, that f(0) = 1. Prove that f ( x ) > 0 for all x in [O,31., , Solution, , Apply the intermediate value theorem (version 2), with c = 0, to f on 10, b] for, each b in (0,3].Since f is continuous on [0,31, it is continuous on [O, b ] ;since, f(0) = 1 > 0, we have f(b) > 0. But b was anything in (0,3],so we have proved, what was asked. A, , - -, , Exercises for Sectlola 3.4, 1. Decide where each of the functions whose graph, is sketched in Fig. 3.1.6 is continuous., , these continuous?, , 2. Which of the functions in Fig. 3.1.7 are continuous at x, = l?, , Figure 3.1.7. Which of, these functions are, continuous at x, = I ?, , (a, , (b), , (c), , Copyright 1985 Springer-Verlag. All rights reserved.

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144, , Chapter 3 Graphing and Maximum-Minimum Problems, , 3. Show that f ( x ) = ( x 2+ 1)(x2- I ) is continuous, at x, = 0., 4. Show that f ( x ) = x 3 + 3 x 2 - 2x is continuous at, x, = 3., 5. Prove that ( x 2- 1 ) / ( x 3 3 x ) is continuous a t, x o = 1., 6. Prove that ( x 4 - 8 ) / ( x 3+ 2 ) is continuous at, x, = 0., 7. Where is ( x 2- l ) / ( x 4 + x 2 + 1 ) continuous?, 8. Where is ( x 4+ I ) / ( x 3 - 8 ) continuous?, 9. Let f ( x ) = ( x 3 + 2 ) / ( x 2- 1). Show that f is continuous on [ - f ,$]., 10. Is the function ( x 3 - 1 ) / ( x 2 - I ) continuous a t, I ? Explain your answer., 11. Let f ( x ) be the step function defined by, , +, , Show that f is discontinuous at 0., 12. Let f ( x ) be the absolute value function: f ( x ) = 1x1;, that is,, , Show that f is continuous a t x , = 0., 13. Using the laws of limits, show that iff and g are, continuous at xo, so is f + g., 14. I f f and g are continuous at x , and g(x,) + 0,, show that f / g is continuous at x,., 15. Where is f ( x ) = 8 x 3 / J n continuous?, 16. Where is f i x ) = 9 x 2 - 3 x / J x 4 - 2 x 2 - 8 continuous?, 17. Show that the equation - s5 + s2 = 2s - 6 has a, solution., 18. Prove that the equation x 3 + 2x - 1 = 7 has a, solution., 19. Prove that f ( x ) = x 8 3x4 - 1 has at least two, distinct zeros., 20. Show that x 4 - 5 x 2 + 1 has at least two distinct, zeros., 21. The roots of f ( x ) = x 3 - 2 x - x 2 + 2 are fl,, - fl, and 1. By evaluating f ( - 3 ) , f(O), f(1.3),, and f(2), determine the sign of f ( x ) on each of, the intervals between its roots., @ 22. Use the method of bisection to approximate fl, to within two decimal places. [Hint: Let f ( x ) =, x 2 - 7. What should you use for a and b?], 23. Find a solution of the equation x 5 - x = 3 to an, accuracy of 0.01., @ 24. Find a solution of the equation x 5 - x = 5 to an, accuracy of 0.1., 25. Suppose that f is continuous on [ - 1, 11 and that, f ( x ) - 2 is never zero on [ - 1, 11. I f f ( 0 ) = 0 ,, show that f ( x ) < 2 for all x in [- 1 , I]., 26. Suppose that f is continuous on [3,5] and that, f ( x ) # 4 for all x in [3,5].If f(3) = 3, show that, f(5) < 4., , 27. Let f ( x ) be 1 if a certain sample of lead is in the, solid state at temperature x ; let f ( x ) be 0 if it is, in the liquid state. Define x , to be the melting, point of this lead sample. Is there any way to, define f(x,) so as to make f continuous? Give, reasons for your answer, and supply a graph, for f., 28. An empty bucket with a capacity of 10 liters is, placed beneath a faucet. At time t = 0, the faucet, is turned on, and water flows from the faucet a t, the rate of 5 liters per minute. Let V ( t ) be the, volume of the water in the bucket at time t., Present a plausible argument showing that V is a, continuous function on ( - w , w ) . Sketch a, graph of V.Is V differentiable on ( - w , w ) ?, 29. Let f ( x ) = ( x 2- 4 ) / ( x - 2), x f 2. Define f(2), so that the resulting function is continuous a t, x = 2., 30. Let f ( x ) = ( x 3- l ) / ( x - I), x 1. Wow should, f ( l ) be defined in order that f be continuous at, each point?, 31. Let f i x ) be defined by, , How can you definef(x) on the interval [ I , 31 in, order to make f continuous on ( - m, m)? (A, geometr; argument will suffice.), Let f ( x ) = x + ( 4 / x ) for x G - and x >, 2., Define f ( x ) for x in (- f ,2) in such a way that, the resulting function is continuous on the whole, real line., Let f ( x ) be defined by f ( x ) = ( x 2- l ) / ( x - 1 ), for x += I . Wow should you define f(1) to make, the resulting function continuous? [Hint: Plot a, graph of f ( x ) for x near 1 by factoring the, numerator.], Let f ( x ) be defined by f ( x ) = l / x for .x # 0. Is, there any way to define f(0) so that the resulting, function will be continuous?, Sketch the graph of a function which is continuous on the whole real line and differentiable, everywhere except at x = 0, 1,2,3,4,5,6,7,8,9,, 10., Is a function which is continuous at xo necessarily differentiable there? Prove or give a counterexample., The function f ( x ) = I / ( x - 1) never takes the, value zero, yet f(0) = - 1 is negative and f(2) = 1, is positive. Why isn't this a counterexample to, the intermediate value theorem?, "Prove" that you were once exactly one meter, tall., , -, , 32., , 33., , +, , 34., , 35., , 36., , 37., , 38., , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.2 Increasing and Decreasing Functions, , In Exercises 39-42, use the method of bisection to find, a root of the given function, on the given interval, to, the given accuracy., F?j 39. x3 - 11 on [ 2 , 3 ] ;accuracy & ., @ 40. x 3 + 7 on [ - 3 , - 11; accuracy &., 41. x5 x2 + 1 on [ - 2 , - 11; accuracy &., @ 42. x4 - 3 x - 2 on [I, 21; accuracy &., +43. In the method of bisection, each estimate of the, solution of f ( x ) = c is approximately twice as, accurate as the previous one. Examining the list, of powers of 2: 2,4,8,16,32,64,128,256,512,, 1024, 2048, 4096, 8192, 16384, 32768, 65536,, 131072, . . . -suggests that we get one more decimal place of accuracy for every three or four, repetitions of the procedure. Explain., *44. Using the result of Exercise 43, determine how, many times to apply the bisection procedure to, guarantee the accuracy A for the interval [ a ,b ] if:, (a) A = & ; [ a , b ] = I3,41., (b) A =&; [ a , b ] = [ - 1 , 3 ] ., (c) A = & ; [ a , b ] = [ 1 1 , 2 3 ] ., (d) A = & ; [ a ,b ] = [0.1,0.2]., *45. Can you improve the method of bisection by, choosing a better point than the one halfway, between the two previous points? [Hint:If f ( a ), and f ( b ) have opposite signs, choose the point, where the line through ( a , f ( a ) ) and ( b , f ( b ) ), crosses the x axis.] Is there a method of division, more appropriate to the decimal system?, Do some experiments to see whether your, method gives more accurate answers then the, , +, , +46., , *47., , *48., , *49., , 145, , bisection method in the same number of steps. If, so, does the extra accuracy justify the extra time, involved in carrying out each step? You might, wish to compare various methods on a competitive basis, either with friends, with yourself by, timing the calculations, or by timing calculations, done on a programmable calculator or microcomputer., prove that iff is continuous on an interval I (not, necessarily closed) and f ( x ) + 0 for all x in I ,, then the sign of f ( x ) is the same for all x in I., Suppose that f is continuous at xo and that in, any open interval I containing xo there are points, x , and x2 such that f ( x , ) < 0 and f ( x 2 ) > 0., Explain why f(xo) must equal 0 ., (a) Suppose that f and g are continuous on the, real line. Show that f - g is continuous. (b) Suppose that f and g are continuous functions on the, whole real line. Prove that if f ( x ) + g ( x ) for all, x , and f(0) > g(O), then f ( x ) > g ( x ) for all x ., Interpret your answer geometrically., Prove that any odd-degree polynomial has a root, by following these steps., 1. Reduce the problem to showing that f ( x ), = x n + an-,xn-', . . . + a , x + a. has a, root, where ai are constants and n is odd., 2. Show that if 1x1 > 1 and 1x1 > 2 x {laol, - . . + lan-,I), then f ( x ) / x n is positive., 3. Conclude that f ( x ) < 0 if x is large negative, and f ( x ) > 0 if x is large positive., 4 . Apply the intermediate value theorem., Show that the polynomial x4 + bx + c = 0 has a, (real) root if 256c3 < 27b4. M;nZ:, , +, , +, , *So., , +b\&, , %Me, , .te,pcYe&, , Eng 'fi k a r t z e n k d, , 4, %, .ikc ~ * W L " &, Uh, , 3.2 Increasing and, Decreasing Functions, The sign of the derivative indicates whether a function is increasing or decreasing., , We begin this section by defining what it means for a function to be, increasing or decreasing. Then we show that a function is increasing when its, derivative is positive and is decreasing when its derivative is negative. Local, maximum and minimum points occur where the derivative changes sign., We can tell whether a function is increasing or decreasing at x, by seeing, how its graph crosses the horizontal line y = f(x,) at x, (see Fig. 3.2.1). This, f i s increasing, , //, , 1s neitller increasing nor, decreasing at . x i, , / is decreasing a t u 2, Y, , =f(x), f is increasing, , the function f is increasing, and decreasing., , I, , I, , ", , I, , I, , .XI, , .Y2, , *$, , p-&pb., , *, .y, , 3, , X, , Copyright 1985 Springer-Verlag. All rights reserved.

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148, , Chapter 3 Graphing and Maximum-Minimum Problems, , geometric picture is the basis of the precise definition of increasing and, decreasing. Note that at the point x , in Fig. 3.2.1, f ( x ) is Iess than f(x,) for x, just to the left of x,, while f ( x ) is larger than f(x,) for x just to the right. We, cannot take x too far to the right, as the figure shows. The following, paragraph gives the technical definition., We say that a function f is increasing at x , if there is an interval ( a ,b ), containing xo such that:, 1. If a < x < x,, then f ( x ) < f(xo)., 2. If x , < x < b, then f ( x ) > f(xo)., , Similarly, f is decreasing at x, if there is an interval ( a ,b ) containing x,, such that:, 1. If a < x < x,, then f ( x ) > f(xo)., 2. If x , < x < b, then f ( x ) < f(xo)., The purpose of the interval ( a ,b ) is to limit our attention to a small region, about x,. Indeed, the notions of increasing and decreasing at x , are local;, they depend only on the behavior of the function near x,. In examples done, "by hand," such as Example 1 below, we must actually find the interval ( a ,b)., We will soon see that calculus provides an easier method of determining, where a function is increasing and decreasing., Example I Show that f ( x ) = x 2 is increasing at x,, , Choose ( a ,b ) to be, say, (1,3). If 1 < x < 2, we have f ( x ) = x 2 < 4 = x i . If, 2 < x < 3, then f ( x ) = x 2 > 4 = x i . We have verified conditions 1 and 2 of, the definition, so f is increasing at x, = 2. A, , Solullon, , Y, , 4, "j''', , P, Y, , 3-2-2. This, changes sign from negative, to positive at x,., , Figure 3.2.3. This function, changes sign from positive, to negative at x,., , = 2., , Of special interest is the case f(xo) = 0. Iff is increasing at such an x,, we say, that f changes sign from negative to positive at x,. By definition this occurs, when f(x,) = 0 and there is an open interval ( a ,b ) containing xo such that, f ( x ) < 0 when a < x < x , and f ( x ) > 0 when x , < x < b. (See Fig. 3.2.2.), Similarly if f(xo) = 0 and f is decreasing at x,, we say that f changes sign from, positive to negative at xo. (See Fig. 3.2.3.), Notice that the chosen interval ( a ,b) may have to be small, since a, function which changes sign from negative to positive may later change back, from positive to negative (see Fig. 3.2.4)., Changes of sign can be significant in everyday life as, for instance, when, the function b = f ( t ) representing your bank balance changes from positive to, negative. Changes of sign will be important for mathematical reasons in the, next few sections of this chapter; it will then be useful to be able to determine, when a function changes sign by looking at its formula., , Figure 3.2.4. This function, changes sign from negative, to positive at x2 and x4 and, from positive to negative at, x , , x3 and x6; it does, neither at x5., , 1, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.2 Increasing and Decreasing Functions, , Example 2, , Solution, , ,,, , ', , = 3.u, , -, , 5, , 147, , Where does f ( x ) = 3x - 5 change sign?, We begin by finding those x for which f ( x ) is negative and those for which it, is positive. First, f is negative when, , 3x-5<0,, 3 x < 5,, x <$., (If you had difficulty following this argument, you may wish to review the, material on inequalities in Section R. 1.) Similarly, f is positive when x > $ . So, f changes sign from negative to positive at x = $. (See Fig. 3.2.5.) Here the, chosen interval ( a ,b ) can be arbitrarily large. A, , .., , Figure 3.2.5. This function, changes sign from negative, to positive at x = 5/3., , Example 3, Solution, , If a function is given by a formula which factors, this often helps us to find, the changes of sign., , Where does f ( x ), , = x2 -, , 5x, , + 6 change sign?, , We write f ( x ) = ( x - 3)(x - 2). The function f changes sign whenever one of, its factors does. This occurs at x = 2 and x = 3. The factors have opposite, signs for x between 2 and 3, and the same sign otherwise, so f changes from, positive to negative at x = 2 and from negative to positive at x = 3. (See Fig., 3.2.6.) A, , Figure 3.2.6. This function, changes sign from positive, to negative at x = 2 and, from negative to positive at, x = 3., , Given a more complicated function, such as x 5 - x3 - 2x2, it may be difficult, to tell directly whether it is increasing or decreasing at a given point. The, derivative is a very effective tool for helping us answer such questions., A linear function I(x) = mx b is increasing everywhere when the slope, rn is positive. If the derivative f ' ( x ) of a function f is positive when x = x,, the, linear approximation to f at x , is increasing, so we may expect f to be, increasing at x, as well. (See Fig. 3.2.7.), , +, , Figure 3.2.7. The function f, is increasing where its, derivative i ? positive., , c, , x, , Copyright 1985 Springer-Verlag. All rights reserved.

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148, , Chapter 3 Graphing and Maximum-Minimum Problems, , We can verify that a function is increasing where its derivative is positive, by using difference quotients. Ect y = f ( x ) and assume that f'(xo) > 0. Then, , for Ax sufficiently small, since Ay/Ax approaches the positive number f'(x,), as Ax approaches zero. Thus, there is an interval (a,b) about x, such that, Ax = x - x, and Ay = f ( x ) - f(x,) have the same sign when x is in the, interval. Thus, when a < x < x,, Ax is negative and so is Ay; hence, f ( x ) < f(xo). Similarly, f ( x ) > f(xo) when x, < x < b. Thus f is increasing at, x,. If f'(x,) < 0, a similar argument shows that f is decreasing at x,., The functions y = x3, y = - x3, and y = x2 show that many kinds of, behavior are possible when f'(xo) = 0 (see Fig. 3.2.8)., , j''(0) = 0 and, j' is increasing, , ( ' ( 0 ) = 0 and, , f is decreasing, a t xo = 0., , j'i0, = 0, and 1 1s, neither increasing, nor decreasing a t ,yo = 0, , Figuve 3.2.8. If f'(x,) = 0,, you cannot tell iff is, increasing or decreasing, without further, information., , Thus we arrive at the following test (see Fig. 3.2.9)., , 1. If f'(x,) > 0, f is increasing at x,., 2. If f'(x,) < 0, f is decreasing at x,., 3. If f'(x,) = 0, the test is inconclusive., , Figure 3.2.9. f ' ( x , ) > 0; f is, increasing at x , .f'(x,) < 0;, f is decreasing at x,., f'(x3) = f'(x4) = 0; f is, neither increasing nor, decreasing at x3 and x,., f'(x5) = 0;f is decreasing at, x 5 . f'(x6) = 0; f is, increasing at x,., , Example 4, , (a) Is x 5 - x 3 - 2x2 increasing or decreasing at -2? (b) Is g(s) = i s 2 - s, increasing or decreasing at s = 2?, (a) Letting f ( x ) = x5 - x3 - 2x2, we have f'(x) = 5x4 - 3x2 - 4 x , a n d Y ( - 2 ), = 5(-2), - 3(-2)2 - 4 ( - 2) = 80 - 12 + 8 = 76, which is positive. Thus x 5 x3 - 2x2 is increasing at -2., (b) By the chain rule, g'(s) = (2s - 1)/2&-,, so g'(2) = 3 / 2 0 > 0. Thus g, is increasing at s = 2. A, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.2 Increasing and Decreasing Functions, , Example 5, Solution, , Let f ( x ) = x 3 - x 2 + x, , + 3. How does f, , change sign at x, , 149, , = - l?, , Notice that f ( - 1) = 0. Also, f'(- 1) = 3(- 1)2 - 2(- 1) + 1 = 6 > 0, so f is, increasing at x = - 1. Thus f changes sign from negative to positive. A, We can also interpret the increasing-decreasing test in terms of velocities and, other rates of change. For instance, if f ( t ) is the position of a particle on the, real-number line at time t , and f'(t,) > 0, then the particle is moving to the, right at time to; if f'(to) < 0, the particle is moving to the left (see Fig. 3.2.10)., , Figure 3.2.10. Positive, velocity means that the, motion is to the right, and, negative velocity means, that the motion is to the, left., , Example 6, Solution, , Direction of, motlo11 a t I,), , \, , ., ., f '(1,) > 0, , f"(to), , <0, , The temperature at time t is given by f ( t ) = ( t + 1)/(t - 1) for t, getting warmer or colder at t = 0?, , < 1., , Is it, , We calculate f'(t) by the quotient rule:, , Since f'(0) = - 2 is negative, f is decreasing, so it is getting colder. A, Instead of focusing our attention on the small intervals used so far, one can, also consider the idea of increasing and decreasing functions on general, intervals, which could be large., Let f be a function defined on an interval I. If f ( x , ) < f(x2) for all, x i < x2 in I , we say that f is increasing on I. If f ( x , ) > f ( x 2 ) for all x , < x2 in, I , we say that f is decreasing on I., It is plausible that i f f is increasing at each point of an interval, then f is, increasing on the whole interval in this new sense. We shall use this important, fact now, deferring the formal proof until Section 3.6., , Example 7, Solution, , On what intervals is f ( x ) = x3 - 2x, , + 6 increasing or decreasing?, , We consider the derivative f'(x) = 3x2 - 2. This is positive when 3x2 - 2 > 0,, i.e., when x2 > 2/3, i.e., either x, or x < Similarly, f'(x) < 0, when x 2 < 2/3, i.e.,, intervals, , (- x - ), , m., , >J2/3, J2/3 < x <J2/3. Thus,, and, , f is increasing on the, ( m , x ) , and f is decreasing on, , (-J2/3,J2/3).A, The result of Example 7 enables us to make a good guess at what the graph, y = f ( x ) looks like. We first plot the points where x =, as in Fig., 3.2.1 l(a). When x = +, we get y = 6 T 4 6 / 9 . We also plot the point, , m,, , + J2/3, , J2/3), , (J2/3,, , x = 0, y = 6. Since f is increasing on (- x , and on, x),, and decreasing elsewhere, the graph must look something like that in Fig., 3.2.11(b). Later in this chapter, we will use techniques like this to study, graphing more systematically., , Copyright 1985 Springer-Verlag. All rights reserved.

