Section 4.2 The Mean Value Theorem
23. Show that the equation has exactly one real solution.
2x + cos x = 0
Solution:
316 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 14. () =
+ 2, [1 4]. is continuous on [1 4] and differentiable on (1 4). 0() = () − ()
− ⇔
2 ( + 2)2 =
2 3−13
4 − 1 ⇔ ( + 2)2= 18 ⇔ = −2 ± 3√
2. −2 + 3√
2 ≈ 224 is in (1 4).
15. () =√, [0 4]. 0() = (4) − (0)
4 − 0 ⇔ 1
2√
= 2 − 0
4 ⇔
1 2√
= 1
2 ⇔ √ = 1 ⇔ = 1. The secant line and the tangent line are parallel.
16. () = −, [0 2]. 0() = (2) − (0)
2 − 0 ⇔ −−=−2− 1
2 ⇔
−=1 − −2
2 ⇔ − = ln1 − −2
2 ⇔
= − ln1 − −2
2 ≈ 08386. The secant line and the tangent line are parallel.
17. () = ( − 3)−2 ⇒ 0() = −2 ( − 3)−3. (4) − (1) = 0()(4 − 1) ⇒ 1 12 − 1
(−2)2 = −2
( − 3)3 · 3 ⇒ 3
4 = −6
( − 3)3 ⇒ ( − 3)3= −8 ⇒ − 3 = −2 ⇒ = 1, which is not in the open interval (1 4). This does not contradict the Mean Value Theorem since is not continuous at = 3.
18. () = 2 − |2 − 1| =
2 − (2 − 1) if 2 − 1 ≥ 0 2 − [−(2 − 1)] if 2 − 1 0 =
3 − 2 if ≥ 12
1 + 2 if 12 ⇒ 0() =
−2 if 12
2 if 12
(3) − (0) = 0()(3 − 0) ⇒ −3 − 1 = 0() · 3 ⇒ 0() = −43 [not ± 2]. This does not contradict the Mean Value Theorem since is not differentiable at = 12.
19.Let () = 2 + cos . Then (−) = −2 − 1 0 and (0) = 1 0. Since is the sum of the polynomial 2 and the trignometric function cos , is continuous and differentiable for all . By the Intermediate Value Theorem, there is a number
in (− 0) such that () = 0. Thus, the given equation has at least one real root. If the equation has distinct real roots and
with , then () = () = 0. Since is continuous on [ ] and differentiable on ( ), Rolle’s Theorem implies that there is a number in ( ) such that 0() = 0. But 0() = 2 − sin 0 since sin ≤ 1. This contradiction shows that the given equation can’t have two distinct real roots, so it has exactly one root.
20.Let () = 3+ . Then (−1) = −1 + 1 0 and (0) = 1 0. Since is the sum of a polynomial and the natural exponential function, is continous and differentiable for all . By the Intermediate Value Theorem, there is a number in (−1 0) such that () = 0. Thus, the given equation has at least one real root. If the equation has distinct real roots and
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28. (a) Suppose that f is differentiable on R and has two roots. Show that f0 has at least one root.
(b) Suppose f is twice differentiable on R and has three roots. Show that f00has at least one real root.
(c) Can you generalize parts (a) and (b)?
Solution:
SECTION 4.2 THE MEAN VALUE THEOREM ¤ 317
with , then () = () = 0. Since is continuous on [ ] and differentiable on ( ), Rolle’s Theorem implies that there is a number in ( ) such that 0() = 0. But 0() = 32+ 0. This contradiction shows that the given equation can’t have two distinct real roots, so it has exactly one root.
21. Let () = 3− 15 + for in [−2 2]. If has two real roots and in [−2 2], with , then () = () = 0. Since the polynomial is continuous on [ ] and differentiable on ( ), Rolle’s Theorem implies that there is a number in ( ) such that 0() = 0. Now 0() = 32− 15. Since is in ( ), which is contained in [−2 2], we have || 2, so 2 4.
It follows that 32− 15 3 · 4 − 15 = −3 0. This contradicts 0() = 0, so the given equation can’t have two real roots in [−2 2]. Hence, it has at most one real root in [−2 2].
