CHAPTER

7

The Exponential Function

7.1 Derivative of b x and the number e Preliminary Questions 1. Which of the following equations is incorrect? 2 (a) 3√ · 35 = 37 (b) ( 5)4/3 = 52/3 (c) 32 · 23 = 1 (d) (2−2 )−2 = 16 2. To which of the following functions does the Power Rule apply? (a) x 2 (b) 2e (c) x e (d) e x 3. For which values of b does b x have a negative derivative? 4. For which values of b is b x concave up? 5. Which point lies on the graph of y = b x for all b? 6. Which of the following statements is not true? (a) (e x ) = e x (b) lim (eh − 1)/ h = 1 h→0

(c) The tangent line to y = e x at x = 0 has slope e. (d) The tangent line to y = e x at x = 0 has slope 1.

1

2

Chapter 7 The Exponential Function

Exercises 1. Rewrite as a whole number (without using a calculator). (a) 70 (b) 102 (2−2 + 5−2 ) (c) (43 )5 /(45 )3 (d) 274/3 (e) 8−1/3 · 85/3 (f) 3 · 41/4 − 12 · 2−3/2 (a) (b) (c) (d) (e) (f)

70 = 1 102 (2−2 + 5−2 ) = 100(1/4 + 1/25) = 25 + 4 = 29 (43 )5 /(45 )3 = 415 /415 = 1 (27)4/3 = (271/3 )4 = 34 = 81 8−1/3 · 85/3 = (81/3 )5 /81/3 = 25 /2 = 24 = 16 3 · 41/4 − 12 · 2−3/2 = 3 · 21/2 − 3 · 22 · 2−3/2 = 0

In Exercises 2–10, solve for the unknown variable. 2. 92x = 98 3. e2x = e x+1 4.

If e2x = e x+1 then 2x = x + 1 if and only if x = 1.

et = e4t−3 5. 3x = ( 13 )x+1 2

3x = (1/3)x+1 is equivalent to 3x = 3−(x+1) . Then x = −(x + 1) if and only if x = −x − 1 if and only if 2x = −1 if and only if x = −1/2. 6. √ ( 5)x = 125 7. 4−x = 2x+1 Rewrite 4−x as (22 )−x = 2−2x . Then 2−2x = 2x+1 or −2x = x + 1. Solving for x gives −3x = 1, or x = −1/3. 8. 4 b = 1012 9. k 3/2 = 27 10.

k 3/2 = 27 is equivalent to (k 1/2 )3 = (91/2 )3 , so k = 9. (b2 )x+1 = b−6

In Exercises 11–16, find the equation of the tangent line at the point indicated. 11. 4e x ,

x0 = 0

Let f (x) = 4e x . Then f (x) = 4e x and f (0) = 4. At x0 = 0, f (0) = 4. So the tangent line is y = 4(x − 0) + 4 = 4x + 4. 12. 4x e , x0 = 0 13. e x+2 , x0 = −1 14.

Let f (x) = e x+2 Then f (x) = e x+2 and f (−1) = e1 . At x0 = −1, f (−1) = e, so the tangent line is y = e(x + 1) + e = ex + 2e. 2

ex ,

x0 = 1

7.1 Derivative of b x and the number e

x0 = 2

15. x 2 e x ,

Let f (x) = x 2 e x Then f (x) = 2xe x + x 2 e x and f (2) = 8e2 . At x0 = 2, f (2) = 4e2 , so the tangent line is y = 8e2 (x − 2) + 4e2 = 8e2 x − 12e2 . 16. −2x 7e + 5e7x , x0 = 1 In Exercises 17–40, find the derivative. 17. 7e2x + 3e4x 18.

d 7e2x dx e−5x −4x+9

+ 3e4x = 14e2x + 12e4x

19. e 20.

d −4x+9 e = −4e−4x+9 dx −x −2x

22.

d ex dx x x 2 e2x 2 x

4e2 + 7e ex 21. x 2

=

−e x x2

2

+

2xe x x

2

=

−e x x2

2

+ 2e x

2

23. x e

x 2 e x = 2xe x + x 2 e x (1 + e x )4 25. (2e3x + 2e−2x )4 24.

26.

d dx

d (2e3x + dx 2 e x +2x−3 1/x

2e−2x )4 = 4(2e3x + 2e−2x )3 (6e3x − 4e−2x

27. e 28.

d 1/x e dx 3 e sin x

−e1/x x2

=

29. e 30.

d sin x e = cos xesin x dx 2 2 e(x +2x+3) x

31. sin(e ) 32.

d sin(e x ) dx √ t

= e x cos e x

e 33.

1 1 − e−3t

34.

d 1 = ddx 1 dt 1−e−3t et+1/2t−1 −2t

35. cos(te

− e−3t )−1 = −1(1 − e−3t )−2 (3e3 t) =

−3e3t (1−e−3t )2

)

cos(te ) = (e−2t − 2te−2t ) sin(te−2t )e−2t (1 − 2t) sin(te−2t ) ex 5−6x 37. tan(e 3x + 1 ) d tan(e5−6x ) = sec2 (e5−6x )(−6e5−6x ) = −6e5=6x sec2 e5−6x 38. d xx+1 ex +x e 39. e2e x −1 x d ex e = e x ee dx 36.

d dt

−2t

3

4

Chapter 7 The Exponential Function

40.

e xe

x

In Exercises 41–46, find the critical points and determine if they are local minima, maxima, or neither. 41. f (x) = e x − x f (x) = e x − 1. Setting equal to zero and solving for x gives e x = 1, which is true if and only if x = 0. f (x) = e x , so f (0) = e0 = 1 > 0. Therefore, x = 0 is a local minimum. 42. f (x) = x + e−x ex for x > 0 43. f (x) = x f (x) = e x−e . Setting equal to zero and solving for x: x2 e x (x−1) = 0 if and only if x = 1 which is our critical point. x2 (e x (x−1)+e x )x 2 +2x(e x (x−1)) . f (1) = (e(0)+e)+2(e(0)) = e > 0. Therefore, f (x) = x4 1 x = 1 is a local minimum. 44. f (x) = x 2 e x et 45. g(t) = 2 t +1 x

g (x) =

x

e x (x−1)2 (x 2 +1)2

and g (x) =

e x (x 4 − 4x 3 + 8x 2 − 4x − 1) (x 2 + 1)3

so second derivative test fails. However, g (x) doesn’t change sign, so x = 1 is not a local extremum. 46. g(t) = (t 3 − 2t)et In Exercises 47–52, find the critical points and points of inflection. Then sketch the graph. 47. xe−x

1 f (x) = e−x − xe−x = (1 − x)e−x = 0 at x = 1, a critical point. f (x) = −e−x − e−x + xe−x = −2e−x + xe−x = 0 at x = 2 which is a point of inflection. 48. −x e + ex 49. e−x cos x on [− π2 , π2 ]

7.1 Derivative of b x and the number e

f (x) = −e−x (sin x + cos x), critical points are 3π/4 and 7π/4. And f (x) = 2 sin xe−x and inflection points are at x = 0, π.

50.

5

e−x 51. e x − x 2

f (x) = e x − 1 = 0 at x = 0, a critical point. f (x) = e does not equal zero anywhere, so there are no points of inflection. 52. 2 −x x e on [0, 10] 53. Use Newton’s Method to find solutions of e x = 5x up to 3 decimal places.

x

20

10

1

2

3

Figure 1 Graphs of e x and 5x.

In other words, we solve for the roots of f (x) = e x − 5x = 0. We use the formula xn+1 = xn − ff (x(xnn)) to find our solutions. f (x) = e x − 5, so xn+1 = xn −

e xn −5x . e xn −5

From the graph, we pick the starting point as x0 = 1/3: x1 = 1/3 − ff (1/3) (1/3) x2 = .25813 − ff (.25813) (.25813) x3 = .25917 − ff (.25917) (.25917)

The root is approximately, x = .259. In Exercises 54–66, evaluate the indefinite integral. 54. x (e + 2) d x

x1 ≈ .25813 x2 ≈ .25917 x3 ≈ .25917

6

Chapter 7 The Exponential Function

4e x d x x 4x 56. 4e4x d x = e + c e d2 x 57. xe x d x x2 x2 58. xe4x d x = e /2 + c (e + 1) d x 59. e−9x d x −9x −9x 60. e x d x−x= −e /9 + c + e ) d x (e 61. (e−x − 4x) d x −x 4x) d x = −e−x − 4x 2 /2 + c = −e−4 − 2x 2 + c 62. (e −10x (7 − e2 ) d x 63. xe−4x d x −4x 2 2 d x = e−4x /(−8) + c 64. xe x ) dx e x cos(e ex 65. dx √ ex + 1 ex d x = 2(e x + 1)1/2 + c √ x +1 e 66. x 2x e (e + 1)3 d x 67. Find an approximation to m 3 using the limit definition and estimate the slope of the tangent line to y = 3x at x = 0 and x = 2. 55.

m 3 ≈ limh→0 3 h−3 = limh→0 3 (3h −1) = 3x . Therefore the slopes of the tangent 68. lines are f (0) ≈ 1 and f (2) ≈ 9 Find the area between the curves y = e xx and y = e2x over [0, 1]. 69. Find the area between the curves y = e and y = e−x over [0, 2]. 2 x −x x x 2 2 70. 0 e − e d x = e + e |0 = 2e − 2 Find the area bounded by the curves y = e2 and y = e x and x = 0. x+h

x

x

h

Further Insights and Challenges 71. Prove that e x is not a polynomial function. Hint: differentiation lowers the degree of a polynomial by one. Letting f (x) = e x , we see that f (x) = e x . The degree x has not changes, so e x 72. cannot be a polynomial. Calculate the first three derivatives of f (x) = xe x . Then guess the formula for 73. Consider theinduction equation etox prove = λx where λ are is a familiar constant.with this method of proof). f (n) (x) (use it if you (a) For which values of λ does the equation have at least one solution? (b) For which values of λ does the equation have a unique solution? For intuition, draw a graph of e x and the line y = λx. (a) For all λ ≥ e and λ ≤ 0 (b) The solution is unique if y = λx is tangent to e x . If they intersect at x = a, this gives: λa = ea and slope = λ = ea . This gives a = 1 and λ = e 74. Prove that the numbers m a and m b satisfy the relation m ab = m a + m b in two different ways. (a) First method: use the limit definition of m b and the relation h a −1 (ab)h − 1 bh − 1 h =b + h h h (b) Second method: use the relation (b x ) = m b b x together with the Product Rule applied to a x b x = (ab)x .

7.2 Inverse Functions

7

7.2 Inverse Functions Preliminary Questions 1. Which of the following functions f (x) coincides with its own inverse? In other words, f −1 (x) = f (x). (a) f (x) = x (b) f (x) = 1 − x (c) f (x) = 1 √ (d) f (x) = x (e) f (x) = |x| (f) f (x) = x −1 2. The graph of a function looks like the track of a roller coaster. Is the function one-to-one? 3. Consider the function f mapping teenagers in the U.S. to their last names. Explain why the inverse of f does not exist. 4. View the following fragment of a train schedule for the New Jersey Transit System as defining a function f from towns to times. Is f one-to-one? What is f −1 (6:27)? Trenton

6:21

Hamilton Township

6:27

Princeton Junction

6:34

New Brunswick

6:38

5. A homework problem asks for a sketch of the graph of the inverse of f (x) = x + cos x. Frank, after trying but failing to find a formula for f −1 (x), says it’s impossible to graph the inverse. Sally hands in an accurate sketch without solving for f −1 . How did Sally do it?

Exercises In Exercises 1–8, find a domain on which f (x) is one-to-one and find a formula for the inverse of f restricted to this domain. Sketch the graph of f (x) and f −1 (x). 1. f (x) = 3x − 2

8

Chapter 7 The Exponential Function

y = 3x − 2 if and only if x = (y + 2)/3 So, f −1 (x) = (x + 2)/3 and its domain is all reals.