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150, , Chapter 3 Graphing and Maximum-Minimum Problems, , ( a ) These points lie on the, g r a p l i } = ~3, ?r + 6., , Figure 3.2.11. First steps in, sketching a graph., , ( b ) We draw f'increasing o n, , and decreas~tig on, , Match each of the functions in the left-hand column of Fig. 3.2.12 with its, derivative in the right-hand column., , Example 8, , Functions, , Derivatrves, , F i p 3.2.12. Matching, , functions and their, derivatives., Solution, , Function (1) is decreasing for x < 0 and increasing for x > 0. The only, functions in the right-hand column which are negative for x < 0 and positive, for x > 0 are (a) and (6). We notice, further, that the derivative of function 1, is not constant for x < 0 (the slope of the tangent is constantly changing),, which eliminates (a). Similar reasoning leads to the rest of the answers, which, are: (l)-(c), (2)-(b), (3)-(e), (4)-(a), (5)-(d). A, We now turn our attention to the points which separate the intervals on which, a function is increasing or decreasing, such as the points x =, in Fig., , + J2/3, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.2 Increasing and Decreasing Functions, , 351, , 3.2.11. In this figure, we see that these points are places where f has a, maximum and a minimum. Here are the formal definitions: A point x, is, called a local minimum point for f if there is an open interval (a, b) around x,, such that f ( x ) 2 f(x,) for all x in (a, b). Similarly, a point x, is called a, local maximum point for f if there is an open internal ( a , b) around x, such, that f ( x ) <f (x,) for all x in ( a , b).* (See Fig. 3.2.13.), , Figure 3.2.13. The graph of, maximum point, , flies above the line, y = f(x,) near a local, , minimum point (a) and, below that line near a local, maximum point (b)., Exarmpie 9, , minimum point, , (a), , (b), , For each of the functions in Fig. 3.2.14, tell whether x, is a local minimum, point, a local maximum point, or neither., , bB;, yo, , Figure 3.2.14. Is no a local, , maximum"!ocai, minimum? Neither?, Solution, , (a), , (hi, , (i.), , (d, , Comparing each graph in Fig. 3.2.14 with the horizontal line through the, heavy dot, we find that x, is a local maximum point in (b), a local minimum, point in (c), and neither in (a) and (d). d,, At a local maximum or minimum point x,, a function f can be neither, increasing nor decreasing, as we see from a comparison of the definitions or, graphic interpretations of these concepts. It follows that the derivative j'(x,), (if it exists) can be neither positive nor negative at such a point; hence, it must, be zero. Points x, where f'(x,) = 0 are called critical points off., The critical point test described in the following display is very important., (Some people remember nothing else after a year of calculus.) A good portion, of this chapter explores the applications of the test and its limitations., , If x, is a local maximum or minimum point of a (differentiable) function, f,then x, is a critical point, i.e., f'(xo) = 0., , In Figs. 3.2.8 and 3.2.9, you will observe that not every critical point is a local, maximum or minimum. In the remainder of this section we shall develop a, test for critical points to be maxima or minima. This will be useful for both, Sometimes the phrase "strict local minimum point" is used when f ( x ) > f ( x o ) for all x in, maximum points. Here, and elsewhere, we use the, term "point" to refer to a number in the domain o f f rather than to a point in the plane., , ( u , h ) other than xo. Likewise for strict local, , Copyright 1985 Springer-Verlag. All rights reserved.

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152, , Chapter 3 Graphing and Maximum-Minimum Problems, , graphing and problem solving. To lead to the test, we ask this question: How, does a function f behave just to the right of a critical point x,? The two, simplest possibilities are that f'(x) > 0 for all x in some interval (x,,6) or that, f ' ( x ) < 0 in such an interval. In the first case, f is increasing on (xo,b), and the, second case is decreasing there. These possibilities are illustrated in Fig. 3.2.15., , Figure 3.2.15. Behavior to, the right of a critical point., , Likewise, the behavior off just to the left of x, can be determined if we know, the sign of f'(x) on an interval (a, x,), as shown in Fig. 3.2.16., , Figure 3.2.16. Behavior to, the left of a critical point., , j"(x), , > 0 on (a. .yo), , The two possibilities in Fig. 3.2.15 can be put together with the two, possibilities in Fig. 3.2.16 to give four different ways in which a function may, behave near a critical point, as shown in Fig. 3.2.17., , j"(x), f"(u), , > 0 on ( u , x o ), > 0 on (x,, b ), , j' is increasing a t x,, , Figure 3.2.17. Four ways in, which a function can, behave near a critical point., , f"(x), j"(x), , <0 on ia,xo), > 0 o n (x,,b), , xo is a local minimum point, , f " ( . ~<) 0 on ( u , s o ), , < 0 on (x,, b ), j'rs decreasing at x o, f'(x), , f"(x), , > 0 on ( a , ~ , ), <, , f'(x) 0 on ixo,b), xo is a local maximum point, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.2 Increasing and Decreasing Functions, , 953, , The situations shown in Fig. 3.2.17 are not the only possibilities, but they, are the most common. For example, f'(x) might be zero on one side of x, or, the other, or it might oscillate wildly. For most of the functions we encounter, in this course, though, the classification of critical points in Fig. 3.2.17 will, suffice. As we have already seen in Fig. 3.2.8, the functions x3, - x3, x2, and, - x 2 provide examples of all four possibilities. Notice that in the case where, x, is a local maximum or minimum point, f' changes sign at x,, while f' has, the same sign on both sides of a critical point where f is increasing or, decreasing., We summarize our analysis of critical points in the form of a test., , Suppose that x, is a critical point for f(x), i.e., f'(x,) = 0., 1. If f '(x) changes sign from negative to positive at x,, then x, is a, (strict) local minimum point for f., 2. If f '(x) changes sign from positive to negative at x,, then x, is a, (strict) local maximum point for f., 3. If ff(x) is negative for all x # x, near x,, then f is decreasing at x,., 4. If f'(x) is positive for all x Z xo near x,, then f is increasing at x,., --, , Sign of f'(x) near x,, to the left, , Sign of f'(x) near x,, to the right, , Behavior off, at x,, , +, , local minimum, local maximum, decreasing, increasing, , -, , +, , This test should not be literally memorized. If you understand Fig. 3.2.17, you, can reproduce the test accurately., Find the critical points of the function f(x), local maximum or minimum points?, , Example 10, , 8x3 + 6x2 - 1. Are they, , We begin by finding the critical points: f f ( x ) = 12x3 - 24x2 + 12x =, 12x(x2- 2x + 1) = 12x(x - 1)2; the critical points are thus 0 and 1. Since, (x - 1)2 is always nonnegative, the only sign change is from negative to, positive at 0. Thus 0 is a local minimum point, and f is increasing at I. A, , Solullon, , Example 11, , Find and classify the critical points of the function, f(x), , Solution, , = 3x4 -, , =, , x 3 + 3x2 - 6x., , The derivative f'(x) = 3x2 + 6x - 6 has roots at - 1 rt fi; it is positive on, + 0,oo) and is negative on ( - 1 - 6 , - 1 + 6 ) ., Changes of sign occur at - 1 (positive to negative) and - 1 +, (negative to positive), so - 1 - 0 is a local maximum point and - 1 + 6 is a, local minimum point. A, , ,(- a,- 1 - 6 ) and (- I, , 0, , Copyright 1985 Springer-Verlag. All rights reserved., , 0

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154, , Chapter 3 Graphing and Maximum-Minimum Problems, , Example 12, Solullon, , Discuss the critical points of y, , = x4 and y =, , - x4., , If f ( x ) = x4,jf'(x) = 4x3, and the only critical point is at x, = 0. We know that, x3 changes sign from negative to positive at 0, so the same is true for 4x3, and, hence x 4 has a local minimum at 0. Similarly, the only critical point of - x4 is, , 0, which is a local maximum point. (See Fig. 3.2.18.) A, , Figure 3.2.18. Critical, behavior of y = x4 and, y = - x4., , Exercises for Section 3.2, 1. Using algebra alone, show that f ( x ) = x 2 is increasing at xo = 3., 2. Using algebra alone, show that f ( x ) = x 2 is decreasing at xo = - 1 ., 3. Show by algebra that f ( x ) = mx b is increasing, for all xo if m > 0., 4. Show by algebra that f ( x ) = mx b is decreasing for all xo if m < 0., Using only algebra, find the sign changes of the functions at the indicated points in Exercises 5-8., 5 . f ( x ) = 2 x - 1 ; xo=' 2 ', 6. f ( x ) = x 2 - I ; x 0 = - 1 ., 7 . f ( x ) = x 5 ; X o = 0., 8. h ( z ) = Z(Z - 2); zo = 2., In Exercises 9-12, determine whether the functions are, increasing, decreasing, or neither at the indicated, points., 9. x3 X + 1; x o = 0 ., 10. x4 + X + 5 ; xo = 0., , Derivatives, , Functions, , +, +, , +, , + 2t2 is the position of a particle, on the real-number line at time t , is it moving to, the left or right at t = I ?, 14. A ball is thrown upward with an initial velocity, of 30 meters per second. The ball's height above, the ground at time t is h ( t ) = 30t - 4.9t2. When, is the ball rising? When is it falling?, 15. The annual inflation rate in Uland during 1968, was approximately r ( t ) = 20[1 ( t 2 - 6t)/500], percent per year, where t is the time in months, from the beginning of the year. During what, months was the inflation rate decreasing? What, are the max-min points of r(t)? Explain their, (political) significance., 13. If f ( t ) = t S- t4, , +, , Figure 3.2.19. Matching, derivatives and functions, (Exercise 2 1)., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.2 Increasing and Decreasing Functions, , 455, , 16. The rate of a chemical reaction between t = 0, and t = 10 is given by r ( t ) = 2 t 3 - 3 t 2 + 1 ., When is the reaction slowing down? When is it, speeding up?, 17. Find the intervals on which f ( x ) = x 2 - 1 is increasing or decreasing., 18. Find the intervals on which x 3 - 3 x 2 + 2 x is, increasing or decreasing., 19. Find the points at which f ( x ) = 2 x 3 - 9 x 2 +, 12x + 5 is increasing or decreasing., 20. Find all points in which f ( x ) = x 2 - 3 x + 2 is, increasing, and at which it changes sign., 21. Match each derivative on the left in Fig. 3.2.19, with the function on the right., 22. Sketch functions whose derivatives are shown in, Fig. 3.2.20., Figure 3.2.21. Are f and f', increasing or decreasing?, , 25. In Fig. 3.2.22, which points are local maxima, and which are local minima?, , Figure 3.2.22. Find the, local maxima and minima., 26. Tell where the function in Fig. 3.2.22 is increasing and where it is decreasing., Find the critical points of the functions in exercises, 27-34 and decide whether they are local maxlma, local, minima, or neither., 27. f ( x ) = x 2 - 2, x2+ 1, 28. f ( x ) = x2- 1, 29. f ( x ) = x 3 + x 2 - 2 30. f ( x ) = 3 x 4, Figinre 3.2.20. Sketch, functions that have these, derivatives., , 23. For each of the functions shown in Fig. 3.2.20, state: (I) where it is increasing: ( 2 ) where it is, decreasing; ( 3 ) its local maximum and minimum, points; ( 4 ) where it changes sign., 24. For each of the functions in Fig. 3.2.21 tell, whether the function is increasing or decreasing, and whether the derivative of the function is, increasing or decreasing., , 35. Is f ( x ) = 1 / ( x 2 + 1 ) increasing or decreasing at, X = 1 , - 3 , 3 , 2 5 , -36?, 36. Let f ( x ) = 4 x 2 + ( I / x ) . Determine whether f is, increasing or decreasing at each of the following, points: (a) 1; (b) - 4; (c) - 5 ; (d) 2 4 ., 37. Describe the change of sign at x = 0 of the, function f ( x ) = mx for m = - 2 , 0 , 2 ., 38. Describe the change of sign at x = 0 of the, function f ( x ) = mx - x 2 form = - 1 , - f , O , + , l ., , Copyright 1985 Springer-Verlag. All rights reserved.

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'156, , Chapter 3 Graphing and Maximum-Minimum Problems, , Find the sign changes (if any) for each of the functions, in Exercises 39-42., 40. x 2 x - 1, 39. x 2 + 3x + 2, x3- 1, 42. 41. x 2 - 4x 4, , +, , +, , x2+ 1, , 43. Using only algebra, determine the sign change of, f(x) = (X - r,)(x - r2) at x = r,, where r, < r2., 44. For an observer standing on the Earth, let f(t), denote the angle from the horizon line to the sun, at time t . When does f(t) change sign?, Find the intervals on which each of the functions in, Exercises 45-48 is increasing or decreasing:, 45. 2~~ - 5~ 7, 46. x5 - x3, , +, , 49. Find a quadratic polynomial which is zero at, x = 1, is decreasing if x < 2, and is increasing if, x > 2., 50. Herring production T (in grams) is related to the, number N of fish stocked in a storage tank by, the equation T = 500N - 5 0 ~ ~ ., (a) Find dT/dN., (b) Unless too many fish are stocked, an increase in the number of fish stocked will, cause an increase in production at the expense of a reduction in the growth of each, fish. (The weight for each fish is T/N.), Explain this statement mathematically in, terms of derivatives and rhe level N* of, stocking which corresponds to maximum, production., A-51. Prove the following assertions concerning the, function f(x) = (x3 - 1)/(x2 - 1):, (a) f can be defined at x = 1 so that f becomes, continuous and differentiable there, but cannot be so defined at x = - 1., (b) f is increasing on ( - co, - 21 and decreasing, on [ - 2, - 1)., (c) If a < - 1, then (a3 - l)/(a2 - 1) d -3., (d) f is increasing on [0, co) and decreasing on, ( - 1,0]. Make up an equality based on this, fact., A-52. Prove that f is increasing at xo if and only if the, function f(x) - f(xo) changes sign from negative, to positive at xo., *53. Using the definition of an increasing function,, prove that iff and g are increasing at xo, then so, is f g., +54. Prove that iff and g are increasing and positive, on an interval I, then fg is increasing on I., A-55. Let f(x) = a, + a l x a2x2+ . . . + a n x n . Under what conditions on the a,'s is f increasing at, xo = O?, , +, , *56. Under what conditions on a , b, c , and d is the, cubic polynomial ax3 + bx2 + cx d strictly increasing or strictly decreasing on ( - co,co)?, (Assume a f 0.), *57. If g and h are positive functions, find criteria, involving gl(x)/g(x) and hf(x)/h(x) to tell when, (a) the product g(x)h(x) and (b) the quotient, g (x)/ h (x) are increasing or decreasing., +58. Let f be a function, and a > 0 a positive real, number. Discuss the relation between the critical, points of f(x), af(x), a + f(x), f(ax), and, f(a + x)., *59. Find a cubic polynomial with a graph like the, one shown in Fig. 3.2.23., , +, , I, , I, , Figure 3.2.23. This is the, graph of what cubic, polynomial?, , *60. (a) Show that there is no quartic polynomial, whose graph is consistent with the information shown in Fig. 3.2.20(e)., (b) Show that if 2 2 is replaced by 2 fl, then, there is a quartic polynomial consistent with, the information in Fig. 3.2.20(e)., *61. Find a relationship between the (positive) values, of a and b which insures that there is a quartic, polynomial with a graph consistent with the information in Fig. 3.2.24., , ~i~~ 3.2.2. when is this, the graph of a quartic?, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.3 The Second Derivative and Concavity, , 157, , 3.3 The Second Derivative, and Concavity, The sign of the second derivative indicates which way the graph of a function is, bending., In the last section we saw that the classification of critical points of a function, f(x) depends on the sign changes of the derivative f'(x). On the other hand,, the sign changes of f ( x ) at a critical point off are determined by the sign of, the derivative of f'(x), i.e., by the sign of the second derivative f "(x). (See Fig., 3.3.1.) After exploring the consequences of this idea, we shall see that the sign, off "(x) is important even when x is not a critical point off., , Figure 3.3.1. f "(x,) > 0 so, f' is increasing at x , and, changes sign from negative, to positive; thus f has a, local minimum at x, Likewise f J ' ( x 2 )< 0 and x,, is a local maximum., , ,., , ( a ) Graph o f t ', , ( b ) Graph off, , Recall that if g(x,) = 0 and g'(x,) > 0, then g(x) changes sign from, negative to positive. Applying this to the case where g is the derivative f ' of a, function f, we find that for a critical point x, of f(x), f'(x) changes sign from, negative to positive iff "(x,) > 0. Thus, by the first derivative test, x, is a local, minimum point. Similar reasoning when f "(x,) < 0 leads to the following test., , Suppose that f'(xo), , = 0., , 1. If f "(x,) > 0, then x, is a (strict) local minimum point., 2. If f "(x,) < 0, then x, is a (strict) local maximum point., 3. If f n ( x 0 ) = 0, the test is inconclusive., , We will discuss the case fu(x0) = 0 shortly. For now, notice that the functions, 3, 3, 4, = x , y = - x , y = x , y = - x 4 (Figs. 3.2.8 and 3.2.18) show that various, things can happen in this case., , y, Example 1, , Use the second derivative test to analyze the critical points of the function, f(x) = x 3 - 6x2 10., , Solution, , Since f'(x) = 3x2 - 12x = 3x(x - 4), the critical points are at 0 and 4. Since, f"(x) = 6x - 12, we find that f"(0) = - 12 < 0 and f"(4) = 12 > 0. By the, second derivative test, 0 is a local maximum point and 4 is a local minimum, point. A, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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158, , Chapter 3 Graphing and Maximum-Minimum Problems, , Example 2, , Solution, , Analyze the critical points of f ( x ) = x3 - x., , f'(x) = 3x2 - 1 and f " ( x )= 6x. The critical points are zeros of f'(x); that is,, x = - + ( 1 / 0 ) ; f " ( - 1/01= - ( 6 / 0 ) < 0 and f U ( l / 6 ) = 6 / 0 > 0. By the, second derivative test, -(I/$?) is a local maximum point and 1/6 is a local, minimum point. A, When f U ( x o )= 0, the second derivative test is inconclusive. We may sometimes use the first derivative test to analyze the critical points, however., , Example 3, Solullon, , + 8x3 + 1 2x2 + 8x + 7., The derivative is f ' ( x ) = 8x3 + 24x2 + 24x + 8, and f'(- 1) = - 8 + 24 24 + 8 = 0,so - 1 is a critical point. Now f " ( x ) = 24x2 + 48x + 24, so f"(- 1), = 24 - 48 + 24 = 0, and the second derivative test is inconclusive. If we factor, f', we find f ' ( x ) = 8(x3+ 3x2 + 3x + 1 ) = 8(x + l13. Thus - 1 is the only root, Analyze the critical point x, = - 1 of f ( x ) = 2x4, , off', f'(- 2) = - 8, and f'(0) = 8, so f' changes sign from negative to positive, at - 1; hence, - 1 is a local minimum point for f. A, Whether or not f'(x,) is zero, the sign of f"(x,) has an important geometric, interpretation: it tells us which way the tangent line to the graph off turns as, the point of tangency moves to the right along the graph (see Fig. 3.3.2). The, two graphs in Fig. 3.3.2 are bent in opposite directions. The graph in part (a), is said to be concave upward; the graph in part (b) is said to be concave, downward., , "t, , "t, , Figu~e3.3.2. (a) The slope, of the tangent line is, increasing; f " ( x ) > 0., (b) The slope of the tangent, line is decreasing;, f "(x) < 0., , Figu~e3.3.3. The function f, is concave upward at xo, and concave downward, at x,., , We can give precise definitions of upward and downward concavity by, considering how the graph off lies in relation to one of its tangent lines. To, accomplish this, we compare f ( x ) with its linear approximation at x,:, I ( x ) = f(xO)+ f'(xO)(x- x,). If there is an open interval ( a ,b ) about x , such, that f ( x ) > I(x) for all x in (a,b ) other than x,, then f is called concave upward, at x,. If, on the other hand, there is an open interval (a,b ) about xo such that, f ( x ) < I(x) for all x in (a,b ) other than x,, then f is called concave downward, at x,., Geometrically, f is concave upward (downward) at x, if the graph of f lies, locally above (below) its tangent line at x,, as in Fig. 3.3.3., Notice that the difference h ( x ) = f ( x ) - I(x) is positive or negative, according to whether f ( x ) > l ( x ) or f ( x ) < I(x). Since h(x,) = f (x,) - I(x,), = 0, we see that f is concave upward or downward at x, according to whether, h ( x ) has a local minimum or maximum at x,. (See Fig. 3.3.4.), If we differentiate h ( x ) = f ( x ) - [f(x,) f'(x,)(x - x,)] twice, we obtain, h l ( x )= f'(x) - f'(x,) and h " ( x ) = f " ( x ) ( x , is treated as a constant). Notice, that h'(xo)= 0, so x, is a critical point for h. Next, observe that ht'(x0), = f M ( x o )SO, , we may conclude from the second derivative test for local, maxima and minima that x, is a local minimum for h iff "(x,) > 0 and a local, maximum if fM(xo)< 0. Thus we have the test in the next box. (Once again,, the functions x3, - x3,x4, - x4 at X , = 0 illustrate the possibilties in case 3.), , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.3 The Second Derivative and Concaviw, , Figure 33.4. The function j, is concave upward at x,, when the difference h(x), between j ( x ) and its linear, approximation at x, has a, local minimum at x,., , 159, , -, , 1. If f u ( x 0 ) > 0 , then f is concave upward at x,., 2. I f f "(x,) < 0, then f is concave downward at x,., 3. If fu(x0) = 0 , then f may be either concave upward at x,, concave, downward at x,, or neither., , Example 4, , Discuss the concavity of f ( x ) = 4 x 3 at the points x = - 1 , x = 0 , and x = 1., , Sol~tlon We have f ' ( x ) = 12x2 and f " ( x )= 24x, so f "(- 1) = -24, f"(0) = 0 , and, f " ( 1 )= 24. Therefore, f is concave downward at - 1 and concave upward at 1., At zero the test is inconclusive; we can see, however, that f is neither concave, upward nor downward by noticing that f is increasing at zero, so that it, crosses its tangent line at zero (the x axis). That is, f is neither above nor, below its tangent line near zero, so f is neither concave up nor concave down, at zero. A, Example 5, , Find the intervals on which f ( x ) = 3 x 3 - 8x + 12 is concave upward and on, which it is concave downward. Make a rough sketch of the graph., , Solution, , Differentiating f, we get f ' ( x ) = 9 x 2 - 8, f " ( x )= 18x. Thus f is concave, upward when 18x > 0 (that is, when x > 0 ) and concave downward when, x < 0. The critical points occur when f ' ( x ) = 0, i.e., at x = + J 8 / 9, = t: 3, Since f "(< 0 , - f @ is a local maximum, and since, f "(5, > 0 , 0 is a local minimum. This information is sketched in Fig., 3.3.5. A, , 0., 0), , Concave, downward, , :a), Concave, upward, , Figure 3.3.5. The critical, points and concavity of, 3x3 - 8x + 12., , We have just seen that a function f is concave upward where Y ( x ) > 0 and, concave downward where f " ( x )< 0 . Points which separate intervals where f, has the two types of concavity are of special interest and are called inflection, points. More formally, we say that the point x, is a inflection point for the, function f if f is twice differentiable near x , and f" changes sign at x,. (See, Fig. 3.3.6.), , Copyright 1985 Springer-Verlag. All rights reserved.