22. () = 4+ 4 + . Suppose that () = 0 has three distinct real roots , , where . Then
() = () = () = 0. By Rolle’s Theorem there are numbers 1and 2with 1 and 2 and 0 = 0(1) = 0(2), so 0() = 0must have at least two real solutions. However
0 = 0() = 43+ 4 = 4(3+ 1) = 4( + 1)(2− + 1) has as its only real solution = −1. Thus, () can have at most two real roots.
23. (a) Suppose that a cubic polynomial () has roots 1 2 3 4, so (1) = (2) = (3) = (4).
By Rolle’s Theorem there are numbers 1, 2, 3with 1 1 2, 2 2 3and 3 3 4and
0(1) = 0(2) = 0(3) = 0. Thus, the second-degree polynomial 0()has three distinct real roots, which is impossible.
(b) We prove by induction that a polynomial of degree has at most real roots. This is certainly true for = 1. Suppose that the result is true for all polynomials of degree and let () be a polynomial of degree + 1. Suppose that () has more than + 1 real roots, say 1 2 3 · · · +1 +2. Then (1) = (2) = · · · = (+2) = 0.
By Rolle’s Theorem there are real numbers 1 +1with 1 1 2 +1 +1 +2and
0(1) = · · · = 0(+1) = 0. Thus, the th degree polynomial 0()has at least + 1 roots. This contradiction shows that () has at most + 1 real roots.
24. (a) Suppose that () = () = 0 where . By Rolle’s Theorem applied to on [ ] there is a number such that
and 0() = 0.
(b) Suppose that () = () = () = 0 where . By Rolle’s Theorem applied to () on [ ] and [ ] there are numbers and with 0() = 0and 0() = 0. By Rolle’s Theorem applied to 0()on [ ] there is a number with such that 00() = 0.
(c) Suppose that is times differentiable on R and has + 1 distinct real roots. Then ()has at least one real root.
25. By the Mean Value Theorem, (4) − (1) = 0()(4 − 1) for some ∈ (1 4). But for every ∈ (1 4) we have
0() ≥ 2. Putting 0() ≥ 2 into the above equation and substituting (1) = 10, we get
(4) = (1) + 0()(4 − 1) = 10 + 30() ≥ 10 + 3 · 2 = 16. So the smallest possible value of (4) is 16.
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35. Use the Mean Value Theorem to prove the inequality
| sin a − sin b| ≤ |a − b| for all a and b
Solution:
318 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
26.If 3 ≤ 0() ≤ 5 for all , then by the Mean Value Theorem, (8) − (2) = 0() · (8 − 2) for some in [2 8].
( is differentiable for all , so, in particular, is differentiable on (2 8) and continuous on [2 8]. Thus, the hypotheses of the Mean Value Theorem are satisfied.) Since (8) − (2) = 60()and 3 ≤ 0() ≤ 5, it follows that
6 · 3 ≤ 60() ≤ 6 · 5 ⇒ 18 ≤ (8) − (2) ≤ 30
27.Suppose that such a function exists. By the Mean Value Theorem there is a number 0 2 with
0() = (2) − (0)
2 − 0 = 5
2. But this is impossible since 0() ≤ 2 52 for all , so no such function can exist.
28.Let = − . Note that since () = (), () = () − () = 0. Then since and are continuous on [ ] and differentiable on ( ), so is , and thus satisfies the assumptions of the Mean Value Theorem. Therefore, there is a number with such that () = () − () = 0()( − ). Since 0() 0, 0()( − ) 0, so
() − () = () 0 and hence () ().
29.Consider the function () = sin , which is continuous and differentiable on R. Let be a number such that 0 2.
Then is continuous on [0 ] and differentiable on (0 ). By the Mean Value Theorem, there is a number in (0 ) such that
() − (0) = 0()( − 0); that is, sin − 0 = (cos )(). Now cos 1 for 0 2, so sin 1 · = . We took to be an arbitrary number in (0 2), so sin for all satisfying 0 2.
