SalasSV_07_04 - 7.4 THE EXPONENTIAL FUNCTION 397 76(a Show...

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Unformatted text preview: 7.4 THE EXPONENTIAL FUNCTION 397 76. (a) Show that f (x) = ln 2x and g (x) = ln 3x have the same derivative. (b) Calculate the derivative of F (x) = ln kx, where k is any positive number. (c) Explain these results in terms of the properties of logarithms. c In Exercises 77–80, use a graphing utility to graph f on the indicated interval. Estimate the x-intercepts of the graph of f and the values of x where f has either a local or absolute extremum (if any). Use four decimal place accuracy in your answers. √ [0, 10]. 77. f (x) = x ln x; 78. f (x) = x3 ln x; 79. f (x) = sin (ln x); 80. f (x) = x2 ln(sin x); [0, 2]. [1, 100]. [0, 2]. 1 and g (x) = −x2 + 4x − 2. x (a) Use a graphing utility to graph f and g together. (b) Use a CAS to find the points of intersection of the graphs. (c) Use a CAS to find the area of the region bounded by the graphs. x−1 c 84. Repeat Exercise 83 with f (x) = and g (x) = |x − 2|. x c 85. Use a CAS to find a one-term expression for f in each of the following: c 83. Let f (x) = (a) (b) (c) f (x) dx = x ln x − x. f (x) dx = 1 x2 ln x − 1 x2 . 2 4 f (x) dx = 1 x3 ln x − 1 x3 . 3 9 c 81. A particle moves along a line with acceleration a(t ) = 4 − 2(t + 1) + 3/(t + 1) feet per second per second from t = 0 to t = 3. (a) Find the velocity v of the particle at each time t during the time interval [0, 3], given that v(0) = 2. (b) Use a graphing utility to graph v and a together. (c) Estimate the time t ∈ [0, 3] at which the particle has maximum velocity and the time at which it has minimum velocity. Use four decimal place accuracy. c 82. Repeat Exercise 81 for a particle whose acceleration is a(t ) = 2 cos 2(t + 1) + 2/(t + 1) feet per second per second from t = 0 to t = 7. Verify your results by differentiating the expressions on the right-hand side. c In Exercise 86–88, use a CAS to find (a) f and f ; (b) where each of f , f , and f is zero; and (c) the intervals on which each of f , f , and f is positive, negative. 86. f (x) = x ln x. 88. f (x) = 1 + 2 ln x . √ 2 ln x 87. f (x) = ln x . x2 7.4 THE EXPONENTIAL FUNCTION Rational powers of e already have an established meaning: by e p/q we mean the qth √ root of e raised to the pth power. But what is meant by e 2 or e π ? Earlier we proved that each rational power e p/q has logarithm p/q: ln e p/q = p . q (7.4.1) The definition of e z for z irrational is patterned after this relation. DEFINITION 7.4.2 If z is irrational, then by e z we mean the unique number that has logarithm z : ln e z = z . √ √ What is e 2 ? It is the unique number that has logarithm 2. What is e π ? It is the unique number that has logarithm π . Note that e x now has meaning for every real value of x: it is the unique number that has logarithm x. 398 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS DEFINITION 7.4.3 The function E (x ) = e x is called the exponential function. for all real x Some properties of the exponential function are listed below. (1) In the first place, (7.4.4) y ex y=x ln e x = x for all real x Writing L(x) = ln x and E (x) = e x , we have L(E (x)) = x for all real x. In x (0, 1) (1, 0) x This says that the exponential function is the inverse of the logarithm function. (2) The graph of the exponential function appears in Figure 7.4.1. It can be obtained from the graph of the logarithm by reflection in the line y = x. (3) Since the graph of the logarithm lies to the right of the y-axis, the graph of the exponential function lies above the x-axis: (7.4.5) Figure 7.4.1 ex > 0 for all real x. (4) Since the graph of the logarithm crosses the x-axis at (1, 0), the graph of the exponential function crosses the y-axis at (0,1): ln 1 = 0 gives e0 = 1. (5) Since the y-axis is a vertical asymptote for the graph of the logarithm function, the x-axis is a horizontal asymptote for the graph of the exponential function: as x → −∞, e x → 0. (6) Since the exponential function is the inverse of the logarithm function, the logarithm function is the inverse of the exponential function; thus (7.4.6) eln x = x for all x > 0. You can verify this equation directly by observing that both sides have the same logarithm: ln (eln x ) = ln x since, for all real t , ln et = t . You know that for rational exponents e ( p / q +r / s ) = e p / q · e r / s . 