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160, , Chapter 3 Graphing and Maximum-Minimum Problems, , Figure3.3.6., T h e inflection, points of / are x, and x2., , I, , .I,, , X2, , To see where f" changes sign, we begin by looking for the points where it, is zero. Then we look at the next derivative, f"'(x,), to see whether a sign, change actually takes place., , An inflection point for f is a point where f" changes sign. If xo is an, inflection point for f, then fu(x0) = 0. I f f " ( x , ) = 0 and f'"(x,) # 0, then, x, is an inflection point for f ., Example 6, , Find the inflection points of f ( x ) = x 2, , Q~:utIoa The first derivative is f'(x), f"(x) = 2, , = 2x, , +( 1 /x)., , - ( l / x 2 ) , and so the second derivative is, , + ( 2 / x 3 ) .The only possible inflection points occur where, , o = f M ( x ) = 2 + -2., , x, That is, x3 = - 1 ; hence, x = - 1 . To test whether this is an inflection point,, we calculate the third derivative: f"'(x) = - 6 / x 4 , so f '(- 1) = - 6 f 0;, hence, - 1 is a inflection point. A, , If f1'(x0)= 0 and f'"(x,) = 0, then x , may or may not be an inflection point., For example, f ( x ) = x4 does not have an inflection point at x , = 0 (since, f " ( x ) = 12x2 does not change sign at 0), whereas f ( x ) = x5 does have an, inflection point at xo = O (since f " ( x ) = 20x3 does change sign at 0). In both, cases, f " ( x , ) = f " ' ( x , ) = 0, so the test in the preceding box fails in this case., We can also detect sign changes of f" by examining the sign of f " ( x ) in, each interval between its roots., Example 7, , Solution, , +9x2+ 1 ., We have f ' ( x ) = 96x3 - 96x2 + 18x, so f " ( x ) = 288x2 - 192x + 18 and f"'(x), Find the inflection points of the function f ( x ), , = 24x4 -, , 32x3, , = 576x - 192. To find inflection points, we begin by solving f " ( x ) = 0; the, quadratic formula gives x = (4, 12. Using our knowledge of parabolas,, we can conclude that f" changes from positive to negative at (4 - fi)/, 12 and, from negative to positive at (4 + f i ) / 1 2 ; thus both are inflection points. One, could also evaluate f"'((4 2 0 ) / 1 2 ) , but this requires more computation., , + n)/, , Some additional insight into the meaning of inflection points can be obtained, by considering the motion of a moving object. If x = f(t) is its position at time, t , then the second derivative d 2 x / d t 2= f n ( t ) is the rate of change of the, velocity d x / d t = f'(t) with respect to time-the acceleration. We assume that, d x / d t > 0, so that the object is moving to the right on the number line. If, d 2 x / d t 2 > 0, the velocity is increasing; that is, the object is accelerating. If, d 2 x / d r 2 < 0, the velocity is decreasing; the object is decelerating. Therefore, a, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.3 The Second Derivative and Concavity, , 161, , point of inflection occurs when the object switches from accelerating to, decelerating or vice versa., , Example 8, , In Fig. 3.3.7, tell whether the points x , through x , are local maxima, local, minima, inflection points, or none of these., , Figwe 3.3.9. Locate the, local maxima and minima, and inflection points., , Solution, , x , is a local maximum point;, x, is an inflection point;, x , is an inflection point;, , x4 is none of these (it is a zero off);, x , is a local minimum point;, x, is a local maximum point. A, , Exercises far Section 3.3, Use the second derivative test to analyze the critical, points of the functions in Exercises 1-8., 1 . f ( x ) = 3x2 2, 2. f ( x ) = x3 - 6x - 3, 3. f ( x ) = 6x5 - x 20 4. f ( x ) = x4 - x 2, x2- 1, x3- 1, 5. f ( x ) = 6. f ( x ) = x2+ 1, x2+ 1, S, 1, 7. g(s) = 8. h ( p ) = p + 1 + s2, P, , +, , +, , Find the intervals on which the functions in Exercises, 9-16 are concave upward and those on which they are, concave downward:, 9. f ( x ) = 3x2 8x + 10, 10. f ( x ) = x3 3x 8, 11. f(xj = x4, , +, + +, , 14. f ( x ) = -L-., 1 x2, 15. f ( x ) = x3 + 4 x 2 - 8x + 1 ., 16. f ( x ) = ( x - 2)3 8., Find the inflection points for the functions in Exercises, 17-24., 18. f ( x ) = x 4 - x 2 + 1, 17. f ( x ) = x3 - x, 20. f ( x ) = x6, 19. f ( x ) = x7, 22., = zx3, 3~, 21. XI = ( X - 1)4, , +, , +, , +, , 25. In each of the graphs of Fig. 3.3.8, tell whether, x, is a local maximum point, a local minimum, point, an inflection point, or none of these., Figure 3.3.8. Is x, a local maximum? A local minimum? An inflection point? Neither?, , Copyright 1985 Springer-Verlag. All rights reserved.

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162, , Chapter 3 Graphing and Maximum-Minimum Problems, , 26. Identify each of the points x , through x6 in Fig., 3.3.9 as a local maximum point, local minimum, point, inflection point, or none of these., , Figure 3.3.9. Classify the, points x,, . . . , x,., Find the local maxima, local minima, and inflection, points of each of the functions in Exercises 27-30. Also, find the intervals on which each function is increasing,, decreasing, and concave upward and downward., 1, 28. 27. a x 2 - 1, x(x - I ), 29. x 3 + 2x2 - 4 x +, 30. x 2 - x 4, 31. Find the inflection points for x n , n a positive, integer. Now does the answer depend upon n?, 32. Is an inflection point always, sometimes, or never, a critical point? Explain., 33. Find a function with inflection points at 1 and 2., [Hint: Start by writing down f"(x). Then figure, out what f'(x) and f(x) should be.], 34. Find a function with local maxima or minima at, 1 and 3 and an inflection point at 2., 35. (a) Relate the sign of the error made in the linear, approximation off to the second derivative off., (b) Apply your conclusion to the linear approximation of l / x at x, = 1., 36. (a) Use the second derivative to compare x 2 with, 9 + 6(x - 3) for x near 3. (b) Show by algebra, that x 2 > 9 + 6(x - 3) for all x # 3., 37. Let f(x) = x 3 - x., (a) Find the linear approximations to f at x,, = - 1,0, and 1., (b) For each such x,, compare the value of, f(xo + Ax) with the linear approximation for, Ax = 2 1, 2 0.1, 2 0.01. Wow does the error, depend upon f "(x,)?, 38. Show that if f'(x,) = f1'(x0) = 0 and /"'jx,) # 0,, then x, is not a local maximum or local minimum point for f., 39. The power output of a battery is given by P, = EI - RI ', where E and R are positive constants., (a) For which current I is the power P a local, maximum? Justify using the second derivative test., (b) What is the maximum power?, 40. A generator of E volts is connected to an inductor of L henrys, a resistor of R ohms, and a, , second resistor of x ohms. Heat is dissipated, from the second resistor, the power P being given, , (a) Find the resistance value xo which makes, the power as large as possible. Justify with, the second derivative test., (b) Find the maximum power which can be, achieved by adjustment of the resistance x ., 41. A rock thrown upward attains a height s = 3 +, 40t - 16t2 feet in t seconds. Using the second, derivative test, find the maximum height of the, rock., 42. An Idaho cattle rancher owns 1600 acres adjacent to the Snake River. He wishes to make a, three-sided fence from 2 miles of surplus fencing,, the enclosure being set against the river to make, a rectangular corral. If x is the length of the, short side of the fence, then A = x(2 - 2x) is the, area enclosed by the fence (assuming the river is, straight)., (a) Show that the maximum area occurs when, x = i, using the second derivative test., (b) Verify that the maximum area enclosed is, 0.5 square mile., (c) Verify that the fence dimensions are f , 1,, miles, when the area enclosed is a maximum., +43. Sketch the graphs of continuous functions on, ( - co, co) with the following descriptions. (If you, think no such function can exist, state that as, your answer.), (a) Three local maxima or minima and two, points of inflection., (b) Two local maxima or minima and three, points of inflection., (c) Four local maxima o r minima and no points, of inflection., (d) Two (strict) local maxima and n o (strict), local minima., *44. Suppose that f'(x,) = fn(x0) = f"'(x,) = 0, but, f""(xo) # 0. Is x, a local maximum point, a local, minimum point, or an inflection point off? Give, examples to show that anything can happen if, S""(xo) = 0., *45. If f(x) is positive for all x, do f ( x ) and l/f(x), have the same inflection points?, n46. Prove that no odd-degree polynomial can be, everywhere concave upward. (As part of your, solution, give a few simple examples and include, a brief discussion of the possibilities for evendegree polynomials.), n47. Prove the following theorem, which shows that, the tangent line at a point of inflection crosses, the graph:, , Copyright 1985 Springer-Verlag. All rights reserved., , +

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3.4 Drawing Graphs, k t x0 be an inj7ection point for f and let, h ( x )= f ( x ) - [f(xo) + f'(xo)(x - x,)] be the difference between f and its linear approximation at, x,. I f f " changes sign from negative to positive [or, , 163, , positive to negative] at xo (for example, i f r f ( x 0 ), > 0 [or < O]), then h changes sign from negative, to positive [orpositive to negative] at x,., The two cases are illustrated in Fig. 3.3.10., , Figure 3.3.10. The graph of, , f crosses its tangent line at, a point of inflection., , ( a ) f""(xo), , >0, , Drawing Graphs, Using calculus to determine the principal features of a graph often produces, better results than simple plotting., , One of the best ways to understand a function is to see its graph. The simplest, way to draw a graph is by plotting some points and connecting them with a, smooth curve, but this method can lead to serious errors unless we are sure, that we have plotted enough points. The methods described in the first three, sections of this chapter, combined with the techniques of differentiation, help, us make a good choice of which points to plot and show us how to connect, the points by a curve of the proper shape., We begin by outlining a systematic procedure to follow in graphing any, function., , To sketch the graph of a function f:, , 1. Note any symmetries of f. Is f(x) = f ( - x), or f (x) = -f (- x), or, neither? In the first case, f is called even; in the second case, f is called, odd. (See Fig. 3.4.1 and the remarks below.), 2. Locate any points where f is not defined and determine the behavior, off near these points. Also determine, if you can, the behavior of f(x), for x very large positive and negative., 3. Locate the local maxima and minima off, and determine the intervals, on which f is increasing and decreasing., 4. Locate the inflection points off, and determine the intervals on which, f is concave upward and downward., 5. Plot a few other key points, such as x and y intercepts, and draw a, small piece of the tangent line to the graph at each of the points you, have plotted. (To do this, you must evaluate j' at each point.), 6. Fill in the graph consistent with the information gathered in steps 1, , Let us examine the graphing procedure beginning with step 1. Iff is eventhat is, f(x) = f(- x)-we may plot the graph for x > 0 and then reflect the, , Copyright 1985 Springer-Verlag. All rights reserved.

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164, , Chapter 3 Graphing and Maximum-Minimum Problems, , result across they axis to obtain the graph for x < 0. (See Fig. 3.4.1 (a).) Iff is, odd, that is, f(x) = -f(- x) then, having plotted f for x > 0, we may reflect, the graph in they axis and then in the x axis to obtain the graph for x & 0., (See Fig. 3.4.1 (b).), , Figure 3.4.1. f is even when, f ( - x ) = f ( x ) and odd, when f ( - x ) = -f ( x ) ., , ( a ) Even function, , ( b ) Odd function, , To decide whether a function is even or odd, substitute - x for x in the, expression for f(x) and see if the resulting expression is the same as f(x), the, negative of f(x), or neither., Example 1, , Classify each of the following functions as even, odd, or neither:, , +, , (a) f(x) = x4 3x2 + 12;, (b) g(x) = x/(l + x2);, (c) h(x) = x/(l + x);, Solution, , +, , +, , +, , (a) f(- x) = ( - x ) +, ~ 3(- x ) ~ 12 = x4 3x2 12 = f(x), so f is even., (b) g(- x) = (- x)/(l + (- x ) ~ =, ) - x/(1 + x2) = -g(x), so g is odd., (c) h(- x) = ( - x)/(l + ( - x)) = - x/(l - x), which does not appear to, equal h(x) or - h(x). To be sure, we substitute x = 2, for which h(x) = f, and h(- x) = 2; thus, h is neither even nor odd. A, Step 2 is concerned with what is known as the asymptotic behavior of the, function f and is best explained through an example. The asymptotic behavior, involves infinite limits of the type limx,,o f(x) = t oo and lirn,,,,, f(x) = I,, as were discussed in Section 1.2., , Example 2, , Find the asymptotic behavior of f(x) = x/(l - x) (f is not defined for x = 1)., , Solution, , For x near 1 and x > 1, 1 - x is a small negative number, so f(x) = x/(l - x), is large and negative; for x near 1 and x < 1, 1 - x is small and positive, so, x/(l - x) is large and positive. Thus we could sketch the part of the graph of, f near x = 1 as in Fig. 3.4.2. The line x = 1 is called a vertical asymptote, for x/(l - x). In terms of limits, we write lirn,,, - [x/(l - x)] = oo and, lim,,,,, [x/(l - x)] = - 00., Next we examine the behavior of x/(l - x) when x is large and positive, and when x is large and negative. Since both the numerator and denominator, also become large, it is not clear what the ratio does. We may note, however,, that, , +, , +, , Figure 3.4.2. Pieces of the, graph y = x / ( l - x) are, plotted near the vertical, asymptote x = 1., , for x 0. As x becomes large (positive or negative), l / x becomes small, and, ( l / x ) - 1 approaches - 1, so l/[(l/x) - 11 approaches I/(- 1) = - 1, i.e.,, lim,,,[x/(l, - x)] = - 1 and lim,,-,[x/(l, - x)] = - 1. Furthermore, when, x is large and positive, 1 / x > 0, so ( I / x ) - 1 > - 1, and therefore, 1/[(l /x) - 11 < 1/(- 1) = - 1, so the graph lies near, but below, the line, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.4 Drawing Graphs, , 465, , = - 1. Similarly, for x large and negative, 1/ x < 0, so (1/ x ) - 1 < - 1, and, therefore l/[(l/x) - 11 > - 1, so the graph lies near, but above, the line, y = - 1. Thus, we could sketch the part of the graph for x large as in Fig., 3.4.3. The line y = - I is called a horizontal asymptote for f. A, , y, , -y=, , 1, , (Hor~zontal, asymptote), , Figare 3.4.3. Pieces of the, graph y = x/(l - x) are, plotted near the horizontal, asymptotey = - 1., , Steps 3 and 4 were described in detail in Sections 3.2 and 3.3; step 5 increases, the accuracy of plotting, and step 6 completes the job. These steps will be, carried out in detail in the examples that follow. The graph y = x/(l - x ), begun above is discussed again in Example 9., Some words of advice: It is important to be systematic; follow the, procedure step by step, and introduce the information on the graph as you, proceed. A haphazard attack on a graph often leads to confusion and, sometimes to desperation. Just knowing steps 1 through 6 is not enough-you, must be able to employ them effectively. The only way to develop this ability, is through practice., Example 3, Solution, , Sketch the graph of f ( x ) = x - -1, x, We carry out the six-step procedure:, , 1. f(-x) = - x + ( l / x ) = -f ( x ) ; f is odd, so we need only study f(x) for, x >, 0., 2. f is not defined for x = 0. For x small and positive, - ( l / x ) is large in, magnitude and negative in sign, so x - ( l / x ) is large and negative as well;, x = 0 is a vertical asymptote. For x large and positive, - ( l / x ) is small and, negative; thus the graph of f(x) = x - ( I / x ) lies below the line y = x ,, approaching the line as x becomes larger. The line y =. x is again called an, asymptote (see Fig. 3.4.4)., , Figure 3.4.4. The lines, x=Oandy=xare, asymptotes., , Copyright 1985 Springer-Verlag. All rights reserved.

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166, , Chapter 3 Graphing and Maximum-Minimum Problems, , 3. f'(x) = 1 + ( 1 / x 2 ) , which is positive for all x # 0 . Thus f is always increasing and there are no maxima or minima., 4. f " ( x ) = - ( 2 / x 3 ) , which is negative for all x > 0 ; f is concave downward, on (0, co)., 5 . The intercept occurs where x - (1 / x ) = 0; that is, x = 1. We have f'(1), = 2 , f ( 2 ) = + , f'(2)=$., , The information obtained in steps 1 through 5 is placed on the graph in Fig., 3.4.5., , > 0 (Fig. 3.4.6). Finally, we use the fact that f is, odd to obtain the other half of the graph by reflecting through the x a n d y, axes. (See Fig. 3.4.7.) A, , 6. We fill in the graph for x, , Figure 3.4.5. The, information obtainec, steps 1 to 5., , II, Figure 3.4.6. The graph for, > 0 is filled in (step 6)., , x, , Figure 3.4.7. The complete, graph is obtained by using, the fact that f is odd., , @ Calculator Remark, While calculators enable one to plot points relatively quickly, and computers, will plot graphs from formulas, the use of calculus is still essential. A, calculator can be deceptive if used alone, as we saw in Example 7, Section, R.6. In Chapter 14 we will see how the computer can help us graph complicated surfaces in space, but it may be unwise to begin expensive computation, before a thorough analysis using calculus. Of course, it may be even quicker to, solve a simple problem by calculus than to go to a machine for plotting. A, Example 4, , Solution, , X, Sketch the graph of f ( x ) = 1+x2', , Again we carry out the six-step procedure:, , I. f ( - x ) = -x/[l + (-x)'] = - x / [ 1 + x2] = -f ( x ) ; f is odd, so itsgraph, must by symmetric when reflected in the x and y axes., 2. Since the denominator 1 + x 2 is never zero, the function is defined everywhere; there are no vertical asymptotes. For x # 0 , we have, , +, , Since l / x becomes small as x becomes large, f ( x ) looks like l / ( x 0 ), = l / x for x large. Thus y = 0 is a horizontal asymptote; the graph is below, y = 0 for x large and negative and above y = 0 for x large and positive., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.4 Drawing Graphs, , 167, , which vanishes when x = zk 1 . To check the sign of f ' ( x ) on (- ao, - I),, ( - 1, l), and ( I , oo), we evaluate it at conveniently chosen points:, f'( - 2 ) = - & , f'(0) = 1 , f'(2) = - & . Thus f is decreasing on ( - oo, - 1), and on (1, ao) and f is increasing on (- 1, I). Hence - 1 is a local minimum, and 1 is a local maximum by the first derivative test., , This is zero when x = 0,0 , and - 0 .Since the denominator of f" is, positive, we can determine the sign by evaluating the numerator. Evaluating at - 2, - 1 , 1 , and 2, we get - 4, 4, - 4, and 4, so f is concave, downward on ( - oo, - 0 )and ( 0 , o ) and concave upward on ( - 0 , 0 ), and ( 0 , a ) ; - 0 ,0,and 0 are points of inflection., , 5., , f(0)= 0;, f(1), , (, , fl(0), = 1,, , =f ;, , 0, , f'(1), , = 0,, , ) y ( O ) =-4., , The only solution of f ( x ) = 0 is x = 0 ., The information obtained in steps 1 through 5 is placed tentatively on, the graph in Fig. 3.4.8. As we said in step 1, we need do this only for x > 0., , Figure 3.4.8. The graph, y = x/(l + x 2 ) after steps I, to 5., , Increasing, concave down, , \, , Decreasing, concave up, , Decreasing, concave down, , 6. We draw the final graph, remembering to obtain the left-hand side by, reflecting the right-hand side in both axes. (You can get the same effect by, rotating the graph 180°, keeping the origin fixed.) The result is shown in, Fig. 3.4.9. A, , Figure 3.4.9. The complete, graph y = x / ( l + x2)., , Example 5, , Solution, , Sketch the graph of f ( x ) = 2 x 3, , + 8 x + 1., , The six steps are as follows:, , 1. f ( - x ) = - 2 x 3 - 8 x, nor odd., , + 1 is not equal to f ( x ) or -f ( x ) , so f, , is neither even, , Copyright 1985 Springer-Verlag. All rights reserved.