30.satisfies the conditions for the Mean Value Theorem, so we use this theorem on the interval [− ]: () − (−)
− (−) = 0() for some ∈ (− ). But since is odd, (−) = −(). Substituting this into the above equation, we get
() + ()
2 = 0() ⇒ ()
= 0().
31.Let () = sin and let . Then () is continuous on [ ] and differentiable on ( ). By the Mean Value Theorem, there is a number ∈ ( ) with sin − sin = () − () = 0()( − ) = (cos )( − ). Thus,
|sin − sin | ≤ |cos | | − | ≤ | − |. If , then |sin − sin | = |sin − sin | ≤ | − | = | − |. If = , both sides of the inequality are 0.
32.Suppose that 0() = . Let () = , so 0() = . Then, by Corollary 7, () = () + , where is a constant, so
() = + .
33.For 0, () = (), so 0() = 0(). For 0, 0() = (1)0= −12and 0() = (1 + 1)0= −12, so again 0() = 0(). However, the domain of () is not an interval [it is (−∞ 0) ∪ (0 ∞)] so we cannot conclude that
− is constant (in fact it is not).
34.Let () = 2 sin−1 − cos−1(1 − 22). Then 0() = 2
√1 − 2 − 4
1 − (1 − 22)2 = 2
√1 − 2 − 4
2√
1 − 2 = 0 [since ≥ 0]. Thus, 0() = 0for all ∈ (0 1). Thus, () = on (0 1). To find , let = 05. Thus,
2 sin−1(05) − cos−1(05) = 2 6
−3 = 0 = . We conclude that () = 0 for in (0 1). By continuity of , () = 0 on [0 1]. Therefore, we see that () = 2 sin−1 − cos−1(1 − 22) = 0 ⇒ 2 sin−1 = cos−1(1 − 22).
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39. Use the method of Example 6 to prove the identity
2 sin−1x = cos−1(1 − 2x2) x ≥ 0
Solution:
318 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
26.If 3 ≤ 0() ≤ 5 for all , then by the Mean Value Theorem, (8) − (2) = 0() · (8 − 2) for some in [2 8].
( is differentiable for all , so, in particular, is differentiable on (2 8) and continuous on [2 8]. Thus, the hypotheses of the Mean Value Theorem are satisfied.) Since (8) − (2) = 60()and 3 ≤ 0() ≤ 5, it follows that
6 · 3 ≤ 60() ≤ 6 · 5 ⇒ 18 ≤ (8) − (2) ≤ 30
27.Suppose that such a function exists. By the Mean Value Theorem there is a number 0 2 with
0() = (2) − (0) 2 − 0 =5
2. But this is impossible since 0() ≤ 2 52 for all , so no such function can exist.
28.Let = − . Note that since () = (), () = () − () = 0. Then since and are continuous on [ ] and differentiable on ( ), so is , and thus satisfies the assumptions of the Mean Value Theorem. Therefore, there is a number with such that () = () − () = 0()( − ). Since 0() 0, 0()( − ) 0, so
() − () = () 0 and hence () ().
29.Consider the function () = sin , which is continuous and differentiable on R. Let be a number such that 0 2.
Then is continuous on [0 ] and differentiable on (0 ). By the Mean Value Theorem, there is a number in (0 ) such that
() − (0) = 0()( − 0); that is, sin − 0 = (cos )(). Now cos 1 for 0 2, so sin 1 · = . We took to be an arbitrary number in (0 2), so sin for all satisfying 0 2.
30.satisfies the conditions for the Mean Value Theorem, so we use this theorem on the interval [− ]: () − (−)
− (−) = 0() for some ∈ (− ). But since is odd, (−) = −(). Substituting this into the above equation, we get
() + ()
2 = 0() ⇒ ()
= 0().
31.Let () = sin and let . Then () is continuous on [ ] and differentiable on ( ). By the Mean Value Theorem, there is a number ∈ ( ) with sin − sin = () − () = 0()( − ) = (cos )( − ). Thus,
|sin − sin | ≤ |cos | | − | ≤ | − |. If , then |sin − sin | = |sin − sin | ≤ | − | = | − |. If = , both sides of the inequality are 0.