Function

Inverse of Function

2.

f (x) = 4 − x 1 3. f (x) = x +1 1 1 if and only if y(x + 1) = 1 if and only if x + 1 = if and only if y= x +1 y 1 1 x = − 1. So, f −1 (x) = − 1 and its domain is (−∞, 0) ∪ (0, ∞) y x

–4

4

4

2

2

–2

2

–4

–2

2

–2

–2

–4

–4

Function 4.

4

4

Inverse

1 f (x) = 1 5. f (x) = 7x2 − 3 x 1 1 1 y = 2 if and only if x 2 y = 1 if and only if x 2 = if and only if x = √ . x y y

9

7.2 Inverse Functions

1 Hence, f −1 (x) = √ and its domain is (0, ∞). x

4

4

2

–4

–2

2

2

4

1

–2

–2

–4

–4

2

Function

4

3

5

Inverse

6.

f (x) = √x12 +1 7. f (x) = x 3 y = x 3 if and only if y 1/3 = x. So, f −1 (x) = (x + 1)1/3 and its domain is all reals.

–4

4

4

2

2

–2

2

4

–4

–2

2

–2

–2

–4

–4

Function

4

Inverse

√ f (x) = x 3 + 9 9. For each function shown in Figure 1, sketch the graph of the inverse (restrict the function’s domain if necessary). 8.

10

Chapter 7 The Exponential Function

(A)

(D)

(B)

(E)

(C)

(F)

Figure 1

(a)

(b)

(c)

(d)

7.2 Inverse Functions

11

(e)

(f) Find the inverse of f (x) = (x − 2)/(x + 3). 11. Let f (x) = x 7 + x + 1. (a) Show that f −1 exists (but do not attempt to find the inverse). Hint: show that f is increasing. (b) What is the domain of f −1 ? (c) Find f −1 (3). Hint: solve f (x) = 3. 10.

(a) Looking at the graph, we see that f (x) is one-to-one. (b) The domain is (−∞, ∞) (c) x = 1, since f (1) = 17 + 1 + 1 = 3 12. Show that the inverse of f (x) = e−x exists (without finding the inverse 2 + 1)−1 isofone-to-one on (−∞, 0] and find a formula for 13. explicitly). Show that fWhat (x) =is(x the domain f −1 ? −1 f for this domain.

12

Chapter 7 The Exponential Function

14.

This is what the graph from (−∞, 0) looks like. It is clearly one to one. 2 y = (x 2 + 1)−1 if and only if y = 1/(x 2 + 1) if and only if x + 1 = 1/y if and only if x 2 = 1y − 1 if and only if x = ± 1y − 1 So, f −1 (x) = − x1 − 1.

Let f (x) = x 2 − 2x. Determine a domain on which f −1 exists (Hint: draw a 15. Show (x) a=formula x + x −1for is one-to-one [1, ∞) and find formula for f −1 on graph)that and ffind f −1 on this on domain (Hint: use athe quadratic −1 this domain. What is the domain of f ? formula).

16.

Looking at the graph, we see that f (x) is one-to-one on the given domain. The domain of f −1 (x) is [2, ∞) since that is the range of f (x). We solve for f −1 (x): 2 y = x + (1/x) if and only if y = x x+1 if and only if y − 1 = 1/x 2 if and only if 2 √ − 1) if and only if x = 1/(y − 1). Therefore, x 2 = 1/(y √ f −1 (x) = 1/(x − 1).

√ Let f (x) = x 2 − 9 (x ≥ 9) and let g(x) be its inverse. 3 + 1.of f and g. 17. Let g(x) be the inverse of f (x) and = xrange (a) Determine the domain (a) Find a formula for g(x). (b) Calculate g(x). (b) Calculate g(x)gin two ways: first using Theorem ?? and then by direct (c) Calculate (x) in two ways: first using Theorem ?? and then by direct calculation. calculation. (a) To find g(x), we solve y = x 3 + 1 for x: y = x3 + 1 y − 1 = x3 1

1

Therefore x = (y − 1) 3 and the inverse is g(x) = (y − 1) 3 . (b) We have f (x) = 3x 2 . According to Theorem ??, g (x) =

1 f

(g(x))

=

1 1 = 2 = 2 3g(x) 3(y − 1) 3

1 3

(y − 1)− 3

This agrees with the answer we get by differentiating directly: d 1 (y − 1) 3 = dx

1 3

(y − 1)− 3 2

In Exercises 18–23, use Theorem ?? to calculate g (x) where g = f −1 .

2

7.2 Inverse Functions

18. 19.

20.

13

√ 7x + 6 3−x −1 Let f (x) = (3 − x)1/2 then f (x) = (1/2)(3 − x)−1/2 (−1) = 2(3−x) 1/2 Then √ solving y = 3 − x for x and switching variables, we obtain the inverse −1 g(x) = 3 − x 2 . Thus, g (x) = 1/ 2(3−3x+x 2 )1/2 = −2x

x −5 21. 4x 3 − 1

Let f (x) = 4x 3 − 2 then f (x) = 12x 2 . Then solving y = 4x 3 − 1 for x and then switching variables, we obtain the inverse g(x) = ( x+1 )1/3 . Thus, 4 x+1 −2/3 g (x) = (1/12)( 4 ) 22. x 23. 2x + + x1−1 let f (x) = 2 + (1/x) then f (x) = −1/x 2 . Then solving y = 2 + x −1 for x and switching variables, we obtain the inverse g(x) = 1/(x − 2). Thus −1 2 g (x) = 1/( (1/((x−2) 2 ) = −1/((x − 2) ) In Exercises 24–29, calculate g(b) and g (b) where g = f −1 (without calculating g(x) explicitly). 24. f (x) = x + cos x, b = 1 25. f (x) = 4x 3 − 2x, b = −2 f (−1) = −2, so g(−2) = −1. . f (x) = 12x 2 − 2 so f (g(−2)) = f (−1) = 12 − 2 = 10. Thus, g (−2) = 1/10 26. √ f (x) = x 2 + 6x, b = 4 27. f (x) = 1/(x + 1), b = 14 −1 f (3) = 1/4, so g(1/4) = 3. f (x) = (x+1) 2 so −1 −1 f (g(1/4)) = f (3) = (3+1)2 = −1/16. Thus, g (x) = 1/( (3+1) 2 ) = −16 28. x f (x) = e , b = e 29. f (x) = cos(x 2 ), b = −1 √ √ f (x) = −2x sin(x 2 ) and f ( π) = −1, so g (−1) = 1/ f ( π). 30.

R&W Use graphical reasoning to determine if the following statements are true or false. If false, modify to make a correct statement. (a) If f (x) is increasing, then f −1 is decreasing. (b) If f (x) is decreasing, then f −1 is decreasing. Further Insights and Challenges (c) If f (x) is concave up, then f −1 is concave up. (d)that If the f (x)inverse is concave then f −1isisagain concave up. function. On the other 31. Show of andown, odd function an odd (e) Linear functions f (x) = ax + b (a = 0) are always hand, explain why an even function does not have an inverse. one-to-one. (f) Quadratic polynomials f (x) = ax 2 + bx + c (a = 0) are always one-to-one. Graphically speaking, f −1 (x) is simply f (x) reflected across the line x = y. (g) sinifx fis(x) notisone-to-one. Therefore, odd, then f −1 (x) is odd as well. On the other hand, an even

32.

function is never one-to-one as it always fails the horizontal line test. Use the Intermediate Value Theorem to show that if f (x) is continuous and one-to-one on an interval [a, b], then f (x) is either an increasing or decreasing function. Give an example of a one-to-one function (necessarily discontinuous) which is neither increasing nor decreasing on [a, b].

14

Chapter 7 The Exponential Function

7.3 Logarithms and their derivatives Preliminary Questions 1. When is ln x negative? 2. What is the logarithm of −3? 3. How do you explain the fact that the functions ln x and ln(2x) have the same derivative? In fact, ln(ax) and ln x have the same derivative for any constant a. 4. Why is the area under the hyperbola y = 1/x between 1 and 3 equal to the area under the same hyperbola between 5 and 15? 5. The slope of the tangent line to the curve y = b x at x = 0 is 4. What is b?

Exercises In Exercises 1–11, calculate directly (without using a calculator). 1. log3 (27) We set equal to x. x x = 37 if and only if x = 3. log2 (83/5 ) 1 ) 3. log5 ( 25

2.

5x = 1/25 if and only if 5x = 5−2 if and only if x = −2 log64 (4) 5. log7 (492 )

4.

7x = 292 if and only if 7x = (72 )2 if and only if 7x = 74 if and only if x = 4. log8 (2) + log4 (2) 7. log25 (30) + log25 ( 56 )

6.

This is equivalent to log25 (30) · (5/6) = log25 25 = 1 log√ 4 (48) − log4 (12) 9. ln( e · e7/5 ) √ ln( e · e7/5 ) = ln(e1/2 · e7/5 ) = ln(e1/2+7/5 ) = ln(e19/10 ) = 19/10 10. ln(e3 ) + ln(e4 ) 11. log2 (4) + log2 (21) − log2 (24) − log2 (7) 8.

)= log2 (4 · 21) − log2 (24) − log2 (7) = log2 (4 · 21) − log2 (24 · 27) = log2 ( 4·21 24·7 log2 (3/6) = log2 (1/2) = −1 12. Find the equation of the tangent line to y = ln x at x = 3. 13. Find the equation of the tangent line to y = 2x at x = 3. y = ln 2 · 2x . At x = 3, the slope is 8 ln 2 and y = 8. The tangent line is y = 8 ln 2(x − 3) + 8 In Exercises 14–31, find the derivative. 14. x ln x

15

7.3 Logarithms and their derivatives

15. (ln x)2 16.

d (ln x)2 dx 2

= 2 ln x(1/x) = (2/x) ln x

ln(x ) x +1 17. ln x3 + 1

We rewrite: ln(x + 1) as ln(x 3 + 1). Thus, 18. ln(x 3 + 3x + 1) 19. ln(2x )

d dx

ln(x + 1) − ln(x 3 + 1) =

1 x+1

2

− x3x 3 +1

We rewrite ln(2x ) as x ln 2 Thus, ddx x ln 2 = ln 2 ln(e x + 1) 21. ln(sin x + 1)

20.

ln(sin x + 1) = ln(ln x) 23. x 2 ln x 22.

d dx

cos x sin x+1

x2 x

x 2 ln x = 2x ln x + ln x 3 25. (ln(ln x x))

24.

26.

d dx

d (ln(ln x))3 dx ln((ln x)3 ) 2x

= 2x ln x + x

= 3(ln(ln x))2 (1/ ln x)(1/x) =

3(ln(ln x))2 x ln x

27. x

We rewrite x 2x as eln x = e2x ln x . Thus d 2x ln x e = (2 ln x + 2x/x)e2x ln x = (2 ln x + 2)e2x ln x 28. d xex x 2 29. x x 2x

x2

We rewrite x x as eln x = e x ln x Thus, 2 2 d x 2 ln x e = (2x ln x + x 2 /x)e x ln x = (2x ln x + x)e x ln x 30. d x2x xx 31. e x 2

x

We rewrite e x as ee

ln x x

2

= ee

x ln x

. Thus,

d e x ln x e dx

= (ln x + 1)e x ln x ee

e x ln x

In Exercises 32–41, find the equation of the tangent line at the point indicated. 32. x √4 ,x x = 3√ 33. ( 2) , x = 2 √ √ √ √ √ √ Then f ( 2) = ( 2) 2 . f (x) = ln 2 · ( 2)x . So, Let√f (x) = (√2)x . √ √ f ( 2) = ln 2√· ( 2) 2 is the slope of the tangent line √ √ √ √ 2 34. y7t= (ln 2)( 2 )(x − 2) + 2 3 , t =2 35. π (3x+9) , x = 1 Let f (x) = π 3x+9 . Then f (1) = π 12 . f (x) = 3 ln π · π 3x+9 so 12 12 12 36. f 2(1) = 3π ln π is the slope of the tangent line y = 3π ln π(x − 1) + π 5 y −2y+9 , y = 1 37. ln t, t = 5 Let f (t) = ln t. Then f (5) = ln 5. f (t) = 1/t so f (5) = 1/5 is the slope of the tangent line y = (1/5)(x − 5) + ln 5.