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Chapter 3 Graphing and Maximum-Minimum Problems, , 168, , 2. f is defined everywhere. We may write, , +, , For x large, the factor 1 (4/x2)+ ( 1 /2x3)is near 1, so f(x) is large, and positive for x large and positive, and large and negative for x large, and negative. There are no horizontal or vertical asymptotes., 3. f'(x)= 6x2 8, which vanishes nowhere and is always positive. Thus f, is increasing on ( - oo, m) and has no critical points., 4. f"(x)= 12x,which is negative for x < 0 and positive for x > 0. Thus f is, concave downward on (- oo,0) and concave upward on (0,oo); zero is a, point of inflection., 5., f(Q)= 1;, f'(0) = 8,, , +, , v, , l, , (1,lI), , f(1)= 1 1 ;, f, , f(- 1), , (0, 1 ) f *, , = - 9;, , f'(1) = 14,, f'(- 1) = 14., , C, , \, , The information obtained so far is plotted in Fig. 3.4.10., Ju, , v, , increasing, concave d o w n, (-1,, , -, , incrras~ng, concave up, , 9), , 6. A look at Fig. 3.4.10 suggests that the graph will be very long and thin. In, fact, f(2) = 33, which is way off the graph. To get a useful picture, we may, stretch the graph horizontally by changing units on the x axis so that a unit, on the x axis is, say, four times as large as a unit on they axis. We add a, couple of additional points by calculating, , Figure 3.4.18. The graph, y = 2x3 + 8 x + 1 after, steps 1 to 5., , Then we draw a smooth curve as in Fig. 3.4.1 1. A, , Figure 3.4.11. The, completed graph, y = 2 x 3 + 8x+ 1 ., , +, , + +, , Any cubic function y = ax3 bx2 cx d may be plotted just as the one in, the preceding example. The critical points are obtained by solving the quadratic equation, which may have one or two roots or none. In Example 5, f'(x) = 0 had no, roots, and f' was always positive. Iff' has two roots, y = f(x) will have one, local maximum and one local minimum. Let us do an example of this type., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.4 Drawing Graphs, , Example 8 Sketch the graph of f ( x ) = 2x3 - 8x, Solullon, , 189, , + 1., , Again, we use the six-step procedure:, , I. There are no symmetries., 2. There are no asymptotes. As in Example 5, f ( x ) is large positive for x large, positive and large negative for x large negative., = 2 1 0 a t. 1.15. Also,, 3. f ( x )= 6 x 2 - 8, which is zero when x =, f ( - 2 ) = f (2) = 16 and f'(0) = - 8, so f is increasing on (- w , - 2 / 6 ], and [ 2 / 6 ,w ) and decreasing on [ - 2 / 6 , 2 / o j . Thus, - 2 / 0 is a local, maximum point and 2 / 6 a local minimum point., 4. f U ( x )= 12x, so f is concave downward on ( - m , 0 ) and concave upward, on (0, m). Zero is an inflection point., 5., f(O)=l;, S'(0) = - 8,, , +, , +, , The data are plotted in Fig. 3.4.12. The scale is stretched by a factor of 4 in, the x direction, as in Fig. 3.4.1 1., 6. W e draw the graph (Fig. 3.4.13). lb,, , Figure 3.4.12. The graph, y = 2x3 - 8x+ 1 after, steps 1 to 5., , Fiwe 3.4.14. The graph, , y=J;;, , Figure 3.4.13. The, , completed graph, y=2~3-8x+1., , Some interesting new features arise when we graph functions which involve, fractional powers. For example, consider the graph of y =& for x > 0., Notice that the slope dy/dx = f x - ' I 2 = 1/(2&) becomes large and positive, as x + 0, while d y / d x approaches 0 as x -+ ca. Thus, the graph appears as in, Fig. 3.4.14., Something similar happens for the cube root j = x ' / ~ so, , that dy/dx, = 1/(3x2I3).This time, the function is defined for all x ; its derivative exists, for x # 0 and is large and positive for x near 0, of both signs. (See Fig. 3.4.15.), We sometimes say that the graph y = x 1 l 3has a vertical tangent at x = 0., You may notice that the graphs y = x ' / 2 and y = x ' / 3 resemble y = x 2, , Copyright 1985 Springer-Verlag. All rights reserved.

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870, , Chapter 3 Graphing and Maximum-Minimum Problems, , I, , tangent line" at the origin., , and y = x3 turned on their sides. This relationship will be explored when we, study inverse functions in Section 5.3., Still more interesting is the graph of y = x2l3,which is also defined for all, x. The derivative is dy/dx = f x - ' / ~= 2/(3 3fi). For x near 0 and positive,, dy/dx is large and positive, whereas for x near Q and negative, dy/dx is large, and negative. Thus, the graph has the appearance shown in Fig. 3.4.16. Again,, , I, , origin., , we can say that the graph has a vertical tangent at x = 0. However, the shape, of the graph near x = Q has not been encountered before. We call x = 0 a, cusp. Note that x = 0 is a minimum point of f(x) = x2l3,but that x2I3is not, differentiable there., In general, a continuous function f is said to have a cusp at x, if f'(x) has, opposite signs on opposite sides of x, but f'(x) "blows up" at x, in the sense, that limx,xo[l /f'(x)] = 0; thus limx,xo, f'(x) = rt co or T co., Example 7, , Let f(x) = (x2 + 1)3/2.(a) Where is f increasing? (b) Sketch the graph off. Are, there any cusps?, , SsBultion, , (a) Using the chain rule, we get f'(x) = 3xJx2 + 1 . Hence f'(x) < O (so f is, decreasing) on ( - co, Q), and f'(x) > Q (so f is increasing) on (0, co)., (b) By the first derivative test, x = Q is a local minimum point. Note that, f(x) is an even function and, , f "(x) = 3, I, , X', , +JX, , Jzi, , so f is concave upward. Thus we can sketch its graph as in Fig. 3.4.17. There, are no cusps. .A, , Figure 3.4.17. The graph, =( x +, ~ 1)3/2., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.4 Drawing Graphs, , Example 8, Solution, , Sketch the graph of (x, , 171, , +, , Letting f(x) = (x + 1)2/3~2,, we have, , + + 1)2/3-2~, + 1)'/3](4x + 3)., , Y(X) = +(x + I ) - ' / ~ X ~(X, = [2x/3(x, , -1, , I, , x, , 0, , 4, , Figure 3.4.18. The graph, y = x2(x + 1)2/3 has a cusp, a t x = -1., , For x near - 1, but x > - 1, f'(x) is large positive, while for x < - 1, f'(x) is, large negative. Since f is continuous at - 1, this is a local minimum and a, cusp., The other critical points are x = 0 and x = - $. From the first derivative, test (or second derivative test, if you prefer), - is a local maximum and zero, is a local minimum. For x > 0, f is increasing since f'(x) > 0; for x < - 1, f is, decreasing since f ' ( x ) < 0. Thus we can sketch the graph as in Fig. 3.4.18. (We, located the inflection points at ( - 33 +. m ) / 4 0 FZ - 0.208 and - 1.442 by, setting the second derivative equal to zero.) A, Sometimes algebraic transformations simplify the job of drawing a graph., , Example 9, , Sketch the graph y = x/(l - x) by (a) the six-step procedure and (b) by, making the transformation u = 1 - x., , Solution, , (a) In Example 1 we carried out steps 1 and 2. To carry out step 3, we, compute using the quotient rule:, , Since dy/dx is always positive (undefined if x = l), the graph has no maxima, and minima and the function is increasing on the intervals ( - oo,1) and, (1, oo). For step 4, we compute that, , so the graph is concave upward on (- rm, 1) and concave downward on (1, XI)., For step 5 we note that y = 0 when x = 0 and y = - 2 when x = 2. At both of, these points dy/dx is 1. The graph is then plotted in Fig. 3.4.19., , Figure 3.4.19. The graph, y = x/(l - x)., , +, , +, , (b) If we let u = x - 1, then x = 1 u, so y = (1 u)/(- u) = - 1 l / u . Letting z = y 1, we get z = - l/u. This graph is easy to sketch and, may already be familiar to you as a hyperbola in the second and fourth, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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172, , Chapter 3 Graphing and Maximum-Minimum Problems, , quadrants of the uz plane. Translating axes oust as we did for parabolas in, Section R.2) produces the same result as in 'Fig. 3.4.19, in which the dashed, lines are the u and z axes. A, , Supplement to Section 3.4, The General Guble, , +, , In Section R.2, we saw how to plot the general linear function, f(x) = ax b,, and the general quadratic function, f(x) = ax2 + bx c (a # 0). The methods, of this section yield the same results: the graph of ax b is a straight line,, while the graph of f(x) = ax2 bx c is a parabola, concave upward if a > 0, and concave downward if a < 0. Moreover, the maximum (or minimum) point, of the parabola occurs when f'(x) = 2ax + b = 0, that is, x = - (b/2a), which, is the same result as was obtained by completing the square., A more ambitious task is to determine the shape of the graph of the, general cubic f(x) = ax3 bx2 + cx d. (We assume that a .f- 0; otherwise,, we are dealing with a quadratic or linear function.) Of course, any specific, cubic can be plotted by techniques already developed, but we wish to get an, idea of what allpossible cubics look like and how their shapes depend on a, b,, c, and d., We begin our analysis with some simplifying transformations. First of all,, we can factor out a and obtain a new polynomial f,(x) as follows:, , +, , + +, , +, , +, , +, , The graphs off and f, have the same basic shape; if a > 0, they axis is just, rescaled by multiplying ally values by a ; if a < 0, they axis is rescaled and, the graph is flipped about the x axis. It follows that we not lose any generality, by assuming that the coefficient of x3 is 1., Therefore, consider the simpler form, , where b, = b/a, c, = c/a, and d, = d/a. In trying to solve cubic equations,, mathematicians of the early Renaissance noticed a useful trick: if we replace x, by x - (b,/3), then the quadratic term drops out; that is,3, , These algebraic ideas are related to a formula for the roots of a general cubic, discovered by, Niccolo Tartaglia (1506-1559) but published by (without Tartaglia's permission) and often, credited to Girolamo Cardano (1501-1576). Namely, the solutions of the cubic equation, x 3 + bx2 + cx + d = 0 are, where S, x,=, , - - (1 S +, 2, , T ) - b- + L m ( s 3 2, !2- ( S -, , =, , V R+, , T),, T ) , Q=-3c 9 b2 ', , There is also a formula for the roots of a quartic equation, but a famous theorem (due to Abel, and Ruffini in the nineteenth century) states that there can be no such algebraic formula for the, general equation of degree > 5. Modern proofs of this theorem can be found in advanced, textbooks on the algebra [such as L. Goldstein, Abstract Algebra, Prentice-Hall, (1973).] These, proofs are closely related to the proof of the impossibility of trisecting angles with ruler and, compass., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.4 Drawing Graphs, , 173, , where c, and d, a;e new constants, depending on b,, c,, and d,. (We leave to, the reader the task of verifying this last statement and expressing c2 and d, in, terms of b,, c,, and d l ; see Exercise 56.), The graph of f,(x - b,/3) = f2(x) is the same as that of f,(x) except that, it is shifted by b,/3 units along the x axis. This means that we lose no, generality by assuming that the coefficient of x 2 is zero-that is, we only need, to graph f2(x) = x3 c2x d2. Finally, replacing f2(x) by f3(x) = f2(x) - d2, just corresponds to shifting the graph d, units along they axis., We have now reduced the graphing of the general cubic to the case of, graphing f3(x) = x 3 + c 2 x For simplicity let us write f(x) for f3(x) and c for, c,. To plot f(x) = x3 + ex, we go through steps 1 to 6:, , +, , +, , 1. f is odd., 2. f is defined everywhere. Since .f(x) = x3(1 + c/x2), f(x) is large and posi-tive (negative) when x is large and positive (negative); there are no, horizontal or vertical asymptotes., 3. f'(x) = 3x2 + C. If c > 0, f'(x) > 0 for all x, and f is increasing everywhere., If c = 0, f'(x) > 0 except at x = 0, so f is increasing everywhere even, though the graph has a horizontal tangent at x = 0. If c < 0, f'(x) has roots, ; f is increasing on ( - w , - J-], and [J T , w ) and, at t, decreasing on [ ,J, T 1. Thus, is a local maximum, point and J- c/3 is a local minimum point., 4. f "(x) = 6x, so f is concave downward for x < 0 and concave upward for, x > 0. Zero is an inflection point., 5. f(0) = 0, f '(0)= c,, , J-c/3, , f(2, , JT, , 6 =), 0,, , JT, , f'(';-C)=, -2c (if c < O ) ;, , 6. We skip the preliminary sketch and draw the final graphs. (See Fig. 3.4.20.), , Figure 3.4.20. The graph of, y = x 3 + CX for: C > O(1);, c = 0 (11); and c < O (111)., , ( a ) Type I, , i b ) Type I1, , (c) Type I l l, , > 0):, Type I: f is increasing at all points, f' > 0 everywhere., , Thus there are three types of cubics (with a, , Type 11: f is increasing at all points; f ' = 0 at one point., Type 111: f has a maximum and a minimum., Type 11 is the transition type between types I and III. You may imagine, the graph changing as c begins with a negative value and then moves toward, zero. As c gets smaller and smaller, the turning points move in toward the, , Copyright 1985 Springer-Verlag. All rights reserved.

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174, , Chapter 3 Graphing and Maximum-Minimum Problems, , origin and the bumps merge at the point where f'(x), become positive, the bumps disappear completely., Example 10, , Solution, , +, , + +, , Convert 2 x 3 3x2 x I to the form x3, cubic is of type I, 11, or 111., , +, , + +, , +, , +(2x3 3x2 x I ) = x3 t x 2, the coefficient of x2 is 9 ) gives, , + cx, , + f x + 4., , = 0., , As c passes zero to, , and determine whether the, , Substituting x - f for x (since, , = X 3 - f X + + ., , Thus, after being shifted along the x and y axes, the cubic becomes x 3 - $ x., Since c < 0, it is of type 111. A, , Exercises for Section 3.4, In Exercises 1-4, is the function even, odd, or neither?, x3 + 6 x, 2. f ( x ) = x, 1 . f ( x ) = ---x2+ 1, X, 4. f ( x ) = x6 + 8 x 2 + 3, 3. f ( x ) = x3+ 1, In Exercises 5-8, describe the behavior of the given, functions near their veitical asymptotes:, , 31. Match the following functions with the graphs in, Fig. 3.4.21., x2+ 1, (4 x2- 1, 1, (b) x + ;, ', 7, , (c) x 3 - 3 x 2 - 9 x, , + 1,, , Sketch the graphs of the functions in Exercises 9-30., , [Hint: Let u, 13. x 4 - x 3, 15. x4 - x 2, , =x, , - 11, , 14. x4 + x 3, 16. x4 - 3 x 2 + 2$!, , [Hint:Sketch the, derivative first.], , Figure 3.4.21. Match these, graphs with the functions in, Exercise 3 1., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.4 Drawing Graphs, , "175, , Figure 3.4.22. Match these, graphs with the functions in, Exercise 32., Match the following functions with the graphs in, Fig. 3.4.22: (a) x 2 l 7 , (b) x 3 l 7 , (c) (1 + x ) ~ -/ 1~,, (d) (1 x ~ ) ' -/ ~1 ., Let f ( x ) = ( x 2 - 3 ) 2 / 3 .(a) Where is f increasing?, Decreasing? (b) Sketch the graph off, noting any, cusps., 34. Show that the graph of, , +, , a46., a47., , +48., has two cusps. Sketch., Sketch the graph of each of the functions in Exercises, 35-40., 36. ( x 2 5 x 4)2/7, 35. ( x ~ 1)'13, 37. ( x - 1 ) 4 / 3 ( ~ 1)2/3 38. x x2/', 39. ( x - 4)'00/yy, 40. ( x 3 2 x 2 x ) ~ / ~ ., , +, , +, , +, , +, +, , +, +, , 41. Let f ( x ) be a polynomial. Show that f ( x ) is an, even (odd) function if only even (odd) powers of, x occur with nonzero coefficients in f ( x ) ., 42. Find a criterion for telling when the quotient, f ( x ) / g ( x ) of two polynomials is even, odd, or, neither., 43. A simple model for the voting population in a, certain district is given by N ( t ) = 30 + 12t2 - t 3 ,, 0 g t g 8, where t is the time in years, N the, population in thousands., (a) Graph N versus t on 0 g t < 8 ., (b) At what time t will the population of voters, increase most rapidly?, (c) Explain the significance of the points t = 0, and t = 8., 44. The population P of mice in a wood varies with, the number x of owls in the wood according to, the formula P = 30 + l o x 2 - x 3 , 0 g x < 10., Graph P versus x ., +45. Suppose that f ( x ) is defined on all of ( - a,oo)., Show that, where e is an even function and o is an odd, function. [Hint: Substitute - x for x , use the fact, , *49., , that e is even and o is odd, and solve for e ( x ), and o ( x ) . ], There is one function which is both even and, odd. What is it?, If f is twice differentiable and x, is a critical, point o f f , must xo be either a local maximum,, local minimum, or inflection point?, What does the graph of [ a x / ( b x c ) ] d look, like if a , b , c , and d are positive constants?, Prove that the graph of any cubic f ( x ) = a x 3, bx2 + cx + d ( a 0) is symmetric about its inflection point in the sense that the function, , + +, , +, , +, , is odd., ~ 5 0 A. drug is injected into a person's bloodstream, and the temperature increase T recorded one, hour later. If x milligrams are injected, then, , (a) The rate of change T with respect to dosage, x is called the sensitivity of the body to the, dosage. Find it., (b) Use the techniques for graphing cubics to, graph T versus x ., (c) Find the dosage at which the sensitivity is, maximum., a 5 1 . Convert x 3 + x 2 3 x 1 to the form x 3 c x, and determine whether the cubic is of type I, 11,, or 111., *52. Convert x 3 - 3 x 2 + 3 x 1 to the form x3 c x, and determine whether the cubic is of type I, 11,, or 111., +53. Suppose that an object has position at time t, given by x = ( t - I)'/'. Discuss its velocity and, acceleration near t = 1 with the help of a graph., ~ 5 4 Consider, ., g ( x ) = f ( x ) ( x - a)p/q, where f is differentiable at x = a , p is even. q is odd and p < q., If f ( a ) # 0 , show that g has a cusp at x = a., [Hint: Look at g l ( x ) for x on either side of a.], , + +, , +, , +, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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178, , Chapter 3 Graphing and Maximum-Minimum Problems, , s 5 5 . For which values of a and p does the cubic, g ( x ) = ax3 px have zero, one, or two critical, points? (Assume a f 0)., *56. This problem concerns the graph of the general, cubic (see the Supplement to this Section):, (a) Find an explicit formula for the coefficient, c2 in f ( x - b , / 3 ) in terms of b , , c , , and d,, and thereby give a simple rule for determining whether the cubic, , +, , (a) One root: (f'has the same sign, at ~ t two, s critical points), , is of type I, 11, or 111., (b) Give a rule, in terms of a , b , c, d , for, determining the type of the general cubic, a x 3 + b x 2 + cx d ., (c) Use the quadratic formula on the derivative, of a x 3 b x 2 c x d to determine, in, terms of a , b , c , and d , how many turning, points there are. Compare with the result in, part (b)., Exercises 57 to 64 concern the graph of the general, quartic:, , +, + +, , +, , *57. Using the substitution x - b / 4 a for x , show, that one can reduce to the case, , (with a new c , d and eij., *58. According to the classification of cubics, f ' ( x ), = 4 x 3 2cx, d can be classified into three, types: I ( c > 0), I 1 ( c = O), and T I 1 ( c < 0), so, we may name each quartic by the type of its, derivative. Sketch the graph of a typical quartic, of type I., s 5 9 . Divide type I 1 quartics into three cases: 1 1 ,, ( d > 0), 11, ( d = 0), II, (d < 0). Sketch their, graphs with e = 0., *60. In case 111 ( c < O), f'fx) has two turning points, and can have one, two, or three roots. By, considering Fig. 3.4.23, show that the sign of, f ' ( x ) at its critical points determines the number of roots. Obtain thereby a classification of, type 111 quartics into five subtypes I I I , , HI2,, HI3, I n 4 , and ITIS. Sketch the graphs for each, case and determine the conditions on c and d, which govern the cases., *61. Using your results from Exercises 59 and 60,, show that the ( c , d ) plane may be divided into, regions, as shown in Fig. 3.4.24, which determine the type of quartic4, Using the results of Exercise 61, classify and sketch the, graphs of the quartics in Exercises 62-64., *62. x4 - 3x2 - 4 x ., , +, , ( h ) Two roots. (f'is zero at one, , of its critical points), , +, , *63. x4 + 4 x 2 + 6 x ., *64. x4 + 7 x ., , For advanced (and sometimes controversial) applications of, this figure, see T. Poston and I. Stuart, Catastrophe Theory and, Its Applications, Pitman (1978)., , possible positions, the, graph of f i n case 111., , (c) Three roots: (f.' has opposite, , signs at its critical p o ~ n t s ), , F i w e 3.4.24. Locating the, value of ( c , d) in this graph, tells the type of quartic., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.5 Maximum-Minimum Problems, , 177, , 3.5 Maximum-Minimum, Problems, A step-by-step procedure aids in the solution of practical maximum-minimum, problems., This section is concerned with maximum-minimum problems of two types., First, we consider the problem of maximizing or minimizing a function on an, interval. Then we apply these ideas to maximum-minimum problems that are, presented in words rather than in formulas. Students are often overwhelmed, by such "word problems," which appear to admit no systematic means of, solution. Fortunately, guidelines do exist for attacking these problems; they, are best learned through practice., The maximum and minimum points discussed in Section 3.2 were local,, since we compared f(xo) with f(x) for x near x,. For many applications of, calculus, however, it is important to find the points where f(x) has the largest, or smallest possible value as x ranges over an entire given interval. In this, section, we show how calculus helps us to locate these global maximum and, minimum points, and we discuss how to translate word problems into calculus, problems involving maxima and minima., Global maxima and minima should be as familiar to you as the daily, weather report. The statement on the 6 P.M. news that "today's high temperature was 26°C" means that:, 1. At no time today was the temperature higher than 26°C., 2. At some time today, the temperature was exactly 26°C., If we let f be the function which assigns to each t in the interval [0, 181 the, temperature in degrees at time t hours after midnight, then we may say that 26, is the (global) maximum value off on [0, 181. It is useful to have a formal, definition., , Let f be a function which is defined on an interval I of real numbers., If M is a real number such that:, 1. f(x) < M f o r a l l x in I, 2. f(xo) = M for at least one x, in I, then we call M the maximum value off on I., If m is a real number such that:, 1. f(x) 2 m for a l l x in I, 2. f(xI) = m for at least one x, in I, , The numbers xo and x , in the definition of maximum and minimum values, represent the points at which these values are attained. They are called, maximum or minimum points for f on I. For the temperature function, discussed above, x, might be 15.5, indicating that the high temperature, , Copyright 1985 Springer-Verlag. All rights reserved.