7.4 THE EXPONENTIAL FUNCTION 399 This property holds for all exponents, including irrational exponents. THEOREM 7.4.7 ea+b = ea · eb for all real a and b. PROOF ln (ea · eb ) = ln ea + ln eb = a + b = ln ea+b . Since the logarithm function is one-to-one, we must have ea+b = ea · eb . We leave it to you to verify that e −b = 1 eb and ea−b = ea . eb (7.4.8) We come now to one of the most important results in calculus. It is marvelously simple. THEOREM 7.4.9 The exponential function is its own derivative: dx (e ) = e x . dx for all real x, The logarithm function is differentiable, and its derivative is never 0. It follows from Section 7.1 that its inverse, the exponential function, is also differentiable. Knowing this, we can show that PROOF dx (e ) = e x dx by differentiating the identity ln e x = x. On the left-hand side, the chain rule gives d 1d (ln e x ) = x (e x ). dx e dx On the right-hand side, the derivative is 1: d (x) = 1. dx Equating these derivatives, we have 1d x (e ) = 1 e x dx and thus dx (e ) = e x . dx 400 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS We frequently run across expressions of the form eu , where u is a function of x. If u is differentiable, then the chain rule gives du du (e ) = e u . dx dx (7.4.10) PROOF d u du du du (e ) = (e ) = eu . dx du dx dx Example 1 d kx (e ) = ekx · k = kekx . dx √d√ √ d √x 1 1√ (b) (e ) = e x ( x ) = e x = √ e x. √ dx dx 2x 2x d −x 2 2d 2 2 (c) ( e ) = e −x (− x2 ) = e−x (− 2x) = −2x e−x . dx dx (a) The relation dx (e ) = e x dx and its corollary d kx (e ) = k ekx dx have important applications to engineering, physics, chemistry, biology, and economics. We discuss some of these applications in Section 7.6. Example 2 Let f (x) = xe−x for all real x. (a) On what intervals does f increase? Decrease? (b) Find the extreme values of f . (c) Determine the concavity of the graph and find the points of inflection. (d) Sketch the graph indicating the asymptotes if any. SOLUTION We have f (x) = xe−x , f (x) = xe−x (− 1) + e−x = (1 − x)e−x , f (x) = (1 − x) e−x (− 1) − e−x = (x − 2) e−x . Since e−x > 0 for all x, we have f (x) = 0 only at x = 1 (critical number). The sign of f and the behavior of f are as follows: f ' : ++++++++++++++++++++ 0 – – – – – – – – – – – – – – – – – – f: 0 increases 1 2 decreases x Thus, f increases on (− ∞, 1] and decreases on [1, ∞). The number f (1) = 1∼ 1 ∼ = = 0. 368 e 2. 72 is a local and absolute maximum. There are no other extreme values. 7.4 THE EXPONENTIAL FUNCTION 401 Now consider f . The sign of f and the behavior of the graph of f are as follows: f": graph: – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 0+++++++ 0 1 concave down 2 concave up x Thus, the graph is concave down on (−∞, 2) and concave up on (2, ∞). The point (2, f (2)) = (2, 2 e−2 ) ∼ 2, = 2 (2. 72)2 ∼ (2, 0. 