16

Chapter 7 The Exponential Function

38.

ln(8 − 4t), t = 1 39. ln(x 2 ), x = 4 Let f (x) = ln x 2 = 2 ln x. Then f (4) = 2 ln 4. f (x) = 2/x so f (4) = 1/2 which is the slope of the tangent line y = (1/2)(x − 4) + 2 ln 4 40. 4 ln(9x + 2), x = 2 41. ln(sin x), x = π4 √ Let f (x) = ln sin x then f (π/4) = ln( 2/2). f (x) = cos x/ sin√ x = cot x so f (π/4) = 1 is the slope of the tangent line y = (x − π/4) + ln( 2/2) In Exercises 42–45, find the local extreme values in the domain {x : x > 0} and use the Second Derivative Test to determine whether these values are local minima or maxima. 42. ln x ln x x 43. 2 x ln x Let f (x) = lnx 2x . Then f (x) = x /x−2x = x−2xx 4 ln x = 1−2x 3ln x . To find critical x4 1−2 ln x points, f (x) = x 3 = 0 at ln x = 1/2 or x = e1/2 . 3 2 2 ln x)3x 2 ) ln x ln x f (x) = (−2/x)x −(1−2 = −5x −6x = −5−6 so x6 x6 x4 −5−6(1/2) 1/2 2 1/2 f (e ) = (e1/2 )4 = −8/e < 0. Thus, x = e is a local maximum 44. x − ln x 45. x ln x 2

Let f (x) = x − ln x. Then f (x) = ln x + 1. To find critical points f (x) = ln x + 1 = 0 at x = 1/e. f (x) = 1/x so f”(e)=1/(1/e)=e¿0.T hus,x=1/eisalocalminimum. In Exercises 46–51, evaluate the definite integral. 46. 2 1 e 1 dx 47. 1 dxx 1 x e 1 d x = ln x|e1 = ln e − ln 1 = 1 x 1 48. 12 e3 1 d x 1 49. 4 x dt t e e3 1 3 dt = ln t|ee = ln e3 − ln e = 3 − 1 = 2 t 50. e −e en 1 dt 1 51. −e2 t dt (any n, m) m t e en 1 n dt = ln t|eem = ln en − ln em = n − m m t e In Exercises 52–69, evaluate the indefinite integral, using substitution if necessary.

7.3 Logarithms and their derivatives

52. 53.

17

dx t dt 2x + 4 t2 + 4

Let u = t 2 + 4. Then du = 2t dt or 12 du = t dt, and 1 1 1 t dt = du = ln t 2 + 4 + C 2 t +4 2 u 2 54. 2 x dx (3x − 1)d x x3 + 2 55. 9 − 2x + 3x 2

56. 57.

58. 59.

60. 61.

62. 63.

64. 65.

Let u = 9 − 2x + 3x 2 then du = −2 + 6x = 6x − 2 = 2(3x − 1)d x. So (3x − 1)d x = (1/2) du = (1/2) ln(9 − 2x + 3x 2 ) + C. u 2 9 − 2x + 3x tan(4x + 1) d x cot x d x x d x. Let u = sin x then du = cos x d x so We rewrite cot x d x as cos sin x cos x du d x = u = ln sin x + c . sin x cos x dx ln x 2 sindxx + 3 x Let u = ln x then du = (1/x)d x. So, 2 ln x (ln x) d x = udu = u 2 /2 = +C 2 x 4 ln x 2+ 5 d x (ln x) x dx x Let u = ln x then du = (1/x)d x. So (ln x)2 (ln x)3 d x = u 2 du = (u 3 /3) = +c x 3 dx x ln x d x (4x − 1) ln(8x − 2) du . Let u = 4x − 1. Then du = 4 d x and the integral becomes 4(u ln(2u)) 1 Further substitute w = ln(2u) = ln 2 + ln u. This makes dw = , so the integral u 1 dw = ln w + C. Back substituting twice, we get becomes 4w 4 1 1 1 ln w + C = ln(ln 2u) + C = ln(ln(8x − 2)) + C. 4 4 4 ln(ln x) d x cotx xlndxx cos x cot x d x = d x. Let u = sin x, so that du = cos x d x and the integral sin x du = ln |u| + C = ln | sin x| + C. becomes u

18

Chapter 7 The Exponential Function

66. x 3 d2x 67. x3x d x

2 Let u = x 2 then du = 2xd x. So x3x d x u x2 = (1/2) 3u du = (1/2) ln3 3 + C = 23ln 3 + C 68. sin x cos x 3 d x 69. ( 12 )3x+2 d x = 3x + 2 then du Let1 u3x+2 = 3d x. So, u (2) d x = (1/3) (1/2)u du = (1/3) (1/2) +C = ln 1/2

(1/2)3x+2 2 ln(1/2)

+ C.

In Exercises 70–75, evaluate the derivative using logarithmic differentiation as in Example ??. 70. x(x + 1)3 x(x 2 + 1) 2 71. √(3x − 1) x +1 √ +1) . Then Let f (x) = x(x x+1 ln f (x) = ln x(x 2 + 1) − ln(x + 1)1/2 = ln x + ln(x 2 + 1) − (1/2) ln(x + 1). (x) d 1 2x ln f (x) = 1/x − 2(x+1) = ff (x) We multiply through by f (x) to get dx + x 2 +1 2

72.

f (x) =

2 x(x √ +1) ( 1 (x+1) x √ 2

+

2x x 2 +1

−

1 ) 2(x+1)

(2x + 1)(4x ) x − 9 x(x + 2) 73. (2x + 1)(2x + 2) x(x+2) Then let f (x) = (2x+1)(2x+2) ln f (x) = (1/2)[ln(x) + ln(x + 2) − ln(2x + 1) − ln(2x + 2)]. (x) d 1 ln f (x) = (1/2)( x1 + x+2 + −2 + −2 ) = ff (x) We multiply through by dx 2x+1 2x+2 x(x+2) 1 1 −2 −2 f (x) to get f (x) = (1/2)( (2x+1)(2x+2) ) · ( x + x+2 + 2x+1 + 2x+2 ) 74. 2 (x + 1)(x 2 + 2)(x 2 + 3)2 x cos x 75. (x + 1) sin x x cos x . Then Let f (x) = (x+1) sin x ln f (x) = ln x cos x − ln(x + 1) sin x = ln x + ln cos x − ln(x + 1) − ln sin x. d sin x 1 x 1 ln f (x) = x1 − cos − x+1 − cos = x1 − tan x − x+1 − cot x. We multiply dx x sin x x cos x 1 1 through by f (x) to get f (x) = (x+1) sin x ( x − tan x − x+1 − cot x) 76. What is the relationship between logb x and log1/b x?

Further Insights and Challenges 77. (a) Show that f (x) g (x) d ln( f (x)g(x)) = + dx f (x) g(x) for any differentiable functions f and g.

(1)

7.3 Logarithms and their derivatives

19

√ (b) Use Eq. (1) to compute the derivative of ln(x cos x) and ln((x − 3) x + 6). (c) Give a new proof of the Product Rule by observing that the right-hand side of Eq. (1) is equal to ( f (x)g(x)) /( f (x)g(x)). (a) (b) (c) 78.

(x) (x) d ln f (x)g(x) = ddx (ln f (x) + ln g(x)) = ff (x) + gg(x) dx d sin x ln x cos x = 1x + −cos and dx x √ d 1 1 1 ln(x − 3) x + 6 = x−3 + 2(x+6)1/21(x+6)1/2 = x−3 + 2(x+6) 1/2 dx f (x) g (x) f (x)g(x)+ f (x)g (x) ( f (x)g(x)) + g(x) = = f (x)g(x) f (x) f (x)g(x)

Show that loga b · logb a = 1. 79. Verify the formula logb x =

loga x loga b

for any positive numbers a, b.

80. 81. 82. 83.

Equivalently, we cross multiply and prove that loga b · logb x = loga x. Let loga b = m and logb x = n then a m = b and bn = x. Substituting, we have (bm )n = x if and only if bmn = x. So, loga x = mn. Since m = loga b and n = logb x, that implies that loga b · logb x = loga x Use Exercise 79 to verify the formula Compute the derivative of log10 x at x = 2. d 1 d 1 x log10 x = ddx lnln10 = x ln1 10 at x =log 2 b⇒x = dx 2 ln 10 dx (ln b)x Find the equation of the tangent line to y = logb x at x = b (any b > 0). Defining ln x as an Integral. Define a function ϕ(x) in the domain x > 0: x 1 ϕ(x) = dt t 1 This exercise proceeds as if we didn’t know that ϕ(x) = ln x and deduces the basic properties of ln x from the integral expression. Prove the following statements: b ab 1 1 (a) dt = dt for all a, b > 0. t t 1 a Hint: use the substitution u = t/a. (b) ϕ(ab) = ϕ(a) + ϕ(b). Hint: break up the integral from 1 to ab into two integrals and use (a). (c) ϕ(1) = 0 and ϕ(a −1 ) = −ϕ(a) for a > 0. (d) ϕ(a n ) = nϕ(a) for all a > 0 and integers n. (e) ϕ(a 1/n ) = n1 ϕ(a) for all a > 0 and integers n. (f) ϕ(ar ) = r ϕ(a) for all a > 0 and rational number r . (g) There exists x such that ϕ(x) > 1. Hint: show that ϕ(a) > 0 for any a > 1. Then take x = a m for m > 1/ϕ(a). (h) Show that ϕ(t) is increasing and use the Intermediate Value Theorem to show that there exists a unique number e such that ϕ(e) = 1. (i) Show that ϕ(x) is a continuous function satisfying ϕ(er ) = r for any rational number r .

20

Chapter 7 The Exponential Function

b (a) Firstly, 1 1t dt = ln t|b1 = ln b − ln 1 = ln b = ϕ(b). Then ab 1/tdt = ln ab − ln a = ln ab = ln b = ϕ(b) a a (b) Using part a: ab a ab a b ϕ(ab) = 1 1/tdt = 1 1/tdt + a 1/tdt = 1 1/tdt + 1 1/tdt. 1 (c) ϕ(1) = 1 1/tdt = 0 and using part a, 1/a 1 ϕ(a −1 ) = ϕ(1/a) = 1 1/tdt = a 1/tdt = −ϕ(a) (d) Using part b: ϕ(a n ) = ϕ(a · a · a · . . . ) a multiplied n times, which is equal to ϕ(a) + ϕ(a) + ϕ(a) + . . . n times, which is equal to nϕ(a) a 1/n (e) ϕ(a 1/n ) = 1 1/tdt = ln(a 1/n = ln(a)/n = (1/n)ϕ(a) (f) Let r = m/n where m and n are integers. Thus ϕ(ar ) = ϕ(a m/n = ϕ(a m · a 1/n ) = ϕ(a m ) + ϕ(a 1/n ) = mϕ(a) + (1/n)ϕ(a) = (m/n)ϕ(a) = r ϕ(a) (g) Firstly, ϕ(x) = ln x is greater than 0 for all values of x > 1. We take x = a m for m > 1/ϕ(a). Then ϕ(x) = ϕ(a m ) = mϕ(a) > 1 since m > 1/ϕ(a). Therefore, such x exists. (h) ϕ(x) is increasing since ϕ (x) = x −1 > 0. Since ϕ(1) = 0 and ϕ(x) > 1 for some x > 0, the Intermediate Value Theorem implies that there exists some number which we call e between 1 and x such that ϕ(e) = 1. It is unique since ϕ is increasing. (i) This follows from part f and part h. 84. Let g(x) be the inverse of a function f (x) satisfying f (x y) = f (x) + f (y). Show that g(x)g(y) = g(x + y).

7.4 Exponential growth and decay Preliminary Questions 1. Two quantities increase exponentially with growth constants k = 1.2 and k = 3.4, respectively. Which quantity doubles more rapidly? 2. If you are given the doubling time and the growth constant of a quantity that increases exponentially, can you determine the initial amount? 3. Does it make sense to speak of the doubling time of a quantity P that satisfies a linear growth law P(t) = P0 t? Why or why not? 4. Referring to his popular book, A Brief History of Time, the renowned physicist Stephen Hawking said “Someone told me that each equation I included in the book would halve its sales.” If n denotes the number of equations (and we treat n as a continuous variable), which differential equation is satisfied by S(n), the number of books sold? 5. Carbon dating is based on the assumption that the ratio R of C12 –C14 in the atmosphere has been constant over the past 50,000 years. If R were actually smaller in the past than it is today, would the age estimates produced by carbon dating be too ancient or too recent?