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178, , Chapter 3 Graphing and Maximum-Minimum Problems, , occurred at 3:30 P.M.Of course, it might be possible that the temperature rose, to 26' at 2 P.M. dipped due to a sudden rain shower, rose again to 26' at 3:30, P.M. and finally decreased toward evening. In that case both 14 and 15.5, would be acceptable values for x,. (See Fig. 3.5.1.) Note that there were local, minima at t = 11 (1 1 A.M.), and t = 15 (3 P.M.)., , 26, 24, , 22, , 20, 18, , Figure 3.5.1. The maximum, temperature (26') occurred, at 2 P.M. ( t = 14) and also, at 3:30 P.M. (t = 15.5)., , I6, , 2, , 4, , 6, , 9, , 12, , 14, , 10, , IX, , f, , Sometimes, to distinguish the points considered here from the local, maxima and minima discussed in Section 3.2, we call them global maximum, and minimum points., We will see shortly that differential calculus provides a powerful technique for locating the maximum and minimum points of functions defined by, formulas. For functions defined by other means, there are sometimes direct, approaches to finding maxima and minima. For example:, 1. The highest ring around a bathtub indicates the maximum water level, achieved since the tub was last scrubbed. (Other rings indicate local, maxima.), 2. On a maximum-minimum thermometer, one can read directly the maximum and minimum temperatures reached since the last time the thermometer was reset., 3. If the graph of a function is available, either from experimental data or by, plotting from a formula as in Sections R.6 and 3.4, the maxima and, minima can be seen as the high and low points on the graph., , Example I, , Solution, Y, , Find the maximum and minimum points and values of f(x) = 1 / ( 1, the interval [ - 2,2]., , f(x) is largest where its denominator is smallest, and vice versa. The maximum point occurs, therefore, at x = 0; the maximum value is 1. The minimum, points occur at - 2 and 2; the minimum value is i., We may verify these statements with the assistance of some calculus:, f'(x) = -2x/(1 + x212, which is positive for x < 0 and negative for x > 0,, so f is increasing on [ - 2,0] and decreasing on [Q,21. It follows that, , + = f(-2), , Figure 3.5.2. The function, f ( x ) = 1/(1, x2) on, [- 2,2] has a maximum, point at x = 0 and, minimum points at, x = 22., , +, , + x2) on, , < f(x) < f(0), , =, , 1, , for, , -2, , <x <0, , and, *f(o)>f(x)>f(2)=i, and we have, , Example 2, , for 0 < x < 2 ,, , 4 < f(x) < 1 for - 2 < x < 2. (See Fig. 3.5.2.), , Find the maximum and minimum points and values, if they exist, for the, function f ( x ) = x2 + 1 on each of the following intervals:, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.5 Maximum-Minimum Problems, , Solution, , 179, , The graphs are indicated in Fig. 3.5.3. In case (a), there is no highest point on, the graph; hence, no maximum. In case (c), one might be tempted to call 2 the, maximum value, but it is not attained at any point of the interval (- 1,l)., Study this example well; it will be a useful test case for the general statements, to be made later in this section. A, , Figure 3.5.3. Solutions to, Example 2., , + means that, , the graph goes off t o i n f ~ n i t y, , 0 means that the endpoint does n o t belong to the graph, , Example 3, , Figure 3.5.4 shows the amount of solar energy received at various latitudes in, the northern hemisphere on June 21 on a square meter of horizontal surface, located at the top of the atmosphere. Find the maximum and minimum points, and values., y = energy In kilowatt-hours, , per square meter, 14, , Figwe 3.5.4. The solar, energy y received on June, 21 at the top of the, atmosphere at various, latitudes x. (See W. 6 ., Kendrew, Climatology,, Oxford University Press,, 1949.), , 40, , Solution, , North Pole, , 20 40, , 60, , $0 90, , x = latitude in degrees, , The maximum value is about 13 kilowatt-hours, attained at x = 90 (the North, Pole); the minimum value is about 9 kilowatt-hours, attained at x = 0 (the, equator). (There are local maximum and minimum points at about x = 30 and, x = 50, but these are not what we are looking for.) A, , Copyright 1985 Springer-Verlag. All rights reserved.

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980, , Chapter 3 Graphing and Maximum-Minimum Problems, , The explanation for the unexpected result in Example 3 is that although, the sun's radiation is weakest at the pole, the summer day is longest there,, resulting in a larger amount of accumulated energy. At the surface of the, earth, absorption of solar energy by the atmosphere is greatest near the pole,, since the low angle of the sun makes the rays pass through more air. If this, absorption is taken into account, the resulting graph is more like the one in, Fig. 3.5.5, which is in better correspondence with the earth's climate., y = energy in k ~ l o w a t t - h o u r s, per square meter, , F i w e 3.5.5. The solar, energy received on June 21, at the surface of the earth,, assuming clear skies. (See, W. 6. Kendrew,, Climatolo~,Oxford, University Press, 1949.), , If I is a closed interval, the problem of finding maxima and minima for a, function f on I is guaranteed to have a solution by the following theoretical, result., , Extreme Value Theorem, , The proof of this theorem is omitted: but the statement should be understood, by everyone. It says that two conditions together are suflcient to ensure that f, has both a maximum and a minimum value on I:, 1. f is continuous on I., 2. I is a closed interval., , Cases (a), (b), (c), (e), and (g) of Example 2 show that condition 1 alone is not, sufficient; that is, if I is not closed, the maxima and minima may not exist., Case (h) shows that the maxima and minima might happen to exist even if I is, not closed; thus these two conditions are not necessary for the existence of, maxima and minima., Notice that, like the intermediate value theorem, the extreme value, theorem is an existence theorem which tells you nothing about how to find the, maxima and minima. However, combining the extreme value theorem with, the critical point test does yield a practical test., To obtain this test, we first note that by the extreme value theorem, the, maximum and minimum points must exist; by the critical point test (Section, 3.2), these points must be either critical points or endpoints. It remains,, therefore, to determine which amongst the critical points and endpoints are the, maximum and minimum points; to do this, it suffices to evaluate f at all of, them and then compare the values. The following box summarizes the, situation., The proof may be found in any of the references listed in the Preface., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.5 Maximum-Minimum Problems, , 181, , To find the maxima and minima for a function f which is continuous on, [ a ,b ] and differentiable on ( a ,b ) :, , 1. Make a list x , , . . . , x, consisting of the critical points off on ( a ,b ), and the endpoints a and b of [ a ,b ] ., 2. Compute the values f ( x , ) , . . . ,f(xn)., The largest of the f ( x , ) is the maximum value of f on [ a , b ] . The, maximum points for f on [ a ,b] are those x, for which f ( x , ) equals the, maximum value., The smallest of the f ( x , ) is the minimum value o f f on [ a ,b]. The, minimum points for f on [a,b] are those x, for which f ( x , ) equals the, , Examp16 4, , Find the maximum and minimum points and values for the function f ( x ) =, ( x 2 - 8 x + 1214 on the interval [ - 10,101., , Solullon The list indicated by the closed interval test consists of, , - 10,10, and the, , critical points. To find the critical points we differentiate:, , = 8(x, , -, , 613(x - 213(x - 4 ), , which is zero when x = 2, 4, or 6 . We compute the value off at each of these, points and put the results in a table:, , x, /(XI, , I, , 1, , -10, , 2, , 4, , (19214 0 (-414, , 6, , 10, , 0, , (3214, , The maximum value is (192)4= 1358954496; the maximum point is the, endpoint - 10. The minimum value is zero; the minimum points are the, critical points 2 and 6 . A, For intervals which are not closed, we may use graphing techniques to decide, which critical points are maxima and minima., , Example 5, , Solution, , Find maximum and minimum points and values forv,, , =, , x4 - x 2 on [ - 1, oo)., , The critical points satisfy the equation 0 = 4 x 3 - 2 x = 2 x ( 2 x 2 - 1). The solutions of the equation are 0,, and all of which lie in the interval, [ - 1, co).We make a table of values of the function, , m,, , J1/2,, , and its derivative, , Figwe 3.5.6. The graph o f, y = ~ 4 - ~ 2 0 n [ - ~ , o o ) . and draw a rough graph (Fig. 3.5.6)., , Copyright 1985 Springer-Verlag. All rights reserved.

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182, , Chapter 3 Graphing and Maximum-Minimum Problems, , fi, , J1/2, , We see that the critical points and, are minimum points;, the minimum value is - The endpoint - 1 and the critical point 0 are not, maximum or minimum points since x 4 - x 2 takes on values both greater and, less than 0 on the interval [ - 1,co).Thus, there are no maximum points (0 is, only a focal maximum). .A, , a., , We now turn our attention to solving maximum-minimum problems given in, words rather than by formulas. To illustrate a general approach to these, problems, we will go through the solution of four sample problems step by, step5, 1. A shepherd lives on a straight coastline and has 500 meters of fencing with, which to enclose his sheep. Assuming that he uses the coastline as one side, of a rectangular enclosure, what dimensions should the rectangle have in, order that the sheep have the largest possible area in which to graze?6, 2. Illumination from a point light source is proportional to the intensity of the, source and inversely proportional to the square of the distance from the, source to the point of observation. Given two point sources 10 meters, apart, with one source four times as intense as thk other, find the darkest, point on the line segment joining the sources., 3. Given four numbers, a, b, c , d , find a number x which best approximates, them in the sense that the sum of the squares of the differences between x, and each of the four numbers is as small as possible., 4. Suppose that it costs (x2/100) + 10x cents to run your car for x days. Once, you sell your car, it will cost you 50 cents a day to take the bus. Wow long, should you keep the car?, , The first thing to do is read the problem carefully. Then ask yourself: "What, is given? What is required?', Sometimes it appears that not enough data are given. In Problem 2, for, instance, one may think: "Illumination is proportional to the intensity, but I'm, not given the constant of proportionality, so the problem isn't workable." It, turns out that in this problem, the answer does not depend upon the proportionality constant. (If it did, you could at least express your answer in terms of, this unknown constant.) On the other hand, some of the data given in the, statement of a problem may be irrelevant. You should do your best at the, beginning of solving a problem to decide which data are relevant and which, are not., Here, in full, is the first step in attacking a maximum-minimum word, problem., , (b) Draw a figure, if one is appropriate., , For a general discussion of how to attack a problem, we enthusiastically recommend Now to, Solve It, by 6. Polya (Princeton University Press, Second Edition, 1957)., This ancient Greek problem is a variant of a famous problem ingeniously solved by Dido, the, daughter of the king of Tyre and founder of Carthage (see M. Kline, Mothernatics: A Cultural, Approach, Addison-Wesley, 1962, p. 114)., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.5 Maximum-Minimum Problems, , Example 6, Solution, , 183, , Carry out step 1 for Problems 1, 2, 3, and 4., , Problem 1 : We draw a picture (Fig. 3.5.7). Let I and w denote the length and, width of the rectangle, and let A be the area enclosed. These quantities are, related by the ecluations Iw = A and I 2w = 500 (since there are 500 meters, of fencing available). We want to maximize A ., , +, , Figure 3.5.7. For which, shape is A largest?, , Problem 2: We place the first source at x = 0 and the second source at x = 10, on the real line (Fig. 3.5.8). Denote by I , and I2 the intensities of the two, sources. Let L,= illumination at x from the first source and L, = illumination, at x from the second source. Then L = L, + L,is the total illumination., The given relations are, kI,, L,= x, , ( k > 0 is a proportionality constant),, , We want to minimize L., Figure 3.5.8. Light sources, of intensities 1, and 1, are, placed at x = 0 and x = 10., The observer is at x., , intensity I,, , &c, -, , +/Tp, o, , Intensity, , -.., , x, , r2, , +'V4, 7;~, 10, , Problem 3: Call the unknown number x. We want to minimize, , Problem 4: x is the number of days we run the car. At this point in solving the, problem, it is completely legitimate to pace around the room, muttering, "What should be minimized?This is not clearly stated in the problem, so we, must determine it ourselves. A reasonable objective is to minimize the total, amount of money to be paid. Wow is this to be done? Well, as soon as the cost, of running the car exceeds 50 cents per day, we should switch to the bus. So, let y = the cost per day of running the car at day x. We want the first x for, which y > 50. The relation between the variables is, , Having set up a problem, we are ready to apply the methods of calculus., , (a) Write the quantity to be maximized or minimized as a function of, one of the other variables in the problem. (This is usually done by, expressing all other variables in terms of the one chosen.), (b) Note any restrictions on the chosen variable., (c) Find the maxima and minima by the methods of this chapter., , Copyright 1985 Springer-Verlag. All rights reserved.

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18.6, , Chapter 3 Graphing and Maximum-Minimum Problems, , The main thing to be mastered in word problems is the technique of translating words into relevant mathematical symbols to which the tools of calculus, can be applied. Once the calculus work is done, the answer must then be, translated back into the terms of the original word problem., Example 7, , Solution, , Carry out step 2 for Problems 1, 2, 3, and 4., Problem 1 : We want to maximize A , so we write it as a function of I or w; we, choose w. Now I + 2w = 500, so 1 = 500 - 2w. Thus A = lw = (500 - 2w)w, =500~, - 2w2., The restriction on w is that 0 < w < 250. (Clearly, only a non-negative w, can be meaningful, and w cannot be more than 250 or else I would be, negative.), To maximize A = 500w - 2w2 on [0,250],we compute dA/dw = 500 4w, which is zero if w = 125. Since d 2 ~ / d w=2 -4 for all w, the second, derivative test tells us that 125 is a maximum point. Hence the maximum, occurs when w = 125 and I = 250., The rectangle should be 250 meters long in the direction parallel to the, coastline and 125 meters in the direction perpendicular to the coastline in, order to enclose the maximal area., Problem 2: We want to minimize, , I', , L = L , 4- L 2 = k l , - +, x2 (10 - x12, , 1., , Since the point is to be between the sources, we must have 0 < x, minimize L on (0, 10), we compute:, , < 10. To, , The critical points occur when, , Hence, , 10 - x, -x, , - 3@ ,, , Thus there is one critical point; we use the first derivative test to determine if, it is a maximum or minimum point. It suffices to check the sign of the, derivative at a point on each side of the critical point., Let x - = 0.0001. Then, , Without calculating this explicitly, we can see that the term - ( 1 / 0 . 0 0 ~ 1 ) is, ~, negative and much larger in size than the other term, so L1(x- ) < 0., , Copyright 1985 Springer-Verlag. All rights reserved.

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3.5 Maximum-Minimum Problems, , Let x, , +, , 1185, , = 9.9999. Then, , Now it is the second term which dominates; since it is positive, L1(x+) > 0., Thus L' changes from negative to positive at x,, so x, = 10/(1 )@) must be, a minimum point., w 3.86 meters, The darkest point is thus at a distance of 10/(1 )fl), from the smaller source. It is interesting to compare the distances of the, darkest point from the two sources. The ratio, , +, , +, , is simply, , '@., , Problem 3: y = (x - a)2 + (x - b)2, strictions on x., , + (x - c12 + ( x - d)2. There, , + +, , are no re-, , which is O only if x = $ ( a + b c d)., Since d$/dx2 = 8 is positive, a ( a b c d ) is the minimum point., The number required is thus the average, or arithmetic mean, of a, b, c, and d., , + + +, , Problem 4: This is not a standard maximum-minimum problem. We minimize our expenses by selling the car at the time when the cost per day,, (x/50)+ 10, reaches the value 50.(See Fig. 3.5.9.) We solve (x/50)+ 10 = 50,, getting x = 2000.Thus the car should be kept for 2000 days. A, , Cost per d a y for car, Cost per d a y for bus, , Figure 3.5.9. When does, the car become more, expensive than the bus?, , (number of days), , In the process of doing a word problem, it is useful to ask general questions, like, "Can I guess any properties of the answer? Is the answer reasonable?", Sometimes a clever or educated guess can carry one surprisingly far toward, the solution of a problem., For instance, consider the problem "Find the triangle with perimeter 1, which has the greatest area." In the statement of the problem, all three sides of, the triangle enter in the same way-there is nothing to single out any side as, special. Therefore, we guess that the answer must have the three sides equal;, that is, the triangle should be equilateral. This is, in fact, correct. Such, reasoning must be used with care: if we ask for the triangle with perimeter 1, and the least area, the answer is not an equilateral triangle; thus, reasoning by, symmetry must often be supplemented by a more detailed analysis., In Problems 1 and 3, the answers have as much symmetry as the data,, , Copyright 1985 Springer-Verlag. All rights reserved.