27) = is the only point of inflection. In Section 10.6, we prove that f (x) = x/e x → 0 as x → ∞. Accepting this result for now, it follows that the x-axis is a horizontal asymptote. The graph is given in Figure 7.4.2. y 1 2 horizontal asymotote y=0 x f (x) = x e –x Figure 7.4.2 Let f (x) = e−x /2 for all real x. (a) Determine the symmetry of the graph and the asymptotes, if any. (b) On what intervals does f increase? Decrease? (c) Find the extreme values. (d) Determine the concavity of the graph and find the points of inflection. (e) Sketch the graph. Example 3 2 NOTE: This function plays a very important role in the mathematical fields of probability and statistics. As you will see after we complete (a)−(e), the graph is the familiar bell-shaped curve. Since f (− x) = e−(− x) /2 = e−x /2 = f (x), f is an even function. Thus the 2 graph is symmetric with respect to the y-axis. As x → ±∞, e−x /2 → 0. Therefore, the x-axis is a horizontal asymptote. There are no vertical asymptotes. Differentiating f , we have SOLUTION 2 2 f ( x ) = e −x 2 /2 (− x) = −xe−x 2 /2 2 /2 f (x) = −x(− xe−x 2 ) − e −x 2 /2 = (x2 − 1)e−x 2 /2 . Since e−x /2 > 0 for all x, we have f (x) = 0 only at x = 0 (critical number). The sign of f and the behavior of f are as follows: f': f: increases ++++++++++++++++++++ 0 – – – – – – – – – – – – – – – – – – 0 decreases x Thus, f is increasing on (−∞, 0] and decreasing on [0, ∞). The number f (0) = e0 = 1 is a local and absolute maximum. There are no other extreme values. 402 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Now consider f (x) = (x2 − 1)e−x of f are as follows: y 2 /2 . The sign of f and the behavior of the graph f": + + + + + + + 0 – – – – – – – – – – – – 0 + + + + + + + graph: (0, 1) –1 concave up 0 1 point of inflection concave up x (–1, e –1/2) –1 f (x) = e – x 2/2 (1, e –1/2) 1 x point concave of down inflection The graph of f is concave up on (− ∞, −1) and on (1, ∞); the graph is concave down on (− 1, 1); (− 1, e−1/2 ) and (1, e−1/2 ) are points of inflection. The graph of f is shown in Figure 7.4.3. The integral counterpart of Theorem 7.4.9 takes the form e x dx = e x + C . Figure 7.4.3 (7.4.11) In practice eg (x) g (x) dx by setting This substitution gives e g (x) g (x) dx = e g (x) + C . u = g (x), is reduced to du = g (x) dx. eu du (7.4.12) Example 4 SOLUTION Find 9 e3x dx. du = 3 dx. eu du = 3 eu + C = 3 e3x + C . Set u = 3x, 9 e3x dx = 3 If you recognize at the very beginning that 3 e 3x = d 3x (e ), dx then you can dispense with the u-substitution and simply write 9 e3x dx = 3 ex √ dx. x √ x, √ √ 3 e3x dx = 3 e3x + C . Example 5 SOLUTION Find Set u = 1 du = √ dx. 2x eu du = 2 eu + C = 2 e √ x ex √ dx = 2 x + C. 7.4 THE EXPONENTIAL FUNCTION 403 If you recognize that 1 2 ex √ x √ = d √x (e ), dx then you can dispense with the u-substitution and integrate directly: ex √ dx = 2 x e 3x dx e 3x + 1 √ 1 2 ex √ x √ dx = 2 e √ x + C. Example 6 SOLUTION Find We can put this integral in the form du u by setting u = e3x + 1, e 3x dx = e 3x + 1 du = 3 e3x dx. Then 1 3 du = u 1 3 ln |u| + C = 1 3 ln (e3x + 1) + C . Intervals involving x e−x statistics. Evaluate Example 7 √ 2 /2 play an important role in probability and 2 ln 3 x e −x 2 /2 dx. 0 Set u = − 1 x2 , du = −x dx. 2 √ At x = 0, u = 0; at x = 2 ln 3, u = −ln 3. Thus SOLUTION √ 2 ln 3 xe−x 2 /2 dx = − 0 −ln 3 eu du = − eu −ln 3 0 = 1 − e−ln 3 = 1 − 0 1 3 = 2. 3 1 Example 8 SOLUTION Evaluate 0 e x (e x + 1)1/5 dx. du = e x dx. Set u = e x + 1, At x = 0, u = 2; at x = 1, u = e + 1. Thus 1 0 e x (e x + 1)1/5 dx = 2 e +1 u1/5 du = 5 6/5 u 6 e +1 2 = 5 [(e + 1)6/5 − 26/5 ]. 6 Remark The u-substitution simplifies many calculations, but you will find that in many cases you can carry out the integration more quickly without it. 404 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS EXERCISES 7.4 Differentiate the given function. 1. y = e 3. 5. 7. 9. y y y y −2x 1 51. 2x + 1 −4x . 2. y = 3e 4. 6. 8. 10. x(e x + 2) dx. 0 2 ln π/4 52. 0 e x sec e x dx. . . =e = e x ln x. = x−1 e−x . = 1 (e x + e−x ). 2 √ √ 11. y = e x ln x. 13. y = (e x2 x2 −1 y = 2e . y = x2 e x . √ y = e x+1 . y = 1 (e x − e−x ). 2 12. y = (3 − 2e−x )3 . 14. y = (e2x − e−2x )2 . 16. y = x2 e x − xe x . 18. y = e2x − 1 . e2x + 1 2 + 1)2 . 15. y = (x2 − 2x + 2)e x . 17. y = ex −1 . ex +1 19. y = e4 ln x . 21. f (x) = sin (e2x ). 23. f (x) = e−2x cos x. 20. y = ln e3x . 22. f (x) = esin 2x . 24. f (x) = ln (cos e2x ). Calculate the following indefinite integrals. 