7.4 Exponential growth and decay

21

Exercises 1. A certain bacteria population P obeys the exponential growth law P(t) = 2000e1.3t (t in hours). (a) How many bacteria are present initially? (b) At what time will there be 10, 000 bacteria?

2.

(a) P(0) = 2000e0 = 2000 bacteria initially. (b) We solve for t. 2000e1.3t = 10, 000 if and only if e1.3t = 5 if and only if ln e1.3t = ln 5 if and only if 1.3t = ln 5. So t = ln 5/1.3 = 1.24 hours.

A quantity P obeys the exponential growth law P(t) = e5t (t in years). 3. A certain colony (a) Atbacteria what time t is doubles P = 10?in size every hour. Find the formula for the number of bacteria present at time t, assuming that initially the colony contains (b) At what time t is P = 20? 1500 bacteria. (c) What is the doubling time for P? The doubling time is given as one hour. Therefore, 1 = ln 2/k. The initial colony is 1500, so P(0) = 1500. Thus we obtain the formula P(t) = 1500eln 2t = 1500(2t) 4. A quantity P obeys the exponential growth law P(t) = Cekt (t in years). 5. The (a) decay constant cobaltconstant is .13. What What is theof growth for Pisifits thehalf-life? doubling time is 7 years? half (b) life=ln 2/.13 = .053 Find the formula for P(t), assuming that the doubling time is 7 years as in 6. andofthat P(0) = Find the (a) decay constant of100. radium, given that its half-life is 1,622 years. 7. (a) Find all solutions to the differential equation y = −5y. (b) Find the solution satisfying the initial condition y(0) = 3.4.

8.

(a) y = −5y, so the solutions are all equations of the form y(t) = ce−5t for some constant c. (b) Initial condition is given as y(0) = 3.4 which determines the constant c. Therefore, y(t) = 3.4e−5t .

The population of a city is P(t) = 2 · e.06t (in millions), where t is measured in 9. The population of Washington state increased from 4.86 million in 1990 to 5.89 years. million 2000. Assuming exponential growth, (a) inCalculate the doubling time of the population. (a) What will the population be in 2010? (b) How long does it take for the population to triple in size? (b) What is the doubling (c) How long does it time? take for the population to quadruple in size? We let 1990 be our starting point. (a) P(0) = 4.86. Therefore, P(t) = 4.86ekt . In 2000, 10 years have gone by, so P(10 = 5.89 = 4.86ek10 . We use this to solve for k. 5.89 = 4.86ek10 if and only if 1.212 = ek10 if and only if ln 1.212 = k10 if and only if k = ln 1.212 = .019. Therefore in 2010, t = 20 and 10 p(20 = 4.86e.019(20) = 7.11 million people. (b) Doubling time = ln 2/.019 ≈ 36 years. 10. Light intensity The intensity of light passing through an absorbing medium decreases exponentially with the distance traveled. Suppose the decay constant for a certain plastic block is 2 when the distance is measured in feet. How thick must the block be to reduce the intensity by a factor of one-third?

22

Chapter 7 The Exponential Function

11. An insect population triples in size after 5 months. Assuming exponential growth, when will it quadruple in size? Suppose we have P(t) = P0 ekt . We know at t = 5 months, p(t + 5) = 3P(t). We solve for k using this. P(t + 5) = 3P(t) if and only if P0 ek(t+5) = 3P0 ekt if and only if P0 ekt+5k = 3P0 ekt if and only if e5k = 3 if and only if ln e5k = ln 3 if and only if k = ln 3/5 = .22. Therefore, P(t) = P0 e.22t . We see how long it takes to quadruple in size, but solving for T. P(t + T ) = 4P(t) if and only if P0 e.22(t+T ) = 4P0 e.22t if and only if e.22t e.22T = 4e.22t if and only if e.22T = 4 if and only if ln e.22T = ln 4 if and only if T = ln 4/.22 if and only if T = 6.301 months. 12. A 10-kg quantity of a radioactive isotope decays to 3 kg after 17 years. Find the 13. Measurements that a sample of sheepskin parchment discovered by decay constant showed of the isotope. archaeologists had a C14 to C12 ratio equal to 40% of that found in the atmosphere. Approximately how old is the parchment? The C 14 − C 12 ratio at time t is Re−.000121t .40 = ln .40 e−.00012t if and only if − .00012t = ln .40 if and only if t = −.00012 = 7635.76 years old. 14. Chauvet Caves In 1994, rock climbers in southern France stumbled upon a 14 15. cave Holocene Ice Age On thecave basispaintings. of diverseCsources of data, a scientist believes containing prehistoric analysis of charcoal specimens that the by Holocene ice age, which stretches to theshowed presentthat time, haveare begun carried French archeologist Helene Valladas theshould paintings between 29,700 10,000 and 32,400 12,000 years old, ago. much She then learns of animals older than that any remains previously known believed to Given have died onset thisof icethe age have been isdiscovered. human art. thatatthetheC14 –C12ofratio atmosphere R = 10−12What , what 14 12 C14 –C12 ratio would she expect to find in the remains if her theory is range of C –C ratio did Valladas find in the charcoal specimens? correct? We have 10−12 e−.000121(10000) = .2981 and 10−12 e−.000121(12000) = .2341, so it is between 23% and 30%. 16. Find the formula for the function f (t) that satisfies the differential equation 17. Atmospheric Let P(h) y(0) be the=atmospheric pressure (in pounds per y = −.7y andPressure the initial condition 10. square inch) at a height h (in miles) above sea level on earth. It can be shown that P satisfies a differential equation P = −k P for some positive constant k. (a) Suppose measurements with a barometer show that P(0) = 14.7 and P(10) = 2.13. What is the decay constant k? (b) Determine the atmospheric pressure 15 miles above sea level.

18.

(a) Since P = −k P for some positive constant k, the P(k) = Ce−kh where C = P(0) = 14.7. Therefore, P(h) = 14.7e−kh . We know that P(10 = 14.7e−k10 = 2.13. Solving for k, 2.13 = 14.7e−k10 if and only if 2.10 = e−k10 if and only if ln( 2.13 = 14.7 14.7 −k10 if and only if k = .193 (b) P(15) = 14.7e−.193(15) = .813 pounds per square inch.

A certain quantity P increases according to the square law P(t) = t 2 . 19. Moore’s Law In Moore that N in ofsize? How (a) Starting at 1965, time t0Gordon = 1, how longpredicted will it take forthe P number to double transistors on a microchip would increase exponentially. long will it take starting at t0 = 2 or 3? (a) Does following tableatoftime datat confirm Moore’s prediction for the period (b) Inthe general, starting 0 , how long will it take for P to double in size? In 1972–2000? If so, estimate the growth constant other words, find such that P(t0 + ) = k. 2P(t0 ). (b) Let (t) be the answer numbertoof(b) transistors tfrom yearsthe after 1972.you Find an obtain if P (c) N How is the different answer would kt approximate formula N (t) ≈ Ce where t is the number of years after 1972. grew exponentially?

7.4 Exponential growth and decay

23

(c) Estimate the doubling time in Moore’s Law for the period 1972–2000. (d) If Moore’s Law continues to hold until the end of the decade, how many transistors will there be in 2010? (e) Can Moore have expected his prediction to hold indefinitely? Transistors 4004 8008 8080 8086 286 386 processor 486 DX processor Pentium processor Pentium II processor Pentium III processor Pentium 4 processor

Year

No. Transistors

1971 1972 1974 1978 1982 1985 1989 1993 1997 1999 2000

2,250 2,500 5,000 29,000 120,000 275,000 1,180,000 3,100,000 7,500,000 24,000,000 42,000,000

45,000,000 40,000,000

No. of transistors

35,000,000 30,000,000 25,000,000 20,000,000 15,000,000 10,000,000 5,000,000 0 1965

1970

1975

1980

1985

1990

1995

2000

2005

Year

Figure 1

20.

(a) Yes, the graph looks like an exponential graph especially towards the latter years. We estimate the growth constant by setting 1972 as our starting point, so P0 = 2250. Therefore, P(t) = 2250ekt . In 2000, t = 28. Therefore, P(28) = 2250e28k = 42000000 if and only if e28k = 1866667 if and only if k = ln 18666.67 = .351. Remark: one can find a better 28 estimate by calculating k for each time t and then averaging the k values. (b) N (t) = 2250e.351t (c) doubling time is ln 2/.351 = 1.97 (d) in 2010, t = 38 years. Therefore, N (38) = 2250e.351(38) = 1, 395, 732, 906 (e) No, you can’t make a microchip smaller than an atom. Economic Model An economist wishes to test the hypothesis that in a certain country, the monthly rate at which jobs increase is proportional to the number of people who already have jobs. Let J (t) be the number of jobs (in millions) at time t (in months). (a) How would he formulate his hypothesis as a differential equation? (b) Suppose there are 15 million jobs at t = 0 and 17 million jobs six months later. If his hypothesis is correct, how many jobs will there be at the end of the year?

24

Chapter 7 The Exponential Function

Further Insights and Challenges 21. Verify that the half-life of a quantity that decays exponentially with decay constant k is equal to ln 2/k. Let y = Ce−kt be an exponential decay function. Let t be the half-life of the C C = Ce−kt , so quantity y, that is, the time t where y = . Solve for t: 2 2 1 = e−kt . Taking logarithms of both sides, we get − ln 2 = −kt, so t = ln 2/k. 22. 2 Isotopes for Dating Which of the following isotopes would be most suitable 23. Two bacteria coloniesold arerocks: cultivated in a laboratory. colony has a for dating extremely Carbon-14 (half-life The 5570first years), Lead-210 doubling 2 hours and the second colony1.3 hasbillion a doubling time of 3 hours. (half-life time 22.26ofyears), Potassium-49 (half-life years)? Explain why. Suppose that initially the first colony contains 1000 bacteria and the second colony contains 3000 bacteria. At what time will the two colonies be of equal size?

24. 25.

26. 27.

Colony 1 has a half life of 2 = ln 2k1 . Therefore, k1 = .347. Thus P1 (t) = 1000e.347t . Colony 2 has a half life of 3 = ln 2/k2 = .231. Therefore, k2 = .231. Thus, P2 (t) = 3000e.231t . We set P1 (t) = P2 (t) and solve for t: 1000e.347t = 3000e.231t if and only if e.347t = 3e.231t if and only if ln e.347t = ln 3e.231t if and only if .347t = ln 3 + .231t if and only if .116t = ln 3 if and only if t = 9.47 hours. Show that if a quantity P grows quadratically (that is, P(t) = P0 t 2 ), then P does Let P = aP(t) be a quantity obeys an there exponential growth with growth not have doubling time. Inthat other words, is no fixed timelaw interval T such constant k. Show that P increases m-fold after an interval of ln m/k-years. that P(t + T ) = 2P(t). P(t + ln m/k) = Cek(t+ln m/k) = Cekt+ln m = Cekt eln m = eln m Cekt = mcekt = m P(t) Inversion of Sugar When cane sugar is dissolved in water, a reaction takes Suppose a quantity increases time 6 hours. place thatthat transforms thePcane sugar exponentially into so-called with invertdoubling sugar. The reaction After place how many hasofPseveral increased by The 50%? takes over ahours period hours. percentage f (t) of unconverted cane sugartime at time known Doubling is 6t=is ln 2/k sotokdecrease = .116. exponentially. P(t) = Ce.116tSuppose Solve forthat t such that fP(t =+−.2 f . What percentage of cane sugar remains unconverted .116t .116T .116t .116t T ) = P(t) + P(t)/2. e e =e + (e /2 if andafter only 5ifhours? e.116T = After 10 hours? 1 + 1/2 if and only if.116T = ln 1.5 if and only if T = 3.49 ≈ 3.5hours

7

The Exponential Function

7.1 Derivative of b x and the number e Preliminary Questions 1. Which of the following equations is incorrect? 2 (a) 3√ · 35 = 37 (b) ( 5)4/3 = 52/3 (c) 32 · 23 = 1 (d) (2−2 )−2 = 16 2. To which of the following functions does the Power Rule apply? (a) x 2 (b) 2e (c) x e (d) e x 3. For which values of b does b x have a negative derivative? 4. For which values of b is b x concave up? 5. Which point lies on the graph of y = b x for all b? 6. Which of the following statements is not true? (a) (e x ) = e x (b) lim (eh − 1)/ h = 1 h→0

(c) The tangent line to y = e x at x = 0 has slope e. (d) The tangent line to y = e x at x = 0 has slope 1.