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488, , Chapter 3 Graphing and Maximum-Minimum Problems, , and indeed these answers might have been guessed before any calculation had, been done. In Problem 2, most people would have a hard time guessing the, answer, but at least one can observe that the final answer has the darkest, point nearer to the weaker of the two sources, which is reasonable., Example 8, , Of all rectangles inscribed in a circle of radius 1, guess which has the largest, area; the largest perimeter. (See Fig. 3.5.10.), , Figure 3.5.10. What shape, gives the rectangle the, largest area? The largest, perimeter?, , Solution, , Since the length and width enter symmetrically into the formulas for area and, perimeter, we may guess that the maximum of both area and perimeter occurs, when the rectangle is a square. A, We can now solve word problems by bringing together all of the preceding, techniques., , Example 9, , Find the dimensions of a rectangular box of minimum cost if the manufacturing costs are 10 cents per square meter on the bottom, 5 cents per square, meter on the sides, and 7 cents per square meter on the top. The volume is to, be 2 cubic meters and the height is to be 1 meter,, , Solution, , Let the dimensions of the base be I and w ;the height is 1. If the total cost is C,, then, C = l0lw 71w + 2 - 5 . (1.1 + w . 1) = 171w + 101 low., , +, , Now 1. w . 1, , =2, , +, , is the total volume. Eliminating w,, , C = 3 4 + 10l+ -20., , I, , We are to minimize C. Let f(1) = 34 + 101 + (2011)on (0,GO). Then, , 20, f ' ( 1 ) = 10 - 7, , I, , =a., , which is 0 when I, (We are concerned only with I > 0.) Since f"(1), = 40/13is positive at I = fi,I = fi is a local minimum point. Since this is the, only critical point, it is also the global minimum. Thus, the dimensions of, minimum cost are fi by fi by 1. A, Example "1 The stiffness S of a wooden beam of rectangular cross-section is proportional, to its breadth and the cube of its thickness. Find the stiffest rectangular beam, that can be cut from a circular log of diameter d., Solution, , Let the breadth be b and thickness be t. Then S is proportional to bt3.From, (0 < t < d). To, Fig. 3.5.11, we have (3 b)2 + (+t)2 = (d/2)', is., b =, we note thatf'(t) = t2(3d2- 4t2)/J=, maximize f(t) = bt3 = t3, = 0 if 3d2 - 4t2= 0, i.e., t = $?d/2. Since f ' ( t ) has the same sign as 3d2 4t2,which changes from positive to negative at t = d/2, t = d/2 is a, maximum point; so the dimensions of the beam should be t = $?d/2 and, , d m ,, , 6, , 6, , b = d/2. A, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.5 Maximum-Minimum Problems, , 687, , Figure 3.5.1 1. Cross section, of a beam being cut from a, log., , Example 11, , The terms marginal revenue a n d marginal cost were defined in Section 2.1., (a) Prove that a t a production level x, that maximizes profit, the marginal, revenue equals the marginal cost (see Fig. 3.5.12)., , Figure 3.5.12. The, production level x, that, maximizes profit has, C1(x,) = R '(x,)., , (b) Determine the number of units of a commodity :hat should be produced, to maximize the profit when the cost and revenue functions are given by, C ( x )= 800 + 30x - 0.01x2 and R ( x ) = 50x - 0.02x2,0 < x < 1500., , Solutlora, , (a) W e have P ( x ) = R ( x ) - C ( x ) . At as maximum point x, for P, we must, have P1(x,) = 0 by the critical point test. Therefore, R'(x,) - C 1 ( x 0=, ) 0 or, R '(x,) = C1(x,)as required., (b) The profit is, , P(x)= R ( x )- C ( X ), , The limits on x are that 0 < x < 1500. The critical points occur where, P 1 ( x )= 0; i.e., 20 - 0 . 0 2 ~= 0, i.e., x = 1000. This is a maximum since P " ( x ), = -0.02 is negative; a t x = 1000, P(1000) = 9200. The endpoint x = 0 is not, a maximum since P(0) = - 800 is negative. Also P(1500) = 6700, so the value, at this endpoint is less than a t ?c = 1000. Thus, the production level should be, set a t x = 1000. A, Example 92, , Solution, , Given a number a, , > 0, find the minimum value of, , W e are to minimize f ( x ) = ( a, , fi, , !'(XI, , Thus, x, , =, , =, , -, , ( a + x ) / G where x, , > 0., , + x ) / G on (0,a).By the quotient rule,, , (a + x ) ( a / 2 f i ), ax, , a is the only critical point in (0,a)., We observe that f ' ( x ) changes, , Copyright 1985 Springer-Verlag. All rights reserved.

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Chapter 3 Graphing and Maximum-Minimum Problems, , 188, , sign from negative to positive at x = a, so this is a minimum point. The, minimum value is f(a) = ( a + a ) / @ = 2. A, The result of this example can be rephrased by saying that, , G, , >2, , for every a, , > 0, x > 0., , Writing b for x, this becomes, , >, , forevery a > O ,, , b>0., , That is, the arithmetic mean ( a + b ) / 2 of a and b is larger than their geometric, mean @. This inequality was proved by using calculus. It can also be proved, by algebra alone. Indeed, since the square of any number is non-negative,, , so 2@, , < a + b. Hence,, , This is sometimes called the arithmetic-geometric mean inequality., --, , Exercises for Section 3.5, Find the maximum and minimum points and values for, each function on the given interval in Exercises 1-4., 1. 2x3 - 5x + 2 on [ l , w ), 2. x 3 - 6x 3 on ( - w , m), x 2 + 2 on (0,51, 3. -, , +, , 10. Figure 3.5.14 shows the temperatures recorded in, Goose Brow during a 24-hour period. When did, the maxima and minima occur, and what were, their values?, , X, , 4. -3x2 + 2x + 1 on ( - co, co), Find the maximum and minimum points for each of the, functions in Exercises 5-8., , 5., , X, , on(-m,~), , 6. x 2 2 2 x + 3 o n ( - 2 , 2 ], x2+5, , Figure 3.5.14. A cold day, in Goose Brow., 9. Figure 3.5.13 shows the annual inflation rate in, Oxbridge for 1970- 1980. Find the approximate, maximum and minimum points and values., , In Exercises 11-30, find the critical points, endpoints,, and global maximum and minimum points and values, for each function on the given interval., 11. f(x) = x 2 - x on [O, 1)., 12. f(x) = x 3 on [ - 1, GO)., 13. f(x) = x4 - 4x2 + 7 on [-4,2]., 14. f(x) = 4x4 - 2x2 1 (a) on [ - 10,201 and, (b) on 1-0.2, - 0.11., , +, , Figure 3.5.13. The annual, inflation rate in Oxbridge., , 17. f(x) = x2 - 3x + 1 on each of the following intervals., (a) ( 2 , ~ ), (c) (-3,21, (b) ( - w , f I, (dl (-5,21, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.5 Maximum-Minimum Problems, (e) (-29 2), (g), (f) ( - o o , m ), (h)[-8,81, 18. f ( x ) = x 3 - 2 x 1 on each of the intervals in, Exercise 17., 19. f ( x ) = 7 x 2 2 x + 4 on (a) [ - 1, 1 1 , (b) ( 0 , a ) ,, (c) I- 4 2 ) ., 20. f ( x ) = x 3 + 6 x 2 - 12x 7 on (a) ( - G O , oo),, (b) ( - Z61, (c) I- 2, I), td) - 4,41., 21. f ( x ) = x 3 - 3 x 2 3 x 1 on ( - oo, ao)., 22. f ( x ) = x 3 3 x 2 - 3 x 1 on [ - l , 2 ] ., 23. f ( x ) = x 3 - x on [ - 2 , 3 ] ., 24. f ( x ) = x 4 8 x 3 3 on [ - 1 , 1 ] ., [ - I 7 1 1, , +, , +, , +, , +, , +, +, , 29. f ( x ), , +, , x, , = -on ( -, , 1, , +, +, , + x2, , oo, oo)., , 30. f ( x ) = Xn, on ( - ao, ao), n a positive integer., 1+x2, Carry out step 1 for Exercises 31-34, (see Example 6 ) ., 31. Of all rectangles with area 1, which has the, smallest perimeter?, 32. Find the point on the arc of the parabola y = x 2, for 0 < x < 1 which is nearest to the point ( 0 , I)., [Hint: Consider the square of the distance between points.], 33. Two point masses which are a fixed distance, apart attract one another with a force which is, proportional to the product of the masses. Assuming that the sum of the two masses is M ,, what must the individual masses be so that the, force of attraction is as large as possible?, 34. Ten miles from home you remember that you left, the water running, which is costing you 10 cents, a n hour. Driving home at speed s miles per hour, costs you 6 ( s / 1 0 ) cents per mile. At what, speed should you drive to minimize the total cost, of gas and water?, , +, , 35., 36., 37., 38., 39., , Carry out step 2 for Exercise 31., Carry out step 2 for Exercise 32., Carry out step 2 for Exercise 33., Carry out step 2 for Exercise 34., A rectangular box, open at the top, is to be, constructed from a rectangular sheet of cardboard 50 centimeters by 80 centimeters by cutting out equal squares in the corners and folding, up the sides. What size squares should be cut out, for the container to have maximum volume?, 40. A window in the shape of a rectangle with a, semicircle on the top is to be made with a, perimeter of 4 meters. What is the largest possible area for such a window?, , 189, , 41. (a) A can is to be made to hold 1 liter (= 1000, cubic centimeters) of oil. If the can is in the, shape of a circular cylinder, what should the, radius and height be in order that the surface area of the can (top and bottom and, curved part) be as small as possible?, (b) What is the answer in part (a) if the total, capacity is to be V cubic centimeters?, (c) Suppose that the surface area of a can is, fixed at A square centimeters. What should, the dimensions be so that the capacity is, maximized?, 42. Determine the number of units of a commodity, that should be produced to maximize the profit, when the cost and revenue functions are given, by C ( x ) = 7 0 0 + 40x - 0 . 0 1 x 2 , R ( x ) = 8 0 x 0.03x2., 43. Determine the number of units of a commodity, that should be produced to maximize the profit, for the following cost and revenue functions:, C ( x ) = 360 8 0 x + . 0 0 2 x 2 . 0 0 0 0 1 x 3 , R ( x ), = lO0x - .0001 x 2 ., 44. If the cost of producing x calculators is C ( x ), = 100, 10x 0.01x2 and the price per calculator at production level x is P ( x ) = 26 - 0 . 1 ~, (this is called the demand equation), what production level should be set in order to maximize, profit?, 45. The U.S. Posi Office will accept rectangular, boxes only if the sum of the length and girth, (twice the width plus twice the height) is at most, 7 2 inches. What are the dimensions of the box of, maximum volume the Post Office will accept?, (You may assume that the width and height are, equal.), 46. Given n numbers, a , , . . . , a,, find a number x, which best approximates them in the sense that, the sum of the squares of the differences between, x and the n numbers is as small as possible., 47. One positive number plus the square of another, equals 48. Choose the numbers so that their, product is as large as possible., 48. Find the point or points on the arc of the parabola y = x 2 for 0 < x < 1 which are nearest to the, point (0, q). Express your answer in terms of q ., [Hint: Minimize the square of the distance between points.], 49. One thousand feet of fencing is to be used to, surround two areas, one square and one circular., What should the size of each area be in order, that the total area be (a) as large as possible and, (b) as small as possible?, 50. A forest can support up to 10,000 rabbits. If, there are x rabbits in the forest, each female can, (10,000 - x ) bunnies in, be expected to bear, a year. What total population will give rise to the, , +, , +, , +, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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190, , Chapter 3 Graphing and Maximum-Minimum Problems, , greatest number of newborn bunnies in a year?, (Assume that exactly half the rabbits are female;, ignore the fact that the bunnies may themselves, give birth to more young during the year; and, remember that the total population including, new bunnies is not to exceed 10,000)., 51. (a) In Fig. 3.5.15, for which value of y does the, line segment PQ have the shortest length?, Express your answer in terms of a and b., [Nint: Minimize the square of the length.], , Figure 3.5.15. The "ladder", PQ just fits into the corner, of the corridor., , (b) What is the length of the longest ladder, which can be slid along the floor around the, corner from a corridor of width a to a, corridor of width b?, 52. One hundred feet of fencing is to be used to, enclose two pens, one square and one triangular., What dimensions should the pens have to enclose the largest possible area?, 53. A conical dunce cap is to be made from a, circular piece of paper of circumference c by, cutting out a pie-shaped piece whose curved, outer edge has length I. What should 1 be so that, the resulting dunce cap has maximum volume?, 54. (a) Suppose that you drive from coast to coast, on Interstate Route 80 and your altitude, above sea level is f ( x ) when you are x miles, from San Francisco. Discuss the critical, points, endpoints, global maximum and, minimum points and values, and local maximum and minimum points for f ( x ) ., (b) Do as in part (a) for a hike to the top of Mt., Whitney, where x is the distance walked, from your starting point., In Exercises 55-59, try to guess the answer, or some, part of the answer, by using some symmetry of the, data., 55. Of all rectangles of area 1, which has the smallest, perimeter?, 56. Of all geometric figures with perimeter 1, which, has the greatest area?, 57. In Problem 1 (see Example 6 ) , suppose that we, allow the fencing to assume any shape, not necessarily rectangular, but still with one side along, the shore requiring no fencing. What shape gives, , the maximum area? A formal proof is not required., 58. What is the answer to Problem 2 (see Example 6 ), if the two intensities are equal? If one of the, intensities is eight times the other?, 59. Of all right triangles of area 1, guess which one, has the shortest perimeter. Which one, if any, has, the longest perimeter?, 60. Use calculus to show that the answer in Example, 8 is correct., *61. Find a function on [ - 1, I] which is continuous, but which is not differentiable at its maximum, point., *62. Find a function defined on [O, 11 which does not, have a maximum value on [O, I]., *63. Find a function defined on [ - 2 , 2 ] which has, neither a maximum value nor a minimum value, on [ - 2,2]., *64. Find a function defined on [ - 3 , - I ] which is, continuous at - 3 and - 1, has a maximum, value on [-3, - 11, but has no minimum value, on [ - 3, - 11., +65. Prove that the maximum value of f on I is, unique; that is,show that if MI and M2 both, satisfy conditions 1 and 2 of the definition at the, beginning of this section, then M, = M2. [Nint:, Show that M I s M2 and M2 < M I.], *66. Let the set I be contained in the set J, and, suppose that f is defined on J (and therefore, defined on I). Show that if m, and m, are the, minimum values of f on I and J, respectively,, then m, s m,., In Exercises 67-70, find the maximum and minimum, values of the given function on the given interval., +67. f ( x ) = px + (q/x), on (0, co)where p and q are, nonzero numbers., *68., , +69. f ( x ) = x / ( l - x 2 ) for - 1 < x s 0 and, f ( x ) = x 3 - x for 0 < x < 2, on ( - l,2)., *70., f(x> =, , ix, , -x4-1, -1, -2, , I, , for - 1 < ~ < 0, for O S X S ~, for l < x < o o, , on ( - 1, oo)., 8*71. Three equal light sources are spaced at x = 0, 1 ,, 2, along a line. At what points between the, sources is the total illumination least? See Problem 2, Example 6 . You should get a cubic equation for ( x - 112. Solve the equation numerically, by using the method of bisection., *72. What happens in Exercise 71 if there are four, light sources instead of three? Can you guess the, correct answer by using a symmetry argument, without doing any calculation?, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.6 The Mean Value Theorem, , 199, , *73. (This minimization problem involves no calcu-, , *74. The cost of running a boat is 10u3 dollars per, , lus.) You ran out of milk the day before your, weekly visit to the supermarket, and you must, pick up a container at the corner grocery store., At the corner store, a quart costs Q cents and a, half gallon costs G cents. At the supermarket,, milk costs q cents a quart and g cents a half, gallon., , mile where u is its speed in still water. What is, the most economical speed to run the boat upstream against a current of 5 miles per hour?, *75. Let f and g be defined on I. Under what conditions is the maximum value off + g on I equal, to the sum of the maximum values off and g, on I ?, *76. Let I be the set consisting of the whole numbers, from 1 to 1000, and let f(x) = 45x - x2. Find the, maximum and minimum points and values for f, on I., *77. Suppose that f is continuous on [a, h ] and is, differentiable and concave upward on the interval (a, h). Show that the maximum point off is, an endpoint., , (a) If Q = 45, G = 80, q = 38, and g = 65, what, size container should you buy at the corner, grocery to minimize your eventual milk expense? (Assume that a quart will get you, through the day.), (b) Under what conditions on Q, 6 , q, and g, should you buy a half gallon today?, , 3.6 The Mean Value, Theorem, If the derivative of a function is everywhere zero, then the function is constant., The mean value theorem is a technical result whose applications are more, important than the theorem itself. We begin this section with a statement of, the theorem, proceed immediately to the applications, and conclude with a, proof which uses the idea of global maxima and minima., The mean value theorem is, like the intermediate value and extreme value, theorems, an exisience theorem. It asserts the existence of a point in an, interval where a function has a particular behavior, but it does not tell you, how to find the point., , Figure 3.6.1. The slope, [f(b) - f(a)]/(b - of the, secant line 1, is equal to the, slope of the tangent line I,., , En physical terms, the mean value theorem says that the average velocity of a, moving object during an interval of time is equal to the instantaneous velocity, at some moment in the interval. Geometrically, the theorem says that a secant, line drawn through two points on a smooth graph is parallel to the tangent, line at some intermediate point on the curve. There may be more than one, such point, as in Fig. 3.6.1. Consideration of these physical and geometric, interpretations should make the theorem believable., We will prove the mean value theorem at the end of this section. For, now, we will concentrate on some applications. These tell us that if we know, something about f ' ( x ) for all x in [ a ,b], then we can conclude something, about the relation between values of f ( x > at different points in [ a ,b]., , Copyright 1985 Springer-Verlag. All rights reserved.

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192, , Chapter 3 Graphing and Maximum-Minimum Problems, , Suppose that f is differentiable on ( a ,b ) and continuous on [a,b]., 1. Assume that there are two numbers A and B such that, A, , < f'(x) < B, , for all x in ( a ,b)., , Then for any two distinct points x , and x , in [a,61,, , 2. If f ' ( x ) = 0 on ( a ,b), then f is constant on [a,b]., 3. Let F ( x ) and G ( x ) be functions such that F'(x) = G f ( x )for all x in, , 2 x , ] = f'(x0) for some, The first consequence holds since [ f(x,) - f ( x I ) ] / [ x x, between x , and x,, by the mean value theorem applied to f on the interval, with endpoints x , and x,. Since x, is in ( a ,b),f'(xo) lies between A and B,so, does the difference quotient. In particular, if f'(x) = 0, we can choose A = B, = 0, which implies that f ( x , ) = f ( x 2 ) Hence f is constant. To obtain the third, consequence, observe that F ( x ) - G ( x ) has zero derivative, so it is a constant., This consequence of the mean value theorem is a fact about antiderivatives, which we used in Section 2.5., Example 1, , Suppose that f is differentiable on the whoie real iine and that f'jx) is, constant. Prove that f is linear., , Solution Let rn be the constant value of f '., Method 1. We may apply the first consequence of the mean value, theorem with x , = 0, x , = x, A = m = B, to conclude that [ f ( x )- f (O)]/(x0) = rn. But then f ( x ) = rnx f (0) for all x, so f is linear., Method 2. Let g ( x ) = f ( x ) - mx. Then g f ( x )= m - rn = 0, so g is a constant., Setting x = 0, g ( x ) = f(0). Thus f ( x ) = mx + f(0) so f is linear. A, , +, , Example 2, , Let f be continuous on [ I , 31 and differentiable on (1,3). Suppose that for all x, in (1,3), 1 < f'(x) < 2. Prove that 2 < f(3) - f(1) < 4., , Solullon, , Apply the first consequence with A = 1, B = 2. Then we have the inequalities, 1 < [ f ( 3 )- f(1)]/(3- 1 ) < 2, and so 2 < f(3) - f ( 1 ) < 4. A, , Example 3, , Let f ( x ) = ( d / d x ) ( x l ., (a) Find f ' ( x ) ., (b) What does the second consequence of the mean value theorem tell you, about f ? What does it not tell you?, , Solullon, , (a) Since 1x1 is linear on ( - cn, 0 ) and (0, oo), its second derivative d21xl/dx2, = f ' ( x ) is identically zero for all x # 0., (b) By consequence 2, f is constant on any open intental on which it is, dyferentiable. I t follows that f is constant on ( - a , @ ) and (0, oo). The, corollary does not say that f is constant on ( - cn, cn). In fact, f ( - 2) = - 1,, while f(2) = 1. A, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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3.6 The Mean Value Theorem, , Example 4, , 193, , Suppose that F1(x) = x for all x and that F(3) = 2. What is F(x)?, , +, , Soiutron, , Let G(x) = x2. Then G'(x) = x = F1(x), so F(x) = G(x) + C = x 2 + C. To, evaluate C, set x = 3 : 2 = F ( 3 ) = + ( 3 2 ) + C = 92 +C.Thus C = 2 - q = - 2, and ~ ( x =) + x 2- 3. A, , Example 5, , The velocity of a train is kept between 40 and 50 kilometers per hour during a, trip of 200 kilometers. What can you say about the duration of the trip?, , 4, , Solution Before presenting a formal solution using the mean value theorem, let us use, common sense. If the velocity is at least 40 kilometers per hour, the trip takes, at most % = 5 hours. If the velocity is at most 50 kilometers per hour, the trip, takes at least ?# = 4 hours. Thus, the trip takes between 4 and 5 hours., To use the mean value theorem, let f(t) be the position of the train at time, t ; let a and b be the beginning and ending times of the trip. By consequence, 1 with A = 40 and B = 50, we have 40 < [f(b) - f(a)]/(b - a) < 50. But, f(b) - f(a) = 200, so, 200 < 50,, 40 Q b-a, 1, -1Q 5 b-a, 5ab-a>4., Hence the trip takes somewhere between 4 and 5 hours, as we found above. A, , Yo, , h, , ', , Figure 3.6.2. Rolle's, theorem: Iff is zero at the, ends of the interval, its, graph must have a, horizontal tangent, ., line, somewhere between., , Our proof of the mean value theorem will use two results from Sections 3.1, and 3.4, which we recall here:, , I. If x, lies in the open interval (a, b) and is a maximum or minimum point, for a function f on an interval [a, b], and iff is differentiable at x,, then, f'(x,) = 0 (critical point test)., 2. I f f is continuous on a closed interval [a, b], then f has a maximum and a, minimum point in [a, b] (extreme value theorem)., We now proceed with the proof in three steps., , Step 1 (Roile's, Let j be continuous on [a, b] and differentiable on (a, b), and assume that, Theorem7) f(a) = f(b) = 0. Then there is apoint xo in (a, b) at which f'(xo) = 0., Indeed, if f(x) = 0 for all x in [a, b], we can choose any x, in (a, b). So, assume that f is not everywhere zero. By the extreme value theorem, f has a, maximum point x , and a minimum point x,. Since f is zero at the ends of the, interval but is not identically zero, at least one of x , , x , lies in (a, b), and not, at an endpoint. Let x, be this point. By the critical point test, f '(x,) = 0, so, Rolle's theorem is proved., Rolle's theorem has a simple geometric interpretation (see Fig. 3.6.2)., Step 2 (Horserace, Theorem), , Suppose that f, and f2 are continuous on [a, b] and differeniiable on (a, b), and, assume that f,(a) = f,(a) and fl(b) = f2(b). Then there exisls a point x, in (a, b), such that f;(xo) = f;(xo>, , ', , Michel Rolle (1652-1719) (pronounced "roll") was actually best known for his attacks on the, calculus. He was one of the critics of the newly founded theory of Newton and Leibniz. It is an, irony of history that he has become so famous for "Rolle's theorem" when he did not even prove, the theorem but used it only as a remark concerning the location of roots of polynomials. (See, D. E. Smith, Source Book in Mathematics, Dover, 1929, pp. 251-260, for further information.), , Copyright 1985 Springer-Verlag. All rights reserved.