25. 27. 29. 31. 33. 35. e2x dx. e dx. xe x dx. e1/x dx. x2 ln e x dx. 4 √ dx. ex √ ex ex +1 dx. 2 26. 28. 30. 32. 34. 36. e−2x dx. e ax+b 53. Let a be a positive constant. (a) Find a formula for the nth derivative of f (x) = eax . (b) Find a formula for the nth derivative of f (x) = e−ax . 54. A particle moves on a coordinate line with its position at time t given by the function x(t ) = Aekt + Be−kt , A > 0, B > 0, k > 0. (a) Find the time t at which the particle is closest to the origin. (b) Show that the acceleration of the particle is proportional to its position. What is the constant of proportionality? 55. A rectangle has one side on the x-axis and the upper two 2 vertices on the graph of y = e−x . Where should the vertices be placed so as to maximize the area of the rectangle? 56. A rectangle has two sides on the positive x- and y-axes and one vertex at a point P that moves along the curve y = e x in such a way that y increases at the rate of 1 unit per minute. 2 How fast is the area of the rectangle changing at the instant when y = 3? 57. The function f (x) = e−x is very important in statistics. (a) What is the symmetry of the graph? (b) On what intervals does the function increase? Decrease? (c) What are the extreme values of the function? (d) Determine the concavity of the graph and find the points of inflection. (e) The graph has a horizontal asymptote. What is it? (f) Sketch the graph. 58. Let R be the region bounded by the graph of y = e x and the x-axis from x = 0 to x = 1. (a) Find the volume of the solid generated by revolving R about the x-axis. (b) Set up the definite integral that gives the volume of the solid generated by revolving R about the y-axis using the shell method. (You will learn how to evaluate this integral in Section 8.2.) 59. Let be the region bounded by the graph of y = e−x and the x-axis from x = 0 to x = 1. (a) Find the volume of the solid generated by revolving about the y-axis. (b) Set up the definite integral that gives the volume of the solid generated by revolving about the x-axis using the disc method. Note that you cannot evaluate this integral by the fundamental theorem of calculus. Sketch the region bounded by the curves and find its area. 60. 61. 62. 63. x y y x = e2y , x = e−y , x = 4. = e x , y = e2x , y = e4 . = e x , y = e, y = x, x = 0. = ey , y = 1, y = 2, x = 2. 2 2 kx dx. xe−x dx. e2 x √ dx. x eln x dx. ex dx. ex +1 xeax dx. eax2 + 1 sin (e−2x ) dx. e2x e−x [1 + cos (e−x )]dx. 2 2 √ 37. 39. 41. 38. 40. 42. 2e2x e2x dx. +3 cos x esin x dx. Evaluate the following definite integrals. 1 1 43. 0 e x dx. ln π 44. 0 1 e−kx dx. xe−x dx. 4−ex dx. ex ex dx. 4−ex 2 45. 0 1 e−6x dx. x 46. 0 1 47. 0 e +1 dx. ex ex e dx. +1 x 48. 0 1 ln 2 49. 0 50. 0 For each of the functions f in Exercises 64–68, find: (a) the domain; (b) the intervals where f increases, decreases; 7.4 THE EXPONENTIAL FUNCTION 405 (c) the extreme values; (d) the concavity of the graph of f and the points of inflection; (e) Sketch the graph, indicating all asymptotes. 64. f (x) = (1 − x)e x . 66. f (x) = x2 e−x . 68. f (x) = (x − x2 )e−x 65. f (x) = e(1/x ) . 67. f (x) = x2 ln x. 2 73. Prove that, if n is a positive integer, then e x > xn for all x sufficiently large. c 69. Draw the graphs of f (x) = xk ln x for various positive values of k . (a) What is lim f (x) when 0 < k < 1? When k ≥ 1? x →0 + (b) f has an absolute minimum if k ≥ 1. What is the xcoordinate of the point where f takes on its absolute minimum? c 70. Draw the graphs of f (x) = xk e−x for various positive values of k . (a) As x → ∞, f (x) →? (b) f has an absolute minimum on [0, ∞). What is the x-coordinate of the point where f takes on its absolute maximum? 71. Take a > 0 and refer to the figure. y HINT: Use Exercise 72. 