1

2

Chapter 7 The Exponential Function

Exercises 1. Rewrite as a whole number (without using a calculator). (a) 70 (b) 102 (2−2 + 5−2 ) (c) (43 )5 /(45 )3 (d) 274/3 (e) 8−1/3 · 85/3 (f) 3 · 41/4 − 12 · 2−3/2 (a) (b) (c) (d) (e) (f)

70 = 1 102 (2−2 + 5−2 ) = 100(1/4 + 1/25) = 25 + 4 = 29 (43 )5 /(45 )3 = 415 /415 = 1 (27)4/3 = (271/3 )4 = 34 = 81 8−1/3 · 85/3 = (81/3 )5 /81/3 = 25 /2 = 24 = 16 3 · 41/4 − 12 · 2−3/2 = 3 · 21/2 − 3 · 22 · 2−3/2 = 0

In Exercises 2–10, solve for the unknown variable. 2. 92x = 98 3. e2x = e x+1 4.

If e2x = e x+1 then 2x = x + 1 if and only if x = 1.

et = e4t−3 5. 3x = ( 13 )x+1 2

3x = (1/3)x+1 is equivalent to 3x = 3−(x+1) . Then x = −(x + 1) if and only if x = −x − 1 if and only if 2x = −1 if and only if x = −1/2. 6. √ ( 5)x = 125 7. 4−x = 2x+1 Rewrite 4−x as (22 )−x = 2−2x . Then 2−2x = 2x+1 or −2x = x + 1. Solving for x gives −3x = 1, or x = −1/3. 8. 4 b = 1012 9. k 3/2 = 27 10.

k 3/2 = 27 is equivalent to (k 1/2 )3 = (91/2 )3 , so k = 9. (b2 )x+1 = b−6

In Exercises 11–16, find the equation of the tangent line at the point indicated. 11. 4e x ,

x0 = 0

Let f (x) = 4e x . Then f (x) = 4e x and f (0) = 4. At x0 = 0, f (0) = 4. So the tangent line is y = 4(x − 0) + 4 = 4x + 4. 12. 4x e , x0 = 0 13. e x+2 , x0 = −1 14.

Let f (x) = e x+2 Then f (x) = e x+2 and f (−1) = e1 . At x0 = −1, f (−1) = e, so the tangent line is y = e(x + 1) + e = ex + 2e. 2

ex ,

x0 = 1

7.1 Derivative of b x and the number e

x0 = 2

15. x 2 e x ,

Let f (x) = x 2 e x Then f (x) = 2xe x + x 2 e x and f (2) = 8e2 . At x0 = 2, f (2) = 4e2 , so the tangent line is y = 8e2 (x − 2) + 4e2 = 8e2 x − 12e2 . 16. −2x 7e + 5e7x , x0 = 1 In Exercises 17–40, find the derivative. 17. 7e2x + 3e4x 18.

d 7e2x dx e−5x −4x+9

+ 3e4x = 14e2x + 12e4x

19. e 20.

d −4x+9 e = −4e−4x+9 dx −x −2x

22.

d ex dx x x 2 e2x 2 x

4e2 + 7e ex 21. x 2

=

−e x x2

2

+

2xe x x

2

=

−e x x2

2

+ 2e x

2

23. x e

x 2 e x = 2xe x + x 2 e x (1 + e x )4 25. (2e3x + 2e−2x )4 24.

26.

d dx

d (2e3x + dx 2 e x +2x−3 1/x

2e−2x )4 = 4(2e3x + 2e−2x )3 (6e3x − 4e−2x

27. e 28.

d 1/x e dx 3 e sin x

−e1/x x2

=

29. e 30.

d sin x e = cos xesin x dx 2 2 e(x +2x+3) x

31. sin(e ) 32.

d sin(e x ) dx √ t

= e x cos e x

e 33.

1 1 − e−3t

34.

d 1 = ddx 1 dt 1−e−3t et+1/2t−1 −2t

35. cos(te

− e−3t )−1 = −1(1 − e−3t )−2 (3e3 t) =

−3e3t (1−e−3t )2

)

cos(te ) = (e−2t − 2te−2t ) sin(te−2t )e−2t (1 − 2t) sin(te−2t ) ex 5−6x 37. tan(e 3x + 1 ) d tan(e5−6x ) = sec2 (e5−6x )(−6e5−6x ) = −6e5=6x sec2 e5−6x 38. d xx+1 ex +x e 39. e2e x −1 x d ex e = e x ee dx 36.

d dt

−2t

3

4

Chapter 7 The Exponential Function

40.

e xe

x

In Exercises 41–46, find the critical points and determine if they are local minima, maxima, or neither. 41. f (x) = e x − x f (x) = e x − 1. Setting equal to zero and solving for x gives e x = 1, which is true if and only if x = 0. f (x) = e x , so f (0) = e0 = 1 > 0. Therefore, x = 0 is a local minimum. 42. f (x) = x + e−x ex for x > 0 43. f (x) = x f (x) = e x−e . Setting equal to zero and solving for x: x2 e x (x−1) = 0 if and only if x = 1 which is our critical point. x2 (e x (x−1)+e x )x 2 +2x(e x (x−1)) . f (1) = (e(0)+e)+2(e(0)) = e > 0. Therefore, f (x) = x4 1 x = 1 is a local minimum. 44. f (x) = x 2 e x et 45. g(t) = 2 t +1 x

g (x) =

x

e x (x−1)2 (x 2 +1)2

and g (x) =

e x (x 4 − 4x 3 + 8x 2 − 4x − 1) (x 2 + 1)3

so second derivative test fails. However, g (x) doesn’t change sign, so x = 1 is not a local extremum. 46. g(t) = (t 3 − 2t)et In Exercises 47–52, find the critical points and points of inflection. Then sketch the graph. 47. xe−x

1 f (x) = e−x − xe−x = (1 − x)e−x = 0 at x = 1, a critical point. f (x) = −e−x − e−x + xe−x = −2e−x + xe−x = 0 at x = 2 which is a point of inflection. 48. −x e + ex 49. e−x cos x on [− π2 , π2 ]

7.1 Derivative of b x and the number e

f (x) = −e−x (sin x + cos x), critical points are 3π/4 and 7π/4. And f (x) = 2 sin xe−x and inflection points are at x = 0, π.

50.

5

e−x 51. e x − x 2

f (x) = e x − 1 = 0 at x = 0, a critical point. f (x) = e does not equal zero anywhere, so there are no points of inflection. 52. 2 −x x e on [0, 10] 53. Use Newton’s Method to find solutions of e x = 5x up to 3 decimal places.

x

20

10

1

2

3

Figure 1 Graphs of e x and 5x.

In other words, we solve for the roots of f (x) = e x − 5x = 0. We use the formula xn+1 = xn − ff (x(xnn)) to find our solutions. f (x) = e x − 5, so xn+1 = xn −

e xn −5x . e xn −5

From the graph, we pick the starting point as x0 = 1/3: x1 = 1/3 − ff (1/3) (1/3) x2 = .25813 − ff (.25813) (.25813) x3 = .25917 − ff (.25917) (.25917)

The root is approximately, x = .259. In Exercises 54–66, evaluate the indefinite integral. 54. x (e + 2) d x

x1 ≈ .25813 x2 ≈ .25917 x3 ≈ .25917

6

Chapter 7 The Exponential Function

4e x d x x 4x 56. 4e4x d x = e + c e d2 x 57. xe x d x x2 x2 58. xe4x d x = e /2 + c (e + 1) d x 59. e−9x d x −9x −9x 60. e x d x−x= −e /9 + c + e ) d x (e 61. (e−x − 4x) d x −x 4x) d x = −e−x − 4x 2 /2 + c = −e−4 − 2x 2 + c 62. (e −10x (7 − e2 ) d x 63. xe−4x d x −4x 2 2 d x = e−4x /(−8) + c 64. xe x ) dx e x cos(e ex 65. dx √ ex + 1 ex d x = 2(e x + 1)1/2 + c √ x +1 e 66. x 2x e (e + 1)3 d x 67. Find an approximation to m 3 using the limit definition and estimate the slope of the tangent line to y = 3x at x = 0 and x = 2. 55.

m 3 ≈ limh→0 3 h−3 = limh→0 3 (3h −1) = 3x . Therefore the slopes of the tangent 68. lines are f (0) ≈ 1 and f (2) ≈ 9 Find the area between the curves y = e xx and y = e2x over [0, 1]. 69. Find the area between the curves y = e and y = e−x over [0, 2]. 2 x −x x x 2 2 70. 0 e − e d x = e + e |0 = 2e − 2 Find the area bounded by the curves y = e2 and y = e x and x = 0. x+h

x

x

h

Further Insights and Challenges 71. Prove that e x is not a polynomial function. Hint: differentiation lowers the degree of a polynomial by one. Letting f (x) = e x , we see that f (x) = e x . The degree x has not changes, so e x 72. cannot be a polynomial. Calculate the first three derivatives of f (x) = xe x . Then guess the formula for 73. Consider theinduction equation etox prove = λx where λ are is a familiar constant.with this method of proof). f (n) (x) (use it if you (a) For which values of λ does the equation have at least one solution? (b) For which values of λ does the equation have a unique solution? For intuition, draw a graph of e x and the line y = λx. (a) For all λ ≥ e and λ ≤ 0 (b) The solution is unique if y = λx is tangent to e x . If they intersect at x = a, this gives: λa = ea and slope = λ = ea . This gives a = 1 and λ = e 74. Prove that the numbers m a and m b satisfy the relation m ab = m a + m b in two different ways. (a) First method: use the limit definition of m b and the relation h a −1 (ab)h − 1 bh − 1 h =b + h h h (b) Second method: use the relation (b x ) = m b b x together with the Product Rule applied to a x b x = (ab)x .

7.2 Inverse Functions

7

7.2 Inverse Functions Preliminary Questions 1. Which of the following functions f (x) coincides with its own inverse? In other words, f −1 (x) = f (x). (a) f (x) = x (b) f (x) = 1 − x (c) f (x) = 1 √ (d) f (x) = x (e) f (x) = |x| (f) f (x) = x −1 2. The graph of a function looks like the track of a roller coaster. Is the function one-to-one? 3. Consider the function f mapping teenagers in the U.S. to their last names. Explain why the inverse of f does not exist. 4. View the following fragment of a train schedule for the New Jersey Transit System as defining a function f from towns to times. Is f one-to-one? What is f −1 (6:27)? Trenton

6:21

Hamilton Township

6:27

Princeton Junction

6:34

New Brunswick

6:38

5. A homework problem asks for a sketch of the graph of the inverse of f (x) = x + cos x. Frank, after trying but failing to find a formula for f −1 (x), says it’s impossible to graph the inverse. Sally hands in an accurate sketch without solving for f −1 . How did Sally do it?

Exercises In Exercises 1–8, find a domain on which f (x) is one-to-one and find a formula for the inverse of f restricted to this domain. Sketch the graph of f (x) and f −1 (x). 1. f (x) = 3x − 2

8

Chapter 7 The Exponential Function

y = 3x − 2 if and only if x = (y + 2)/3 So, f −1 (x) = (x + 2)/3 and its domain is all reals.