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194, , Chapter 3 Graphing and Maximum-Minimum Problems, , Let f ( x ) = f l ( x ) - f 2 ( x ) . Since fl and f2 are differentiable on ( a , b ) and, continuous on [ a , b ] , so is f . By assumption, f ( a ) = f ( b ) = 0 , so from Step 1,, f ' ( x , ) = 0 for some x , in ( a , b ) . Thus f { ( x , ) = f;(x,) as required (see Fig., 3.6.3)., We call this the "horserace theorem" because it has the following interpretation. Suppose that two horses run a race starting together and ending in a, tie. Then, at some time during the race, they must have had the same velocity., , Fiere 3.6.3. The curves y, = f, ( x ) and y = f 2 ( x ) have, , parallel tangent lines when, Y = f , ( x ) - f 2 ( x ) has a, , horizontal tangent line., We apply Step 2 to a given function f and a linear function I that matches f at, its endpoints, namely,, , I ( x ) =f ( a ), , +(x - a), , b-a, , Note that I(a) = f ( a ) , I(b) = f ( b ) , and I 1 ( x )= [f ( b ) - f ( a ) ] / ( b - a ) . By Step, 2, f'(x,) = l1(x0)= [f ( b ) - f ( a ) ] / ( b - a ) for some point x , in ( a ,b ) ., This completes our proof of the mean value theorem., Example 6, , Let f ( x ) = x4 - 9 x 3 + 26x2 - 24x. Note that f(0) = 0 and f(2) = 0 . Show, without calculating that 4 x 3 - 27x2 + 52x - 24 has a root somewhere strictly, between 0 and 2., , Solution, , Since f(0) = 0 and f(2) = 0 , Rolle's theorem shows that f' is zero at some x , in, ( 0 , 2 ) ; that is, 0 < x , < 2. A, , Example 7, , Suppose that f is a differentiable function such that f(0) = 0 and f ( 1 ) = 1., Show that j'(xo) = 2xo for some xo in ( 0 , l ) ., , Solution, , Use the horserace theorem with f , ( x ) = f ( x ) , f2(x) = x2, and [ a ,bj = 10, I]. A, , Exercises for Section 3.6, 1. Suppose that f is continuous on [O,+],and differentiable on (0,+), and that 0.3 < f'(x) s I for, 0 < x <+.Prove that 0.15 < [ f ( + )- f(0)] < 0.5., 2. Suppose that f is continuous on [ 3 , 5 ] and differentiable on ( 3 , 5 ) , and that f < f'(x) < f for, 3 < x < 5 . Show that 1 < [ j ( 5 )- f(3)] s $., 3. Suppose that ( d / d x ) [f ( x ) - 2 g ( x ) ] = 0 . What, can you say about the relationship between f and, , 8, , 4. If f " ( x ) = 0 on ( a ,b), what can you say about f?, , 5. Let f ( x ) = x S + 8 x 4 - 5 x 2 + 15. Prove that, somewhere between - 1 and 0 the tangent line to, the graph off has slope - 2 ., 6. Let f ( x ) = 5 x 4 + 9 x 3 - 1 l x 2 + 10. Prove that the, graph off has slope 9 somewhere between - 1, and 1., 7. Let f ( x ) =/=., Show that somewhere between 2 and 3 the tangent line to the graph off, , has slope, , m., , Copyright 1985 Springer-Verlag. All rights reserved.

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Review Exercises for Chapter 3, , 8. Letf(x) = x 7 - x 5 - x 4 + 2 x + 1. Prove that the, graph o f f has slope 2 somewhere between - 1, and 1., 9. Suppose that a n object lies at x = 4 when t = 0, and that the velocity dx/dt is 35 with a possible, error of -t 1, for all t in [0, 21. What can you say, about the object's position when t = 2?, 10. The fuel consumption of an automobile varies, between 17 and 23 miles per gallon, according to, the conditions of driving. Let f(x) be the number, of gallons of fuel in the tank after x miles have, been driven. If f(100) = 15, give upper and lower, estimates for f(200)., 11. Directly verify the validity of the mean value, theorem for f ( x ) = x 2 - x + 1 o n [ - 1,2] by, finding the point(s) x,. Sketch., 12. Let f(x) = x 3 on the interval [-2,3]. Find explicitly the value(s) of x, whose existence is guaranteed by the mean value theorem. Sketch., 13. Suppose that F'(x) = - ( l / x 2 ) for all x f: 0. Is, F ( x ) = I / x + C, where C is a constant?, 14. Suppose that f ( x ) = x 2 and f(1) = 0. What is, f(x>?, 15. Let f(x) = 1x1 - 1. Then f ( - 1) = f(1) = 0, but, f ( x ) is never equal to zero on [ - 1, 11. Does this, contradict Rolle's theorem? Explain., 16. Suppose that the horses in a race cross the finish, line with equal velocities. Must they have had the, same acceleration at some time during the race?, Find the antiderivatives of the functions in Exercises, 17-20., 17. f(x) = ) x - 4 x 2 + 21., 18. XI = 6 2 - ax3, + 1 5 -~ 11., 1 + 2x., 19. f(x) = x2, 20. f(x) = 2 x ( x 2 7)'O0., Find the antiderivative F(x) for the given function f(x), satisfying the given condition in Exercises 2 1-24., ) 2 ~ 4 ~; ( 1 =) 2., 21. f ( ~ =, 22. f(x) = 4 - x ; F(2) = 1., 23. f(x) = x 4 + x 3 + x2; F ( l ) = I ., , +, , 995, , 25. Let f be twice differentiable on (a, b), and suppose that f(x) = 0 at three distinct points in, (a, b). Prove that there is a point xo in (a, b) at, which fn(x,) = 0., a26. Use the mean value theorem to prove the increasing function theorem: I f f is continuous on [a, b], and differentiable on (a, b), and f'(x) > 0 for x, in (a, b), then f is increasing on [a. b]., a27. Suppose that f and g are continuous on [a, b] and, that f' and g' are continuous on (a, b). Assume, that, Prove that there is a number c in ( a , b) such that, the line tangent to the graph of f at (c, f(c)) is, parallel to the line tangent to the graph of g, - at, (c, g(c))., +28. Let f be a polynomial. Suppose that f has a, double zero at a and b. (A polynomial f has a, doubje zero at x = a if f(x) = ( x - a)2g(x) for, some polynomial g). Show that f'(x) has at least, three roots in [ a ,b]., a29. The coyote population in Nevada was the same, at three consecutive times t l , t 2 ,t3. Assume that, the population N ( t ) is a differentiable function, of time t and is nonconstant on I t , , t2] and on, [t,, t,]. Establish by virtue of the mean value, theorem the existence of two times T , T* in, ( I , , r2: and ( t 2 ,r,), respectively, for which the, coyote population decreases., *30. (a) Let f be differentiable on (a, b) [and continuous on [a, b]]. Suppose that, for all x in the, open interval (a, b), the derivative f'(x) belongs to a certain set S of real numbers., Show that for any two distinct points x , and, x2 in (0,b) [in [ a ,b]], the difference quotient, , belongs io S as well., (b) Use (a) to prove consequence 1 of the mean, value theorem., (c) What does (a) tell you if S = [a, b)? If S, = (0, oo)?, , Review Exercises 808. Chapter 3, On what intervals are the functions in Exercises 1-4, continuous?, I . f(x ) =, , 1, , for x, , --------, , J x V, , 3. f ( x ), , =, , 1, x-l, , -, , 5. Explain why the function h given by, , 1, +x-2, , >2, , is continuous at 2. Is h continuous on the whole, real line?, 6. Let f(x) = [ I / x ] + [(x2- I)/x]. Can you define f(0) so that the resulting function is continuous at all x?, 7. Find a function which is continuous on the, whole real line and differentiable for all x except 1, 2, and 3. ( A sketch will do.), , Copyright 1985 Springer-Verlag. All rights reserved.

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Chapter 3 Graphing and Maximum-Minimum Problems, , 7198, , 8. Sketch the graph of a function that has one, vertical asymptote at .w = I . two maximum, points, and no other critical points., 9. Show that f ( x ) = x% x 2 + x - 2 has a root, between x = 0 and x = 1., 10. Show that f i x ) = x 5 + x 3 + 2x + I has a root, between x = - 1 and x = 0., 11. Find a solution to the equation in Exercise 9 to, within two decimal places using the method of, bisection., 8 12. Find a solution to the equation in Exercise 10 to, within two decimal places using the method of, bisection., In Exercises 13-16, determine the intervals on which, the given function is increasing and decreasing., 13. 8 x 3 - 3 x 2 + 2., 14. 5x3 + 2x2 - 3x + 10., X, , 15., 1, , + ( x - 1)2 ', X, , ,, Xf1,2., x2 - 3x 2, 17. Suppose that the British inflation rate (in percent per year) from 1990 to 2000 is given by the, function B ( t ) = 10(t3/100 - t2/20 - t / 4 + I),, where i is the time in years from January 1,, 1993. When is the inflation rate increasing?, Now might a politician react to minimum, points? Maximum points? Inflection points'?, 18. A paint can is kicked off the roof of a 156-foot, building, its distance S from the ground being, given by S = 156 + 4t - 16t2 ( S in feet, t in, seconds)., (a) How high up does the can go before it falls, downward?, (b) At what time does the can return to roof, level?, (c) Find the velocity of the can when it collides with the ground., 19. The rate of growth of a tree between t = 0 and, t = 100 y e a r s is given by t h e f o r m u l a, r ( t ) = 1 0 - ~ ( -r ~75t2) + 10 inches per year., When is the growth slowing down? speeding, up?, 20. A bicycle is moving at ( A t3 t 4t2 - 31 + 2), miles per hour, where t is the time in hours,, 0 < t < 2. Does the bicycle change direction, during this period?, In Exercises 21-24, find the critical points of the given, function and determine if they are local maxima. minima, or neither., + 3 8x2 + 2x - 3, 21. 2 ~ ~ - 5 ~ ~ + 4 ~ 22., @ 24. ( x 3 - 12x +, 23. ( x 2 + 2x - 3)', For each of the functions in Exercises 25-30. answer, the questions below and draw a graph:, (i) Where isfcontinuous?, (ii) Where is f differentiable?, (iii) On which intervals is f increasing? Decreasing?, (iv) Where is./ concave upward or downward?, 16., , +, , ( v ) What are the critical points, endpoints, local maximum and minimum points, and inflection points for f ?, 25. f ( x ) = - 7 x 3 + 2x2 + 15 on ( - cu, oc)., 26. f ( s ) = 6 x 2 + 3 x + 4 on ( - oc,m ) ., 27. f ( x ) = x 3 - 2x + 1 on [-1,2]., 28. /(x) = x 4 - 3 x 3 + x 2 on [-6,6]., , 3x on its domain, 2-5x, x, 30. f ( x ) = ---- on ( - m , c u ) , n is a positive, 1 + x2", integer., Explain why each statement in Exercises 31-40 is true, or false: (Justify if true, give a counterexample if false.), 3 1. Every continuous function is differentiable., 32. If f is a continuous function on [0,2], and, f(1) = - 1 and f(2) = 1, then there must be a, point x in [0, 11 where f(x) = 0., 33. If a function is increasing at x = 1 and at x = 2,, it must be increasing at every point between 1, and 2., 34. If a differentiable function f on ( - oo,oo) has a, local maximum point at x == 0,then f'(0) = 0., 35. f(x) = x 4 - x 3 is increasing at zero., 36. f(x) = x 3 + 3 x takes its largest value on [ - 1, 11, at an endpoint., 37. Parabolas never have inflection points., 38. All cubic functions y = a x 3 + bx2 cx + d,, u + 0 have exactly one inflection point., 39. f(x) = 3/(5x2 + 1) on ( - oo,oof has a !oca!, maximum a t x = 0., 40. y = x 5 has an inflection point at x = 0., Sketch the graphs of the functions in Exercises 41-50., 41. x3 + 3 x + 2, 42. x 5 - 3x4, 29. f ( x ) =, , --------, , +, , 49. X ( X- 1l3j2, SO. X(X- I ) ~ ' ~, In Exercises 51-56, find the maximum and minimum, value of each function on the designated interval., 51. f ( x ) = 3x3 + x 2 - x + 5; [-2,2], 52. /(x) = 2 ~ +3 5 ~ -2 4 ~ [ ; 2.21, , x3, :(-2.2), 55. /( x ) = -------x2-4, 56. f ( x ) = - x4 + 8x' + 2; ( - cu, m ), In Exercises 57-62, sketch the graph of the indicated, function on the designated interval., 57. The function in Exercise 5 1., 58. The function in Exercise 52., 59. The function in Exercise 53., 60. The function in Exercise 54., 61. The function in Exercise 55., 62. The function in Exercise 56., , Copyright 1985 Springer-Verlag. All rights reserved.

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Review Exercises for Chapter 3, , 63. A wooden picture frame is to be 2 inches wide, on top and bottom and 1 inch wide on the, sides. Assuming the cost of a frame to be proportional to its front surface area, find the dimensions of the cheapest frame which will surround an area of 100 square inches., 64. At time t, a rectangle has sides given by I(t), =2, I and w(t) = 1 + t2, for - 1 g t g I. (a), When does this rectangle have minimum area?, (b) When is the area shrinking the fastest?, 65. Find the dimensions of the right circular cylinder of greatest volume that can be inscribed in a, given right circular cone. Express your answer, in terms of the height h of the cone and the, radius r of the base of the cone., 66. A rectangular box with square bottom is to, have a volume of 648 cubic centimeters. The, top and bottom are to be padded with foam, and pressboard, which costs three times as, much per square centimeter as the fiberboard, used for the sides. Which dimensions produce, the box of least cost?, 67. A box company paints its open-top squarebottom boxes white on the bottom and two, sides and red on the remaining two sides. If red, paint costs 50% more than white paint, what are, the dimensions of the box with volume V which, costs least to paint? In what sense is the, "shape" of this box independent of V?, 68. Find the maximum area an isosceles triangle, can have if each of its equal sides has a length, of 10 centimeters., 69. The material for the top, bottom, and lateral, surface of a tin can costs of a cent per square, centimeter. The cost of sealing the top and, bottom to the lateral surface is p cents times the, total length in centimeters of the rims (see Fig., 3.R.1) which are to be sealed. Find (in terms of, p) the dimensions of the cheapest can which will, hold a volume of V cubic centimeters. Express, your answer in terms of the solution of a cubic, equation; d o not solve it., , +, , +", , 70. We quote from V. Belevitch, Classical Network, Theory (Holden-Day, 1968, p. 159): "When a, generator of e.m.f. e and internal positive resistance R, is connected to a positive load resis-, , 197, , tance R, the active power dissipated in the load, , +, , e a constant,, is maximum with respect to R for R = R,, that, is, when the load resistance is matched to the, internal resistance of the generator.", Verify this statement. (You do not need to, know any electrical engineering.), 71. A fixed-frequency generator producting 6 volts, is connected to a coil of 0.05 henry, in parallel, with a resistor of 0.5 ohm and a second resistor, of x ohms. The power P is given by, w = l e j 2 ~ / ( ~R, )2,, , Find the maximum power and the value of x, which produces it., 72. The bloodstream drug concentration C ( t ) in a, certain patient's bloodstream t hours after injection is given by, , Find the maximum concentration and the number of hours after injection at which it occurs., 73. In a drug-sensitivity problem, the change T in, body temperature for an x-milligram injection, is given by, , (a) Find the sensitivity T'(x) when x = 2., (b) Graph T versus x., 74. The power P developed by the engine of an, aircraft flying at a constant (subsonic) speed c, is given by, , where c > 0, d > 0., (a) Find the speed no which minimizes the, power., (b) Let Q ( t ) denote the amount of fuel (in, gallons) the aircraft has at time t. Assume, the power is proportional to the rate of fuel, consumption. At what speed will the flight, time from takeoff to fuel exhaustion be, maximized?, 75. Prove that for any positive numbers a. h. c., , by minimizing f ( . u ) = ( n + h + x ) / '$&., 76. Find the number x which best approximates 1 ,, 2. 3, and 5 in the sense that the sum of the, fourth powers of the differences between .w and, each number is minimized. Compute x to, within 0.1 by using the bisection method., 77. A manufacturer of hand calculators can produce up to 50,000 units with a wholesale price, , Copyright 1985 Springer-Verlag. All rights reserved.