2 74. Let f (x) = e−x and g (x) = x2 . (a) Use a graphing utility to sketch the graphs of f and g . (b) Estimate the x-coordinates of the points of intersection of the two curves. Use four decimal place accuracy. (c) Estimate the area of the region bounded by the two curves. 75. Repeat Exercise 74 for the functions f (x) = e x and g (x) = 4 − x2 . c In Exercises 76–78 use a graphing utility to graph the functions f and g . Your graphs should suggest that f and g are inverses of each other. Confirm this by using the methods in Section 7.1. √ g (x) = ln x. 76. f (x) = e2x ; √ 2 77. f (x) = e x ; g (x) = ln x. 78. f (x) = e x−2 ; g (x) = 2 + ln x. c 79. Use a graphing utility to graph f (x) = sin (e x ). (a) Estimate the zeros of f . (b) Use a CAS to obtain a general formula for the zeros of f . c 80. Repeat Exercise 79 with f (x) = esin x − 1. c 81. Let f (x) = e−x and g (x) = ln x. (a) Use a graphing utility to draw the graphs of f and g together. (b) Estimate the x-coordinate of a point of intersection of the two graphs. (c) Find the slope of the tangent lines for f and g at the point found in part (b). (d) Are these tangent lines perpendicular to each other? c 82. (a) Use a graphing utility to draw the graphs of f (x) = 10e−x and g (x) = 7 − e x together. (b) Use a CAS to find the points of intersection of the two graphs. (c) Use a CAS to find the area of the region between the graphs. c 83. Use a CAS to find the following indefinite integrals. (a) (b) 1 dx. 1−ex e−x 1−ex ex 4 A II B y = eax I y = e–a x x tangent tangent (a) Find the points of tangency, marked A and B. (b) Find the area of region I. (c) Find the area of region II. 72. Prove that for all x > 0 and all positive integers n ex > 1+x + x3 x2 xn + + ··· + . 2! 3! n! Recall, n! = n(n − 1)(n − 2) · · · 3 · 2 · 1. x x dx. HINT: ex = 1+ 0 x et dt > 1 + 0 x dt = 1 + x (c) (1 + t ) dt 0 ex = 1+ 0 et dt > 1 + x2 , 2 etan x dx. cos2 x = 1+x+ and so on. c 84. Use a CAS to solve the equation 0xn eu du = n for xn where n is a positive integer. How much area is in the region bounded by the x-axis and y = e x from x = 0 to x = xn ? 406 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS PROJECT 7.4 Estimating the Number e Since e is an irrational number, we cannot hope to express e as a repeating or terminating decimal. In this project, we derive a numerical estimate for e from the numerical representation of the natural logarithm function: x by showing: a. ln b. ln 1+ 1 n 1 1+ n ≤ ≥ 1 n (HINT : 1 ≤ 1 on[1, 1 + 1/n]. ) t ln x = 1 1 dt , t x > 0. Let n be a positive integer. Then Problem 2. 1 ln 1 + n = 1 1+1/n 1 n+1 1 1 (HINT : ≥ on [1, 1 + 1/n]. ) t 1 + 1/n n n+1 Show that 1+ 1 n ≤e ≤ 1+ 1 n . 1 dt t (1) y The result in Problem 2 is an elegant characterization of e but not a very efficient tool for calculating e. For example, 1+ 1 100 100 ≈ 2. 7048138 and 1+ 1 101 100 ≈ 2. 7318619 gives e rounded off to one decimal place: e ≈ 2. 7. Problem 3. Evaluate (1) for n = 1, 000, n = 10, 000, n = 100, 000, and n = 1, 000, 000. What is the accuracy of these estimates for e? y= 1 1+ 1 n 1 t x Based on these results of Problem 2, we conjecture that (2) 1+ 1 x x → e as x → ∞. Problem 1. Show that 1 1 ≤ ln 1 + n+1 n ≤ 1 n Problem 4. Prove that (2) holds. ( HINT: At x = 1, the logarithm function has derivative h→ 0 lim ln (1 + h) − ln 1 =1 h . 7.5 ARBITRARY POWERS; OTHER BASES Arbitrary Powers: The Function f (x) = xr The elementary notion of exponent applies only to rational numbers. Expressions such as 105 , 21/3 , 7−4/5 , π −1/2 make sense, but so far we have attached no meaning to expressions such as 10 2 , √ 2π , 7− 3 , √ π e. The extension of our sense of exponent to allow for irrational exponents is conveniently done by making use of the logarithm function and the exponential function. The heart of the matter is to observe that for x > 0 and p/q rational xp/q = e ( p/q) ln x . ...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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