Function

Inverse of Function

2.

f (x) = 4 − x 1 3. f (x) = x +1 1 1 if and only if y(x + 1) = 1 if and only if x + 1 = if and only if y= x +1 y 1 1 x = − 1. So, f −1 (x) = − 1 and its domain is (−∞, 0) ∪ (0, ∞) y x

–4

4

4

2

2

–2

2

–4

–2

2

–2

–2

–4

–4

Function 4.

4

4

Inverse

1 f (x) = 1 5. f (x) = 7x2 − 3 x 1 1 1 y = 2 if and only if x 2 y = 1 if and only if x 2 = if and only if x = √ . x y y

9

7.2 Inverse Functions

1 Hence, f −1 (x) = √ and its domain is (0, ∞). x

4

4

2

–4

–2

2

2

4

1

–2

–2

–4

–4

2

Function

4

3

5

Inverse

6.

f (x) = √x12 +1 7. f (x) = x 3 y = x 3 if and only if y 1/3 = x. So, f −1 (x) = (x + 1)1/3 and its domain is all reals.

–4

4

4

2

2

–2

2

4

–4

–2

2

–2

–2

–4

–4

Function

4

Inverse

√ f (x) = x 3 + 9 9. For each function shown in Figure 1, sketch the graph of the inverse (restrict the function’s domain if necessary). 8.

10

Chapter 7 The Exponential Function

(A)

(D)

(B)

(E)

(C)

(F)

Figure 1

(a)

(b)

(c)

(d)

7.2 Inverse Functions

11

(e)

(f) Find the inverse of f (x) = (x − 2)/(x + 3). 11. Let f (x) = x 7 + x + 1. (a) Show that f −1 exists (but do not attempt to find the inverse). Hint: show that f is increasing. (b) What is the domain of f −1 ? (c) Find f −1 (3). Hint: solve f (x) = 3. 10.

(a) Looking at the graph, we see that f (x) is one-to-one. (b) The domain is (−∞, ∞) (c) x = 1, since f (1) = 17 + 1 + 1 = 3 12. Show that the inverse of f (x) = e−x exists (without finding the inverse 2 + 1)−1 isofone-to-one on (−∞, 0] and find a formula for 13. explicitly). Show that fWhat (x) =is(x the domain f −1 ? −1 f for this domain.

12

Chapter 7 The Exponential Function

14.

This is what the graph from (−∞, 0) looks like. It is clearly one to one. 2 y = (x 2 + 1)−1 if and only if y = 1/(x 2 + 1) if and only if x + 1 = 1/y if and only if x 2 = 1y − 1 if and only if x = ± 1y − 1 So, f −1 (x) = − x1 − 1.

Let f (x) = x 2 − 2x. Determine a domain on which f −1 exists (Hint: draw a 15. Show (x) a=formula x + x −1for is one-to-one [1, ∞) and find formula for f −1 on graph)that and ffind f −1 on this on domain (Hint: use athe quadratic −1 this domain. What is the domain of f ? formula).

16.

Looking at the graph, we see that f (x) is one-to-one on the given domain. The domain of f −1 (x) is [2, ∞) since that is the range of f (x). We solve for f −1 (x): 2 y = x + (1/x) if and only if y = x x+1 if and only if y − 1 = 1/x 2 if and only if 2 √ − 1) if and only if x = 1/(y − 1). Therefore, x 2 = 1/(y √ f −1 (x) = 1/(x − 1).

√ Let f (x) = x 2 − 9 (x ≥ 9) and let g(x) be its inverse. 3 + 1.of f and g. 17. Let g(x) be the inverse of f (x) and = xrange (a) Determine the domain (a) Find a formula for g(x). (b) Calculate g(x). (b) Calculate g(x)gin two ways: first using Theorem ?? and then by direct (c) Calculate (x) in two ways: first using Theorem ?? and then by direct calculation. calculation. (a) To find g(x), we solve y = x 3 + 1 for x: y = x3 + 1 y − 1 = x3 1

1

Therefore x = (y − 1) 3 and the inverse is g(x) = (y − 1) 3 . (b) We have f (x) = 3x 2 . According to Theorem ??, g (x) =

1 f

(g(x))

=

1 1 = 2 = 2 3g(x) 3(y − 1) 3

1 3

(y − 1)− 3

This agrees with the answer we get by differentiating directly: d 1 (y − 1) 3 = dx

1 3

(y − 1)− 3 2

In Exercises 18–23, use Theorem ?? to calculate g (x) where g = f −1 .

2

7.2 Inverse Functions

18. 19.

20.

13

√ 7x + 6 3−x −1 Let f (x) = (3 − x)1/2 then f (x) = (1/2)(3 − x)−1/2 (−1) = 2(3−x) 1/2 Then √ solving y = 3 − x for x and switching variables, we obtain the inverse −1 g(x) = 3 − x 2 . Thus, g (x) = 1/ 2(3−3x+x 2 )1/2 = −2x

x −5 21. 4x 3 − 1

Let f (x) = 4x 3 − 2 then f (x) = 12x 2 . Then solving y = 4x 3 − 1 for x and then switching variables, we obtain the inverse g(x) = ( x+1 )1/3 . Thus, 4 x+1 −2/3 g (x) = (1/12)( 4 ) 22. x 23. 2x + + x1−1 let f (x) = 2 + (1/x) then f (x) = −1/x 2 . Then solving y = 2 + x −1 for x and switching variables, we obtain the inverse g(x) = 1/(x − 2). Thus −1 2 g (x) = 1/( (1/((x−2) 2 ) = −1/((x − 2) ) In Exercises 24–29, calculate g(b) and g (b) where g = f −1 (without calculating g(x) explicitly). 24. f (x) = x + cos x, b = 1 25. f (x) = 4x 3 − 2x, b = −2 f (−1) = −2, so g(−2) = −1. . f (x) = 12x 2 − 2 so f (g(−2)) = f (−1) = 12 − 2 = 10. Thus, g (−2) = 1/10 26. √ f (x) = x 2 + 6x, b = 4 27. f (x) = 1/(x + 1), b = 14 −1 f (3) = 1/4, so g(1/4) = 3. f (x) = (x+1) 2 so −1 −1 f (g(1/4)) = f (3) = (3+1)2 = −1/16. Thus, g (x) = 1/( (3+1) 2 ) = −16 28. x f (x) = e , b = e 29. f (x) = cos(x 2 ), b = −1 √ √ f (x) = −2x sin(x 2 ) and f ( π) = −1, so g (−1) = 1/ f ( π). 30.

R&W Use graphical reasoning to determine if the following statements are true or false. If false, modify to make a correct statement. (a) If f (x) is increasing, then f −1 is decreasing. (b) If f (x) is decreasing, then f −1 is decreasing. Further Insights and Challenges (c) If f (x) is concave up, then f −1 is concave up. (d)that If the f (x)inverse is concave then f −1isisagain concave up. function. On the other 31. Show of andown, odd function an odd (e) Linear functions f (x) = ax + b (a = 0) are always hand, explain why an even function does not have an inverse. one-to-one. (f) Quadratic polynomials f (x) = ax 2 + bx + c (a = 0) are always one-to-one. Graphically speaking, f −1 (x) is simply f (x) reflected across the line x = y. (g) sinifx fis(x) notisone-to-one. Therefore, odd, then f −1 (x) is odd as well. On the other hand, an even

32.

function is never one-to-one as it always fails the horizontal line test. Use the Intermediate Value Theorem to show that if f (x) is continuous and one-to-one on an interval [a, b], then f (x) is either an increasing or decreasing function. Give an example of a one-to-one function (necessarily discontinuous) which is neither increasing nor decreasing on [a, b].

14

Chapter 7 The Exponential Function

7.3 Logarithms and their derivatives Preliminary Questions 1. When is ln x negative? 2. What is the logarithm of −3? 3. How do you explain the fact that the functions ln x and ln(2x) have the same derivative? In fact, ln(ax) and ln x have the same derivative for any constant a. 4. Why is the area under the hyperbola y = 1/x between 1 and 3 equal to the area under the same hyperbola between 5 and 15? 5. The slope of the tangent line to the curve y = b x at x = 0 is 4. What is b?

Exercises In Exercises 1–11, calculate directly (without using a calculator). 1. log3 (27) We set equal to x. x x = 37 if and only if x = 3. log2 (83/5 ) 1 ) 3. log5 ( 25

2.

5x = 1/25 if and only if 5x = 5−2 if and only if x = −2 log64 (4) 5. log7 (492 )

4.

7x = 292 if and only if 7x = (72 )2 if and only if 7x = 74 if and only if x = 4. log8 (2) + log4 (2) 7. log25 (30) + log25 ( 56 )

6.

This is equivalent to log25 (30) · (5/6) = log25 25 = 1 log√ 4 (48) − log4 (12) 9. ln( e · e7/5 ) √ ln( e · e7/5 ) = ln(e1/2 · e7/5 ) = ln(e1/2+7/5 ) = ln(e19/10 ) = 19/10 10. ln(e3 ) + ln(e4 ) 11. log2 (4) + log2 (21) − log2 (24) − log2 (7) 8.

)= log2 (4 · 21) − log2 (24) − log2 (7) = log2 (4 · 21) − log2 (24 · 27) = log2 ( 4·21 24·7 log2 (3/6) = log2 (1/2) = −1 12. Find the equation of the tangent line to y = ln x at x = 3. 13. Find the equation of the tangent line to y = 2x at x = 3. y = ln 2 · 2x . At x = 3, the slope is 8 ln 2 and y = 8. The tangent line is y = 8 ln 2(x − 3) + 8 In Exercises 14–31, find the derivative. 14. x ln x

15

7.3 Logarithms and their derivatives

15. (ln x)2 16.

d (ln x)2 dx 2

= 2 ln x(1/x) = (2/x) ln x

ln(x ) x +1 17. ln x3 + 1

We rewrite: ln(x + 1) as ln(x 3 + 1). Thus, 18. ln(x 3 + 3x + 1) 19. ln(2x )

d dx

ln(x + 1) − ln(x 3 + 1) =

1 x+1

2

− x3x 3 +1

We rewrite ln(2x ) as x ln 2 Thus, ddx x ln 2 = ln 2 ln(e x + 1) 21. ln(sin x + 1)

20.

ln(sin x + 1) = ln(ln x) 23. x 2 ln x 22.

d dx

cos x sin x+1

x2 x

x 2 ln x = 2x ln x + ln x 3 25. (ln(ln x x))

24.

26.

d dx

d (ln(ln x))3 dx ln((ln x)3 ) 2x

= 2x ln x + x

= 3(ln(ln x))2 (1/ ln x)(1/x) =

3(ln(ln x))2 x ln x

27. x

We rewrite x 2x as eln x = e2x ln x . Thus d 2x ln x e = (2 ln x + 2x/x)e2x ln x = (2 ln x + 2)e2x ln x 28. d xex x 2 29. x x 2x

x2

We rewrite x x as eln x = e x ln x Thus, 2 2 d x 2 ln x e = (2x ln x + x 2 /x)e x ln x = (2x ln x + x)e x ln x 30. d x2x xx 31. e x 2

x

We rewrite e x as ee

ln x x

2

= ee

x ln x

. Thus,

d e x ln x e dx

= (ln x + 1)e x ln x ee

e x ln x

In Exercises 32–41, find the equation of the tangent line at the point indicated. 32. x √4 ,x x = 3√ 33. ( 2) , x = 2 √ √ √ √ √ √ Then f ( 2) = ( 2) 2 . f (x) = ln 2 · ( 2)x . So, Let√f (x) = (√2)x . √ √ f ( 2) = ln 2√· ( 2) 2 is the slope of the tangent line √ √ √ √ 2 34. y7t= (ln 2)( 2 )(x − 2) + 2 3 , t =2 35. π (3x+9) , x = 1 Let f (x) = π 3x+9 . Then f (1) = π 12 . f (x) = 3 ln π · π 3x+9 so 12 12 12 36. f 2(1) = 3π ln π is the slope of the tangent line y = 3π ln π(x − 1) + π 5 y −2y+9 , y = 1 37. ln t, t = 5 Let f (t) = ln t. Then f (5) = ln 5. f (t) = 1/t so f (5) = 1/5 is the slope of the tangent line y = (1/5)(x − 5) + ln 5.