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498, , Chapter 3 Graphing a n d Maximum-Minimum P r o b l e m s, of $16 for a fixed cost of $9000 plus $1 1.50 for, each unit produced. Let n- stand for the number, of units produced., (a) Explain why the total revenue R , total cost, C and profit P must satisfy the equations, , 82. Saveway checkers have to memorize the sale, prices from the previous day's newspaper advertisement. A reasonable approximation for the, percentage P of the new prices memorized after, t hours of checking is P ( t ) = 96t - 24t2, 0 s i, < 3., , (b) The break-even point is the production level, x for which the profit is zero. Find it., (c) Determine the production level x which, corresponds to a $4500 profit., (d) If more than 50,000 units are produced. the, revenue is R = 16x - $ x 10 "x2, but the, cost formula is unchanged. Find the x that, maximizes profits., 78. A manufacturer sells x hole punchers per week., The weekly cost and revenue equations are, , What is the maximum percentage of the, list memorized ba)~"rn$i, . P A ~ Q L o3, . shceaes?, ~, If f"'(x) = 0 on ( - m , m), show that there are, constants A, B, and C such that f(x) = .4x2 +, Bx + C., Verify the horserace theorem for f l ( x ) = x 2 +, x - 2 and f2(x) = x i + 3 x 2 - 2x - 2 on [0, I]., Persons between 30 and 75 inches in height h, have average weight W = f (h/ 10)' Ibs., (a) What is the average weight of a person 5, feet 2 inches in height?, (b) A second grade child grows from 48 inches, to 50 inches. Use the linear approximation, to estimate his approximate weight gain, and compare with a direct calculation., A storage vessel for a chemical bleach mixture, is manufactured by coating the inside of a thin, hollow plastic cube with fiberglass. The cube, has 12-inch sides, and the coating is 4 inch, thick. Use the volume formula V = x3 and the, mean value theorem to approximate the volume, of the fiberglass coating. [Hint: The fiberglass, volume is V(12) - V(11.8).], (a) Show that the quotient f / g has a critical, point when the ratios f'/f and g ' / g are, equal., (b) Find a similar criterion for a product fg to, have a critical point., (Refer to Review Problem 30 of Chapter I and, Fig. 1.R.3.) Let P, with coordinates ( x , , y,),be, inside the parabola y = x 2 ; that is, y , > x:., Show that the path consisting of two straight, lines joining P to a point (x, x2) on the parabola, and then ( x , x " to (0,:) has minimum length, when the first segment is vertical., The graph of a factored polynomial (such as, v = X ( X - l)*(.x - 213(.x - 3)'(x - 4)12) near a, root r appears similar to that of the function, p = C ( X - r)", where c and n are chosen appropriately for the root r., (a) Let y = x ( x + I ) ~ ( x 2)'. For values of x, near 2, the factor x ( x 1)* is approximately 2(2 I ) =~ 18, so y is approximately 18(x - 2)'. Sketch the graph near, x = 2., (b) Argue that near the roots 0, 1 , 3 the equation y = lOx(x - l ) ' ( ~- 3)2 looks geometrically like -v = - 90x, y = 40(x - I ) ~y,, = 240(x - 3)2, respectively. Use this information to help sketch the graph from x = 0, to x = 4., , x2, 0 s x 4 8000, R ( x ) = IOx - 1000 ', (a) Find the minimum cost., (b) Find the maximum revenue., (c) Define the profit P by the formula P ( x ), = R(x) - C(x). Find the maximum profit., 79. A homeowner plans to construct a rectangular, vegetable garden with a fence around it. The, garden requires 800 square feet, and one edge is, on the property line. Three sides of the fence, will be chain-link costing $2.00 per linear foot,, while the property line side will be inexpensive, screening costing $.50 per linear foot. Which, dimensions will cost the least?, 80. A rental agency for compact cars rents 96 cars, each day for $16.00 per day. Each dollar increase in the rental rate results in four fewer, cars being rented., (a) Wow should the rate be adjusted to maximize the income?, (b) What is the maximum income?, 81. The Smellter steel works and the Green Copper, Corporation smelter are located about 40 miles, apart. Particulate matter concentrations in parts, per million theoretically decrease by an inverse, square law, giving, for example,, , (l, , sx, , ,< 39),, , k a constant,, , as the concentration x miles from Smellter. This, model assumes that Green emits three times, more particulate matter than Smellter., (a) Find C f ( x ) all critical points of C., (b) Assuming you wished to build a house, between Smellter and Green at the point of, least particulate concentration, how far, would you be from Smellter?, , 83., , 84., 85., , 86., , 87., , 88., , 89., , +, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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Review Exercises for Chapter 3, , +, , +90. The astroid x213 y 2 / 3= 1 is a planar curve, which admits no self-intersections, but it has, four cusp points., (a) Apply symmetry methods to graph the, astroid., (b) Divide the astroid into two curves, each of, which is the graph of a continuous function. Find equations for these functions., (c) Find the points where dy/dx is not defined, and compare with the cusp points. Use, implicit differentiation., (d) Explain why no tangent line exists at the, cusp points., +91. A function f is said to majorize a function g on, [a, b] if f(x) > g(x) for all a < x < b., (a) Show by means of a graph that x majorizes, x 2 on [0, I]., (b) Argue that "f majorizes g on [a, b y means, that the curve .y = -f(x), a G x < b, lies on, or above the curve y = g(x), a G x G b., (c) If m > n > 0, then x n majorizes x m on, [O, 11. Explain fully., (d) If m > n > 0, ( x - a)" majorizes ( x - a)", on [ a , a I]. Why?, (e) Given that m j. n, m > 0, and n > 0, determine on which intervals ( x - a)" majorizes, (X - a)" (or conversely)., (f) Graph the functions y = x - 1. y = ( x = (X - I ) ~ , = (X - i)4, j, = ( X - 1)', on the same set of axes for - 1 < x < 3., +92. Let f(x) = 1/(1 x2)., (a) For which values of c is the function, f(x) + cx increasing on the whole real line?, Sketch the graph of f(x) + cx for one such, C., (b) For which values of c is the function, f(x) - cx decreasing on the whole real, line? Sketch the graph of f(x) - cx for one, such c., (c) HOWare your answers in parts (a) and (b), related to the inflection points o f f ?, *93. Let f be a nonconstant polynom~alsuch that, f(0) = f ( I). Prove that f has a e+ti.ct C o d mhe\l*aurn, , +, , +, , +94. Prove that, given any n numbers a , , . . . , a,,, there is a uniquely determined number x for, which the sum of the fourth powers of the, differences between x and the a,'s is minimized., [Hint: Use the second derivative.], +95. Prove the following intermediate value theorem, for derivatives: Iff is differentiable at all points, , 189, , of [a, b], and if f'(a) and f'(b) have opposite, signs, then there is a point x, E ( a ,b) such that, f'(x,) = 0. [The example given in Review Exercise 84, Chapter 5, shows that this theorem does, not follow from the intermediate value theorem, of Section 3.1 .], +96. Let f be differentiable function on (0,m) such, that all the tangent lines to the graph off pass, through the origin. Prove that f is linear. [Hint:, Consider the function f(x)/x.], +97. Let f be continuous on [3,5] and differentiable, on (3,5), and suppose that f(3) = 6 and f(5), = 10. Prove that, for some x, in the interval, (3,5), the tangent line to the graph of f at x,, passes through the origin. [Hint: Consider the, function f(x)/x.] Illustrate your result with a, sketch., +98. A function defined on (a, b) is called convex, when the following inequality holds for x, y, in, (a, b) and t in [0, I]:, Demonstrate by the following graphical argument that if f " is continuous and positive on, (a, b), then f is convex:, (a) Show that f is concave upward at each, point of (a, b)., (b) By drawing a graph, convince yourself that, the straight line joining two points on the, graph lies above the graph between those, points., (c) Use the fact in (b) to deduce the desired, inequality., +99. Suppose that f is continuous on [a, b]. f ( a ), = f ( b ) = 0. and x2fU(x)+ 4xfr(x) 2f(x) > 0, for x in (a,h). Prove that f ( x ) < 0 for x in, [a, b]., *loo. A rubber cube of incompressible material is, pulled on all faces with a force T. The material, stretches by a factor v in two directions and, contracts by a factor c - 2 in the other. By, balancing forces, one can establish RivlinS equation :, , +, , where a is a constant (analogous to the spring, constant for a spring). Show that Rivlin's equation has one (real) solution if T < 33,/? a and, has three solutions if T > 3 '6 a ., , Copyright 1985 Springer-Verlag. All rights reserved.

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11.2 ~ ' ~ 6 p i t a lRule, 's, , +, , < 1, which in turn implies that I f(x)I < I LI 1, (since I f(x)I = I f ( x ) - L + LI < I f ( x ) - LI +, I LI < 1 + I LI). Finally, choose 6, > 0 such that, I g ( x ) - MI < &/[2(ILI l)]whenever Ix - xol, < S 3 , x # X O . Let E be the smallest of 6 , , 6,, and, 6,. If Ix - xol < 6 , x # xo, then Ix - xol < S l ,, I X - xol < 8 2 , and Ix - xol < S 3 , SO by the, choice of 6 , , S,, S3, we have, , +, , *78., *79., *80., and so I f ( x ) g ( x ) - LMI < E ., Now prove the quotient rule for limits., *77. Study the following proof of the one-sided comf ( x ) = L and g, posite function rule: If lim,,,,,+, is continuous at L , then g ( f ( x ) ) is defined for all, x in some interval of the form ( x o , b ) , and, lim,, +, , ,, g ( f ( x ) )= g ( L ) ., Proof: Let E > 0. We must find a positive, , *81., , 521, , number 6 such that whenever xo < x < x, + 6 ,, g ( f ( x ) ) is defined and I g ( f ( x ) ) - g(L)I < E ., Since g is continuous at L , there is a positive, number p such that whenever 1 y - LI < p, g ( y ), is defined and I g ( y ) - g ( L ) ( < E . Now since, lim,,xo+ f ( x ) = L, we can find a positive number 6 such that whenever x , < x < x , + 6 ,, I f ( x ) - LI < p. For such x , we apply the previously obtained property of p, with y = f ( x ) , to, conclude that g ( f ( x ) ) is defined and that, I g ( f ( x ) ) - g(L)I < E ., Now prove the composite function rule., Use the E-A definition to prove the sum rule for, limits at infinity., Use the B-6 definition to prove the reciprocal, test for infinite limits., Suppose that a function f is defined on an open, interval I containing x,, and that there are numbers m and K such that we have the inequality, I f ( x ) - f(xo) - m ( x - xo)l < K I X - x0I2 for all, x in I. Prove that f is differentiable at xo with, derivative f'(xo) = m ., Show that lim,,,, f ( x ) = 1 if and only if, limyjo+ f ( l / y ) = I. (This reduces the computation of limits at infinity to one-sided limits at, zero.), , LyH6pital'sRule, Differentiation can b e used t o evaluate limits., , L'HGpital's rule2 is a very efficient way of using differential calculus to, evaluate limits. It is not necessary to have mastered the theoretical portions of, the previous sections to use YHGpital's rule, but you should review some of the, computational aspects of limits from either Section 11.1 or Section 1.3., L'HGpital's rule deals with limits of the form lim,,,.f ( x ) / g ( x ) ] , where, lim,,,o f ( x ) and lim,,,og(x) are both equal to zero or infinity, so that the, quotient rule cannot be applied. Such limits are called indeterminate forms., (One can also replace x , by CQ, x , + , or x , - .), Our first objective is to calculate lirn,,,o[ f ( x ) / g ( x ) ] if f ( x , ) = 0 and, g ( x , ) = 0. Substituting x = x , gives us 8, so we say that we are dealing with, an indeterminate form of type 8. Such forms occurred when we considered the, derivative as a limit of difference quotients; in Section 1.3 we used the limit, rules to evaluate some simple derivatives. Now we can work the other way, around, using our ability to calculate derivatives in order to evaluate quite, complicated limits: l'H6pitaYs rule provides the means for doing this., The following box gives the simplest version of YHGpital's rule., In 1696, Guillaume F. A. l'H6pital published in Paris the first calculus textbook: Analyse des, Infiniment Petits (Analysis of the infinitely small). Included was a proof of what is now referred to, as SHSpital's rule; the idea, however, probably came from J. Bernoulli. This rule was the subject, of some work by A. Cauchy, who clarified its proof in his Cours d'Analyse (Course in analysis) in, 1823. The foundations were in debate until almost 1900. See, for instance, the very readable, article, "The Law of the Mean and the Limits $, z," by W. F. Osgood, Annals of Mathematics,, Volume 12 (1898-1899), pp. 65-78., , Copyright 1985 Springer-Verlag. All rights reserved.

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522, , Chapter 111 Limits, ~ ' ~ 6 p i t a lRule,, 's, and Numerical Methods, , L'HGpital's Rule: Preliminary Version, , I, , Let f and g be differentiable in an open interval containing x,; assume, that f(x,) = g(x,) = 0. If g'(x,) # 0, then, f ( x ) - f'(x0), lim - - -., g(x), gf(xo), , X+Xo, , To prove this, we use the fact that f(x,), , =0, , and g(x,) = 0 to write, , As x tends to x,, the numerator tends to f'(x,), and the denominator tends to, gf(xo)Z 0, so the result follows from the quotient rule for limits., Let us verify this rule on a simple example., Example 1, Solution, , Find lim, x+l, , [a, 1., X-1, , Here we take x, = 1, f(x) = x3 - 1, and g(x) = x - 1. Since g'(1) = 1, the, preliminary version of l'H6pital's rule applies to give, , We know two other ways (from Chapter 1) to calculate this limit. First, we can, factor the numerator:, , Letting x + 1, we again recover the limit 3. Second, we can recognize the, function (x3 - l)/(x - 1) as the different quotient [h(x) - h(l)]/(x - 1) for, h(x) = x3. As x + 1, this different quotient approaches the derivative of h at, x = 1, namely 3. A, The next example begins to show the power of l'H6pital's rule in a more, difficult limit., Example 2, , Find 1im, X+O, , Solution, , 'OSX, , -, , sinx, , ., , We apply l'Hdpital's rule with f(x) = cosx - 1 and g(x) = sinx. We have, f(0) = 0, g(0) = 0, and gf(0) = 1 # 0, so, lim, X+O, , COSX, , f'(0), gf(0), , sinx, , This method does not solve all, find, lim sinx - x, x+o, , -sin(O), cos(0), , - -= 0. g, - 1 = --, , 8, , problems. For example, suppose we wish to, , x3, , If we differentiate the numerator and denominator, we get (cosx - 1)/3x2,, which becomes g when we set x = 0. T h ~ suggests, s, that we use I'H6pital's rule, , Copyright 1985 Springer-Verlag. All rights reserved.

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11.2 ~ ' ~ b p i t a lRule, 's, , 523, , again, but to do so, we need to know that lim,,,, [f(x)/g(x)] is equal to, limx+xo[f'(x)/gf(x)], even when f'(xo)/gf(xo) is again indeterminate. The, following strengthened version of l'H6pital's rule is the result we need. Its, proof is given later in the section., , Let f and g be differentiable on an open interval containing x,, except, perhaps at x, itself. Assume:, , Example 3, , Solution, , Calculate lim, x+o, , This is in, , 'OSX x2, , ., , i form, so by l'H6pital's, , rule,, , lim cos x - 1 = lim - sinx, x2, x-to, 2x, , x+o, , if the latter limit can be shown to exist. However, we can use l'H6pital's rule, again to write, , - sinx =, - cosx, lim 2x, x+o, 2, , x+o, , Now we may use the continuity of cos x to substitute x = 0 and find the last, limit to be - f ; thus, , To keep track of what is going on, some students like to make a table:, form, , f, -, , type, , limit, , -, , g, , cosx - 1, x2, , 0 indeterminate, 0, , ?, , f', -, , - sin x, , 0, , ?, , g', , 2x, , - cosx, , 0, , indeterminate, , determinate, , Each time the numerator and denominator are differentiated, we must check, the type of limit; if it is $, we proceed and are sure to stop when the limit, becomes determinate, that is, when it can be evaluated by substitution of the, limiting value., , Copyright 1985 Springer-Verlag. All rights reserved.

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524, , Chapter 11 Limits, ~ ' ~ 6 p i t a lRule,, 's, and Numerical Methods, , Warning, , Exarnple 4, Solution, , If l'H6pital's rule is used when the limit is determinate, incorrect answers can, result. For example, limX,,[(x2 + l)/x] = oo but l'H6pital's rule would lead, to limX,,(2x/l) which is zero (and is incorrect)., Find lirn sinx - ., X-+O tanx - x, This is in $ form, so we use l'H6pital's rule:, form, g, , l, , type, , limit, , sinx - x, tanx - x, cosx - I, sec2x - 1, - sinx, 2 sec x (sec x tan x), -, , COSX, , 4 sec2xtan2.-c+ 2 sec4x, ~h~~ lim ~ i n x- x, X+O, tanx - x, , determinate, , = --, , :.A, , L'HGpital's rule also holds for one-sided limits, limits as x + co, or if we have, indeterminates of the form g.To prove the rule for the form $ in case, x + m , weusea trick: s e t t = l / x , s o t h a t x = l / t a n d t - + O + a s x + + c o ., Then, f'(x> = lim f'(l/t>, lirn x++w gf(x) t+o+ gf(l/t), = lirn, "0', , =, , lirn, , t+o+, , =, , - t"f'l/t), - t2g'(l/t), , (d/dt)f ( l / t ), (d/dt)g(l/t), , f(l/t), lirn t+o+ g ( l / t ), , = lirn, , x++m, , (by the chain rule), (by l'Hbpital's rule), , f (XI, -, , g(x> ', It is tempting to use a similar trick for the g form as x + x,, but it does, not work. If we write, , which is in the $ form, we get, f(.), lirn - ,im -gf(x>/E g(x)12, "+" g(x), -f'(x)/[f(x)I2, which is no easier to handle. For the correct proof, see Exercise 42., The use of l'H6pital's rule is summarized in the following display., , Copyright 1985 Springer-Verlag. All rights reserved.

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11.2 ~ ' ~ 6 ~ i t a, Rule, l's, , 525, , and take the limit of the new fraction; repeat the process as many times, as necessary, checking each time that l'H6pital's rule applies., If limx,xo f(x) = limx+xog(x) = 0 (or each is + co), then, , (x, may be replaced by, , + co or xo +- )., , The result of the next example was stated at the beginning of Section 6.4. The, solution by l'H6pital's rule is much easier than the one given in Review, Exercise 90 of Chapter 6., Example 5, Solution, , Find lim, x+m, , h,, where p > 0., XP, , This is in the form g.Differentiating the numerator and denominator, we, find, , since p, , > 0. A, , Certain expressions which do not appear to be in the form f(x)/g(x) can be, put in that form with some manipulation. For example, the indeterminate, form co . 0 appears when we wish to evaluate limx,xo f(x)g(x) where, limx,xo f(x) = co and limx,xog(x) = 0. This can be converted to 8 or g form, by writing, , Example 6, Solution, , Find limxjo+ x ln x., We write x lnx as (lnx)/(l/x), which is now in g form. Thus, , - lim (- x) = O. A, lim x l n x = lim -!!E = lim -x+o+ 1/x x+o+ - 1/x2 x+o+, , x+o+, , Indeterminate forms of the type 0' and 1" can be handled by using logarithms:, Example 7, Solution, , Find (a) limxjo+ x x and (b) lim,,,x'/(', , -"), , (a) This is of the form 0°, which is indeterminate because zero to any power is, zero, while any number to the zeroth power is 1. To obtain a form to which, l'H6pital's rule is applicable, we write x x as exp(x ln x). By Example 6, we, have lirn,,, + x ln x = 0. Since g(x) = exp(x) is continuous, the composite, function rule applies, giving lim,,, + exp(x ln x) = e ~ p ( l i r n ~, + ,x~, ln x) = e0, = 1, so lim,,,,, xx = 1. (Numerically, O.lO.' = 0.79, O.OO1°.OO1= 0.993, and, O.OOOO1°~OOOO1, = 0.99988.), , Copyright 1985 Springer-Verlag. All rights reserved.

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526, , Chapter 11 Limits, ~ ' ~ 6 p i t a lRule,, 's, and Numerical Methods, , (b) This has the indeterminate form 1". We have xi/("- ') = e('" ")/("-I);, applying l'H6pital's rule gives, , limx'/(x-l), x+ 1, , =, , lim e ( l n ~ ) / ( ~ 1)- = elim.T-t[(lnx)/(x- ')I, , = el =, , e., , x-+ 1, , If we set x = 1 +(l/n), then x + 1 when n+co; we have l / ( x - I ) = n, so, I/n)". Thus l'H6pital's rule gives, the limit we just calculated is lim,,,(l, another proof of the limit formula lim,,,(l, + l/n)" = e. A, , +, , The next example is a limit of the form co - co., Example 8, Solbtron, , Find, We can convert this limit to $ form by bringing the expression to a common, denominator:, form, , type, , limit, , 1, 1, xsinx, X2, x - sinx, x 2sinx, , co-co, , ?, , I - cosx, 2x sinx x2cosx, , +, , sin x, , 2sinx, , + 4xcosx - x2sinx, , cos x, 6 cos x - 6x sin x - x2cosx, , ~ h u X+O, ls i m (x-sinx, -1- l ) = x2, l.A, , 0, , 0, , ?, , determinate, , 6, , Finally, we shall prove l'H6pital's rule. The proof relies on a generalization of, the mean value theorem., Gauchy's mean, value theorem, , Suppose that f and g are continuous on [a, b] and differentiableon (a, b) and that, g(a) # g(b, in (a, b) such that, g'(c), , Proof, , f(b) - f(a), = f '@)., g(b) - g(a), , First note that if g(x) = x, we recover the mean value theorem in its usual, form. The proof of the mean value theorem in Section 3.6 used the function, l(x) = f(a), , + (x - a) f (b)b --af.(a), , For the Cauchy mean value theorem, we replace x - a by g(x) - g(a) and, look at, , Copyright 1985 Springer-Verlag. All rights reserved.