32.Suppose that 0() = . Let () = , so 0() = . Then, by Corollary 7, () = () + , where is a constant, so
() = + .
33.For 0, () = (), so 0() = 0(). For 0, 0() = (1)0= −12and 0() = (1 + 1)0= −12, so again 0() = 0(). However, the domain of () is not an interval [it is (−∞ 0) ∪ (0 ∞)] so we cannot conclude that
− is constant (in fact it is not).
34.Let () = 2 sin−1 − cos−1(1 − 22). Then 0() = 2
√1 − 2 − 4
1 − (1 − 22)2 = 2
√1 − 2 − 4
2√
1 − 2 = 0 [since ≥ 0]. Thus, 0() = 0for all ∈ (0 1). Thus, () = on (0 1). To find , let = 05. Thus,
2 sin−1(05) − cos−1(05) = 2 6
−3 = 0 = . We conclude that () = 0 for in (0 1). By continuity of , () = 0 on [0 1]. Therefore, we see that () = 2 sin−1 − cos−1(1 − 22) = 0 ⇒ 2 sin−1 = cos−1(1 − 22).
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1
42. Fixed points A number a is called a fixed point of a function f if f (a) = a. Prove that if f0(x) 6= 1 for all real numbers x, then f has at most one fixed point.
Solution:
SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 319 35. Let () = arcsin
− 1
+ 1
− 2 arctan√
+2. Note that the domain of is [0 ∞). Thus,
0() = 1
1 −
− 1
+ 1
2
( + 1) − ( − 1) ( + 1)2 − 2
1 + · 1
2√= 1
√ ( + 1)− 1
√ ( + 1)= 0.
Then () = on (0 ∞) by Theorem 5. By continuity of , () = on [0 ∞). To find , we let = 0 ⇒ arcsin(−1) − 2 arctan(0) +2 = ⇒ −2 − 0 +2 = 0 = . Thus, () = 0 ⇒
arcsin
− 1
+ 1
= 2 arctan√
−2.
36. Let () be the velocity of the car hours after 2:00PM. At 2:10PM, =10 60 =1
6. Then(16) − (0)
1
6− 0 =65 − 50
1 6
= 90. By the Mean Value Theorem, there is a number such that 0 16 with 0() = 90. Since 0()is the acceleration at time , the acceleration hours after 2:00PMis exactly 90 kmh2.
37. Let () and () be the position functions of the two runners and let () = () − (). By hypothesis,
(0) = (0) − (0) = 0 and () = () − () = 0, where is the finishing time. Then by the Mean Value Theorem, there is a time , with 0 , such that 0() = () − (0)
− 0 . But () = (0) = 0, so 0() = 0. Since
0() = 0() − 0() = 0, we have 0() = 0(). So at time , both runners have the same speed 0() = 0().
38. Assume that is differentiable (and hence continuous) on R and that 0() 6= 1 for all . Suppose has more than one fixed point. Then there are numbers and such that , () = , and () = . Applying the Mean Value Theorem to the function on [ ], we find that there is a number in ( ) such that 0() = () − ()
− . But then 0() = −
− = 1, contradicting our assumption that 0() 6= 1 for every real number . This shows that our supposition was wrong, that is, that
cannot have more than one fixed point.
4.3 How Derivatives Affect the Shape of a Graph
1. (a) is increasing on (1 3) and (4 6). (b) is decreasing on (0 1) and (3 4).
(c) is concave upward on (0 2). (d) is concave downward on (2 4) and (4 6).
(e) The point of inflection is (2 3).
2. (a) is increasing on (0 1) and (3 7). (b) is decreasing on (1 3).
(c) is concave upward on (2 4) and (5 7). (d) is concave downward on (0 2) and (4 5).
(e) The points of inflection are (2 2), (4 3), and (5 4).
3. (a) Use the Increasing/Decreasing (I/D) Test. (b) Use the Concavity Test.
(c) At any value of where the concavity changes, we have an inflection point at ( ()).
4. (a) See the First Derivative Test.
(b) See the Second Derivative Test and the note that precedes Example 7.
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