16

Chapter 7 The Exponential Function

38.

ln(8 − 4t), t = 1 39. ln(x 2 ), x = 4 Let f (x) = ln x 2 = 2 ln x. Then f (4) = 2 ln 4. f (x) = 2/x so f (4) = 1/2 which is the slope of the tangent line y = (1/2)(x − 4) + 2 ln 4 40. 4 ln(9x + 2), x = 2 41. ln(sin x), x = π4 √ Let f (x) = ln sin x then f (π/4) = ln( 2/2). f (x) = cos x/ sin√ x = cot x so f (π/4) = 1 is the slope of the tangent line y = (x − π/4) + ln( 2/2) In Exercises 42–45, find the local extreme values in the domain {x : x > 0} and use the Second Derivative Test to determine whether these values are local minima or maxima. 42. ln x ln x x 43. 2 x ln x Let f (x) = lnx 2x . Then f (x) = x /x−2x = x−2xx 4 ln x = 1−2x 3ln x . To find critical x4 1−2 ln x points, f (x) = x 3 = 0 at ln x = 1/2 or x = e1/2 . 3 2 2 ln x)3x 2 ) ln x ln x f (x) = (−2/x)x −(1−2 = −5x −6x = −5−6 so x6 x6 x4 −5−6(1/2) 1/2 2 1/2 f (e ) = (e1/2 )4 = −8/e < 0. Thus, x = e is a local maximum 44. x − ln x 45. x ln x 2

Let f (x) = x − ln x. Then f (x) = ln x + 1. To find critical points f (x) = ln x + 1 = 0 at x = 1/e. f (x) = 1/x so f”(e)=1/(1/e)=e¿0.T hus,x=1/eisalocalminimum. In Exercises 46–51, evaluate the definite integral. 46. 2 1 e 1 dx 47. 1 dxx 1 x e 1 d x = ln x|e1 = ln e − ln 1 = 1 x 1 48. 12 e3 1 d x 1 49. 4 x dt t e e3 1 3 dt = ln t|ee = ln e3 − ln e = 3 − 1 = 2 t 50. e −e en 1 dt 1 51. −e2 t dt (any n, m) m t e en 1 n dt = ln t|eem = ln en − ln em = n − m m t e In Exercises 52–69, evaluate the indefinite integral, using substitution if necessary.

7.3 Logarithms and their derivatives

52. 53.

17

dx t dt 2x + 4 t2 + 4

Let u = t 2 + 4. Then du = 2t dt or 12 du = t dt, and 1 1 1 t dt = du = ln t 2 + 4 + C 2 t +4 2 u 2 54. 2 x dx (3x − 1)d x x3 + 2 55. 9 − 2x + 3x 2

56. 57.

58. 59.

60. 61.

62. 63.

64. 65.

Let u = 9 − 2x + 3x 2 then du = −2 + 6x = 6x − 2 = 2(3x − 1)d x. So (3x − 1)d x = (1/2) du = (1/2) ln(9 − 2x + 3x 2 ) + C. u 2 9 − 2x + 3x tan(4x + 1) d x cot x d x x d x. Let u = sin x then du = cos x d x so We rewrite cot x d x as cos sin x cos x du d x = u = ln sin x + c . sin x cos x dx ln x 2 sindxx + 3 x Let u = ln x then du = (1/x)d x. So, 2 ln x (ln x) d x = udu = u 2 /2 = +C 2 x 4 ln x 2+ 5 d x (ln x) x dx x Let u = ln x then du = (1/x)d x. So (ln x)2 (ln x)3 d x = u 2 du = (u 3 /3) = +c x 3 dx x ln x d x (4x − 1) ln(8x − 2) du . Let u = 4x − 1. Then du = 4 d x and the integral becomes 4(u ln(2u)) 1 Further substitute w = ln(2u) = ln 2 + ln u. This makes dw = , so the integral u 1 dw = ln w + C. Back substituting twice, we get becomes 4w 4 1 1 1 ln w + C = ln(ln 2u) + C = ln(ln(8x − 2)) + C. 4 4 4 ln(ln x) d x cotx xlndxx cos x cot x d x = d x. Let u = sin x, so that du = cos x d x and the integral sin x du = ln |u| + C = ln | sin x| + C. becomes u

18

Chapter 7 The Exponential Function

66. x 3 d2x 67. x3x d x

2 Let u = x 2 then du = 2xd x. So x3x d x u x2 = (1/2) 3u du = (1/2) ln3 3 + C = 23ln 3 + C 68. sin x cos x 3 d x 69. ( 12 )3x+2 d x = 3x + 2 then du Let1 u3x+2 = 3d x. So, u (2) d x = (1/3) (1/2)u du = (1/3) (1/2) +C = ln 1/2

(1/2)3x+2 2 ln(1/2)

+ C.

In Exercises 70–75, evaluate the derivative using logarithmic differentiation as in Example ??. 70. x(x + 1)3 x(x 2 + 1) 2 71. √(3x − 1) x +1 √ +1) . Then Let f (x) = x(x x+1 ln f (x) = ln x(x 2 + 1) − ln(x + 1)1/2 = ln x + ln(x 2 + 1) − (1/2) ln(x + 1). (x) d 1 2x ln f (x) = 1/x − 2(x+1) = ff (x) We multiply through by f (x) to get dx + x 2 +1 2

72.

f (x) =

2 x(x √ +1) ( 1 (x+1) x √ 2

+

2x x 2 +1

−

1 ) 2(x+1)

(2x + 1)(4x ) x − 9 x(x + 2) 73. (2x + 1)(2x + 2) x(x+2) Then let f (x) = (2x+1)(2x+2) ln f (x) = (1/2)[ln(x) + ln(x + 2) − ln(2x + 1) − ln(2x + 2)]. (x) d 1 ln f (x) = (1/2)( x1 + x+2 + −2 + −2 ) = ff (x) We multiply through by dx 2x+1 2x+2 x(x+2) 1 1 −2 −2 f (x) to get f (x) = (1/2)( (2x+1)(2x+2) ) · ( x + x+2 + 2x+1 + 2x+2 ) 74. 2 (x + 1)(x 2 + 2)(x 2 + 3)2 x cos x 75. (x + 1) sin x x cos x . Then Let f (x) = (x+1) sin x ln f (x) = ln x cos x − ln(x + 1) sin x = ln x + ln cos x − ln(x + 1) − ln sin x. d sin x 1 x 1 ln f (x) = x1 − cos − x+1 − cos = x1 − tan x − x+1 − cot x. We multiply dx x sin x x cos x 1 1 through by f (x) to get f (x) = (x+1) sin x ( x − tan x − x+1 − cot x) 76. What is the relationship between logb x and log1/b x?

Further Insights and Challenges 77. (a) Show that f (x) g (x) d ln( f (x)g(x)) = + dx f (x) g(x) for any differentiable functions f and g.

(1)

7.3 Logarithms and their derivatives

19

√ (b) Use Eq. (1) to compute the derivative of ln(x cos x) and ln((x − 3) x + 6). (c) Give a new proof of the Product Rule by observing that the right-hand side of Eq. (1) is equal to ( f (x)g(x)) /( f (x)g(x)). (a) (b) (c) 78.

(x) (x) d ln f (x)g(x) = ddx (ln f (x) + ln g(x)) = ff (x) + gg(x) dx d sin x ln x cos x = 1x + −cos and dx x √ d 1 1 1 ln(x − 3) x + 6 = x−3 + 2(x+6)1/21(x+6)1/2 = x−3 + 2(x+6) 1/2 dx f (x) g (x) f (x)g(x)+ f (x)g (x) ( f (x)g(x)) + g(x) = = f (x)g(x) f (x) f (x)g(x)

Show that loga b · logb a = 1. 79. Verify the formula logb x =

loga x loga b

for any positive numbers a, b.

80. 81. 82. 83.

Equivalently, we cross multiply and prove that loga b · logb x = loga x. Let loga b = m and logb x = n then a m = b and bn = x. Substituting, we have (bm )n = x if and only if bmn = x. So, loga x = mn. Since m = loga b and n = logb x, that implies that loga b · logb x = loga x Use Exercise 79 to verify the formula Compute the derivative of log10 x at x = 2. d 1 d 1 x log10 x = ddx lnln10 = x ln1 10 at x =log 2 b⇒x = dx 2 ln 10 dx (ln b)x Find the equation of the tangent line to y = logb x at x = b (any b > 0). Defining ln x as an Integral. Define a function ϕ(x) in the domain x > 0: x 1 ϕ(x) = dt t 1 This exercise proceeds as if we didn’t know that ϕ(x) = ln x and deduces the basic properties of ln x from the integral expression. Prove the following statements: b ab 1 1 (a) dt = dt for all a, b > 0. t t 1 a Hint: use the substitution u = t/a. (b) ϕ(ab) = ϕ(a) + ϕ(b). Hint: break up the integral from 1 to ab into two integrals and use (a). (c) ϕ(1) = 0 and ϕ(a −1 ) = −ϕ(a) for a > 0. (d) ϕ(a n ) = nϕ(a) for all a > 0 and integers n. (e) ϕ(a 1/n ) = n1 ϕ(a) for all a > 0 and integers n. (f) ϕ(ar ) = r ϕ(a) for all a > 0 and rational number r . (g) There exists x such that ϕ(x) > 1. Hint: show that ϕ(a) > 0 for any a > 1. Then take x = a m for m > 1/ϕ(a). (h) Show that ϕ(t) is increasing and use the Intermediate Value Theorem to show that there exists a unique number e such that ϕ(e) = 1. (i) Show that ϕ(x) is a continuous function satisfying ϕ(er ) = r for any rational number r .

20

Chapter 7 The Exponential Function

b (a) Firstly, 1 1t dt = ln t|b1 = ln b − ln 1 = ln b = ϕ(b). Then ab 1/tdt = ln ab − ln a = ln ab = ln b = ϕ(b) a a (b) Using part a: ab a ab a b ϕ(ab) = 1 1/tdt = 1 1/tdt + a 1/tdt = 1 1/tdt + 1 1/tdt. 1 (c) ϕ(1) = 1 1/tdt = 0 and using part a, 1/a 1 ϕ(a −1 ) = ϕ(1/a) = 1 1/tdt = a 1/tdt = −ϕ(a) (d) Using part b: ϕ(a n ) = ϕ(a · a · a · . . . ) a multiplied n times, which is equal to ϕ(a) + ϕ(a) + ϕ(a) + . . . n times, which is equal to nϕ(a) a 1/n (e) ϕ(a 1/n ) = 1 1/tdt = ln(a 1/n = ln(a)/n = (1/n)ϕ(a) (f) Let r = m/n where m and n are integers. Thus ϕ(ar ) = ϕ(a m/n = ϕ(a m · a 1/n ) = ϕ(a m ) + ϕ(a 1/n ) = mϕ(a) + (1/n)ϕ(a) = (m/n)ϕ(a) = r ϕ(a) (g) Firstly, ϕ(x) = ln x is greater than 0 for all values of x > 1. We take x = a m for m > 1/ϕ(a). Then ϕ(x) = ϕ(a m ) = mϕ(a) > 1 since m > 1/ϕ(a). Therefore, such x exists. (h) ϕ(x) is increasing since ϕ (x) = x −1 > 0. Since ϕ(1) = 0 and ϕ(x) > 1 for some x > 0, the Intermediate Value Theorem implies that there exists some number which we call e between 1 and x such that ϕ(e) = 1. It is unique since ϕ is increasing. (i) This follows from part f and part h. 84. Let g(x) be the inverse of a function f (x) satisfying f (x y) = f (x) + f (y). Show that g(x)g(y) = g(x + y).

7.4 Exponential growth and decay Preliminary Questions 1. Two quantities increase exponentially with growth constants k = 1.2 and k = 3.4, respectively. Which quantity doubles more rapidly? 2. If you are given the doubling time and the growth constant of a quantity that increases exponentially, can you determine the initial amount? 3. Does it make sense to speak of the doubling time of a quantity P that satisfies a linear growth law P(t) = P0 t? Why or why not? 4. Referring to his popular book, A Brief History of Time, the renowned physicist Stephen Hawking said “Someone told me that each equation I included in the book would halve its sales.” If n denotes the number of equations (and we treat n as a continuous variable), which differential equation is satisfied by S(n), the number of books sold? 5. Carbon dating is based on the assumption that the ratio R of C12 –C14 in the atmosphere has been constant over the past 50,000 years. If R were actually smaller in the past than it is today, would the age estimates produced by carbon dating be too ancient or too recent?

7.4 Exponential growth and decay

21

Exercises 1. A certain bacteria population P obeys the exponential growth law P(t) = 2000e1.3t (t in hours). (a) How many bacteria are present initially? (b) At what time will there be 10, 000 bacteria?

2.

(a) P(0) = 2000e0 = 2000 bacteria initially. (b) We solve for t. 2000e1.3t = 10, 000 if and only if e1.3t = 5 if and only if ln e1.3t = ln 5 if and only if 1.3t = ln 5. So t = ln 5/1.3 = 1.24 hours.

A quantity P obeys the exponential growth law P(t) = e5t (t in years). 3. A certain colony (a) Atbacteria what time t is doubles P = 10?in size every hour. Find the formula for the number of bacteria present at time t, assuming that initially the colony contains (b) At what time t is P = 20? 1500 bacteria. (c) What is the doubling time for P? The doubling time is given as one hour. Therefore, 1 = ln 2/k. The initial colony is 1500, so P(0) = 1500. Thus we obtain the formula P(t) = 1500eln 2t = 1500(2t) 4. A quantity P obeys the exponential growth law P(t) = Cekt (t in years). 5. The (a) decay constant cobaltconstant is .13. What What is theof growth for Pisifits thehalf-life? doubling time is 7 years? half (b) life=ln 2/.13 = .053 Find the formula for P(t), assuming that the doubling time is 7 years as in 6. andofthat P(0) = Find the (a) decay constant of100. radium, given that its half-life is 1,622 years. 7. (a) Find all solutions to the differential equation y = −5y. (b) Find the solution satisfying the initial condition y(0) = 3.4.

8.

(a) y = −5y, so the solutions are all equations of the form y(t) = ce−5t for some constant c. (b) Initial condition is given as y(0) = 3.4 which determines the constant c. Therefore, y(t) = 3.4e−5t .

The population of a city is P(t) = 2 · e.06t (in millions), where t is measured in 9. The population of Washington state increased from 4.86 million in 1990 to 5.89 years. million 2000. Assuming exponential growth, (a) inCalculate the doubling time of the population. (a) What will the population be in 2010? (b) How long does it take for the population to triple in size? (b) What is the doubling (c) How long does it time? take for the population to quadruple in size? We let 1990 be our starting point. (a) P(0) = 4.86. Therefore, P(t) = 4.86ekt . In 2000, 10 years have gone by, so P(10 = 5.89 = 4.86ek10 . We use this to solve for k. 5.89 = 4.86ek10 if and only if 1.212 = ek10 if and only if ln 1.212 = k10 if and only if k = ln 1.212 = .019. Therefore in 2010, t = 20 and 10 p(20 = 4.86e.019(20) = 7.11 million people. (b) Doubling time = ln 2/.019 ≈ 36 years. 10. Light intensity The intensity of light passing through an absorbing medium decreases exponentially with the distance traveled. Suppose the decay constant for a certain plastic block is 2 when the distance is measured in feet. How thick must the block be to reduce the intensity by a factor of one-third?

22

Chapter 7 The Exponential Function

11. An insect population triples in size after 5 months. Assuming exponential growth, when will it quadruple in size? Suppose we have P(t) = P0 ekt . We know at t = 5 months, p(t + 5) = 3P(t). We solve for k using this. P(t + 5) = 3P(t) if and only if P0 ek(t+5) = 3P0 ekt if and only if P0 ekt+5k = 3P0 ekt if and only if e5k = 3 if and only if ln e5k = ln 3 if and only if k = ln 3/5 = .22. Therefore, P(t) = P0 e.22t . We see how long it takes to quadruple in size, but solving for T. P(t + T ) = 4P(t) if and only if P0 e.22(t+T ) = 4P0 e.22t if and only if e.22t e.22T = 4e.22t if and only if e.22T = 4 if and only if ln e.22T = ln 4 if and only if T = ln 4/.22 if and only if T = 6.301 months. 12. A 10-kg quantity of a radioactive isotope decays to 3 kg after 17 years. Find the 13. Measurements that a sample of sheepskin parchment discovered by decay constant showed of the isotope. archaeologists had a C14 to C12 ratio equal to 40% of that found in the atmosphere. Approximately how old is the parchment? The C 14 − C 12 ratio at time t is Re−.000121t .40 = ln .40 e−.00012t if and only if − .00012t = ln .40 if and only if t = −.00012 = 7635.76 years old. 14. Chauvet Caves In 1994, rock climbers in southern France stumbled upon a 14 15. cave Holocene Ice Age On thecave basispaintings. of diverseCsources of data, a scientist believes containing prehistoric analysis of charcoal specimens that the by Holocene ice age, which stretches to theshowed presentthat time, haveare begun carried French archeologist Helene Valladas theshould paintings between 29,700 10,000 and 32,400 12,000 years old, ago. much She then learns of animals older than that any remains previously known believed to Given have died onset thisof icethe age have been isdiscovered. human art. thatatthetheC14 –C12ofratio atmosphere R = 10−12What , what 14 12 C14 –C12 ratio would she expect to find in the remains if her theory is range of C –C ratio did Valladas find in the charcoal specimens? correct? We have 10−12 e−.000121(10000) = .2981 and 10−12 e−.000121(12000) = .2341, so it is between 23% and 30%. 16. Find the formula for the function f (t) that satisfies the differential equation 17. Atmospheric Let P(h) y(0) be the=atmospheric pressure (in pounds per y = −.7y andPressure the initial condition 10. square inch) at a height h (in miles) above sea level on earth. It can be shown that P satisfies a differential equation P = −k P for some positive constant k. (a) Suppose measurements with a barometer show that P(0) = 14.7 and P(10) = 2.13. What is the decay constant k? (b) Determine the atmospheric pressure 15 miles above sea level.

18.

(a) Since P = −k P for some positive constant k, the P(k) = Ce−kh where C = P(0) = 14.7. Therefore, P(h) = 14.7e−kh . We know that P(10 = 14.7e−k10 = 2.13. Solving for k, 2.13 = 14.7e−k10 if and only if 2.10 = e−k10 if and only if ln( 2.13 = 14.7 14.7 −k10 if and only if k = .193 (b) P(15) = 14.7e−.193(15) = .813 pounds per square inch.

A certain quantity P increases according to the square law P(t) = t 2 . 19. Moore’s Law In Moore that N in ofsize? How (a) Starting at 1965, time t0Gordon = 1, how longpredicted will it take forthe P number to double transistors on a microchip would increase exponentially. long will it take starting at t0 = 2 or 3? (a) Does following tableatoftime datat confirm Moore’s prediction for the period (b) Inthe general, starting 0 , how long will it take for P to double in size? In 1972–2000? If so, estimate the growth constant other words, find such that P(t0 + ) = k. 2P(t0 ). (b) Let (t) be the answer numbertoof(b) transistors tfrom yearsthe after 1972.you Find an obtain if P (c) N How is the different answer would kt approximate formula N (t) ≈ Ce where t is the number of years after 1972. grew exponentially?

7.4 Exponential growth and decay

23

(c) Estimate the doubling time in Moore’s Law for the period 1972–2000. (d) If Moore’s Law continues to hold until the end of the decade, how many transistors will there be in 2010? (e) Can Moore have expected his prediction to hold indefinitely? Transistors 4004 8008 8080 8086 286 386 processor 486 DX processor Pentium processor Pentium II processor Pentium III processor Pentium 4 processor

Year

No. Transistors

1971 1972 1974 1978 1982 1985 1989 1993 1997 1999 2000

2,250 2,500 5,000 29,000 120,000 275,000 1,180,000 3,100,000 7,500,000 24,000,000 42,000,000

45,000,000 40,000,000

No. of transistors

35,000,000 30,000,000 25,000,000 20,000,000 15,000,000 10,000,000 5,000,000 0 1965

1970

1975

1980

1985

1990

1995

2000

2005

Year

Figure 1

20.

(a) Yes, the graph looks like an exponential graph especially towards the latter years. We estimate the growth constant by setting 1972 as our starting point, so P0 = 2250. Therefore, P(t) = 2250ekt . In 2000, t = 28. Therefore, P(28) = 2250e28k = 42000000 if and only if e28k = 1866667 if and only if k = ln 18666.67 = .351. Remark: one can find a better 28 estimate by calculating k for each time t and then averaging the k values. (b) N (t) = 2250e.351t (c) doubling time is ln 2/.351 = 1.97 (d) in 2010, t = 38 years. Therefore, N (38) = 2250e.351(38) = 1, 395, 732, 906 (e) No, you can’t make a microchip smaller than an atom. Economic Model An economist wishes to test the hypothesis that in a certain country, the monthly rate at which jobs increase is proportional to the number of people who already have jobs. Let J (t) be the number of jobs (in millions) at time t (in months). (a) How would he formulate his hypothesis as a differential equation? (b) Suppose there are 15 million jobs at t = 0 and 17 million jobs six months later. If his hypothesis is correct, how many jobs will there be at the end of the year?

24

Chapter 7 The Exponential Function

Further Insights and Challenges 21. Verify that the half-life of a quantity that decays exponentially with decay constant k is equal to ln 2/k. Let y = Ce−kt be an exponential decay function. Let t be the half-life of the C C = Ce−kt , so quantity y, that is, the time t where y = . Solve for t: 2 2 1 = e−kt . Taking logarithms of both sides, we get − ln 2 = −kt, so t = ln 2/k. 22. 2 Isotopes for Dating Which of the following isotopes would be most suitable 23. Two bacteria coloniesold arerocks: cultivated in a laboratory. colony has a for dating extremely Carbon-14 (half-life The 5570first years), Lead-210 doubling 2 hours and the second colony1.3 hasbillion a doubling time of 3 hours. (half-life time 22.26ofyears), Potassium-49 (half-life years)? Explain why. Suppose that initially the first colony contains 1000 bacteria and the second colony contains 3000 bacteria. At what time will the two colonies be of equal size?

24. 25.

26. 27.

Colony 1 has a half life of 2 = ln 2k1 . Therefore, k1 = .347. Thus P1 (t) = 1000e.347t . Colony 2 has a half life of 3 = ln 2/k2 = .231. Therefore, k2 = .231. Thus, P2 (t) = 3000e.231t . We set P1 (t) = P2 (t) and solve for t: 1000e.347t = 3000e.231t if and only if e.347t = 3e.231t if and only if ln e.347t = ln 3e.231t if and only if .347t = ln 3 + .231t if and only if .116t = ln 3 if and only if t = 9.47 hours. Show that if a quantity P grows quadratically (that is, P(t) = P0 t 2 ), then P does Let P = aP(t) be a quantity obeys an there exponential growth with growth not have doubling time. Inthat other words, is no fixed timelaw interval T such constant k. Show that P increases m-fold after an interval of ln m/k-years. that P(t + T ) = 2P(t). P(t + ln m/k) = Cek(t+ln m/k) = Cekt+ln m = Cekt eln m = eln m Cekt = mcekt = m P(t) Inversion of Sugar When cane sugar is dissolved in water, a reaction takes Suppose a quantity increases time 6 hours. place thatthat transforms thePcane sugar exponentially into so-called with invertdoubling sugar. The reaction After place how many hasofPseveral increased by The 50%? takes over ahours period hours. percentage f (t) of unconverted cane sugartime at time known Doubling is 6t=is ln 2/k sotokdecrease = .116. exponentially. P(t) = Ce.116tSuppose Solve forthat t such that fP(t =+−.2 f . What percentage of cane sugar remains unconverted .116t .116T .116t .116t T ) = P(t) + P(t)/2. e e =e + (e /2 if andafter only 5ifhours? e.116T = After 10 hours? 1 + 1/2 if and only if.116T = ln 1.5 if and only if T = 3.49 ≈ 3.5hours