Section 4.1 Maximum and Minimum Values

(1)

Section 4.1 Maximum and Minimum Values

45. Find the critical numbers of the function. f (θ) = 2 cos θ + sin²θ.

Solution:

SECTION 4.1 MAXIMUM AND MINIMUM VALUES ¤ 305 35. () =  − 1

²−  + 1 ⇒

⁰() = (²−  + 1)(1) − ( − 1)(2 − 1)

(²−  + 1)² =²−  + 1 − (2²− 3 + 1)

(²−  + 1)² = −²+ 2

(²−  + 1)² = (2 − ) (²−  + 1)².

⁰() = 0 ⇒  = 0, 2. The expression ²−  + 1 is never equal to 0, so ⁰()exists for all real numbers.

The critical numbers are 0 and 2.

36. () =  − 1

²+ 4 ⇒ ⁰() = (²+ 4)(1) − ( − 1)(2)

(²+ 4)² = ²+ 4 − 2²+ 2

(²+ 4)² = −²+ 2 + 4 (²+ 4)² .

⁰() = 0 ⇒  = −2 ±√ 4 + 16

−2 = 1 ±√

5. The critical numbers are 1 ±√

5. [⁰()exists for all real numbers.]

37. () = ^34− 2^14 ⇒ ⁰() = ³₄^−14−²4^−34=¹₄^−34(3^12− 2) =3√

 − 2 4√⁴

³ .

⁰() = 0 ⇒ 3√

 = 2 ⇒ √

 =²₃ ⇒  =⁴9. ⁰()does not exist at  = 0, so the critical numbers are 0 and⁴₉.

38. () =√³

4 − ²= (4 − ²)^13 ⇒ ⁰() = ¹₃(4 − ²)^−23(−2) = −2

3(4 − ²)^23. ⁰() = 0 ⇒  = 0.

⁰(±2) do not exist. Thus, the three critical numbers are −2, 0, and 2.

39.  () = ^45( − 4)² ⇒

⁰() = ^45· 2( − 4) + ( − 4)²·⁴5^−15= ¹₅^−15( − 4)[5 ·  · 2 + ( − 4) · 4]

=( − 4)(14 − 16)

5^15 = 2( − 4)(7 − 8) 5^15

⁰() = 0 ⇒  = 4,⁸7. ⁰(0)does not exist. Thus, the three critical numbers are 0,⁸₇, and 4.

40. () = 4 − tan  ⇒ ⁰() = 4 − sec². ⁰() = 0 ⇒ sec² = 4 ⇒ sec  = ±2 ⇒ cos  = ±¹2 ⇒

 =^₃ + 2, ^5₃ + 2, ^2₃ + 2, and ^4₃ + 2are critical numbers.

Note: The values of  that make ⁰()undeﬁned are not in the domain of .

41.  () = 2 cos  + sin² ⇒ ⁰() = −2 sin  + 2 sin  cos . ⁰() = 0 ⇒ 2 sin  (cos  − 1) = 0 ⇒ sin  = 0 or cos  = 1 ⇒  =  [ an integer] or  = 2. The solutions  =  include the solutions  = 2, so the critical numbers are  = .

42. () = 3 − arcsin  ⇒ ⁰() = 3 − 1

√1 − ². ⁰() = 0 ⇒ 3 = 1

√1 − ² ⇒ √

1 − ²=¹₃ ⇒ 1 − ²=¹₉ ⇒ ²= ⁸₉ ⇒  = ±²3

√2 ≈ ±094, both in the domain of , which is [−1 1].

43.  () = ²^−3 ⇒ ⁰() = ²(−3^−3) + ^−3(2) = ^−3(−3 + 2). ⁰() = 0 ⇒  = 0,²3

[^−3is never equal to 0]. ⁰()always exists, so the critical numbers are 0 and²₃.

44.  () = ⁻²ln  ⇒ ⁰() = ⁻²(1) + (ln )(−2⁻³) = ⁻³− 2⁻³ln  = ⁻³(1 − 2 ln ) = 1 − 2 ln 

³ .

⁰() = 0 ⇒ 1 − 2 ln  = 0 ⇒ ln  =¹2 ⇒  = ^12≈ 165. ⁰(0)does not exist, but 0 is not in the domain of , so the only critical number is√

.

50. A formula for the derivative of a function f is given. How many critical numbers does f have?

f⁰(x) = 100 cos²x 10 + x² − 1 Solution:

306 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION

45.The graph of ⁰() = 5^{−01||}sin  − 1 has 10 zeros and exists everywhere, so  has 10 critical numbers.

46.A graph of ⁰() = 100 cos²

10 + ² − 1 is shown. There are 7 zeros between 0 and 10, and 7 more zeros since ⁰is an even function.

⁰exists everywhere, so  has 14 critical numbers.

47. () = 3²− 12 + 5, [0 3]. ⁰() = 6 − 12 = 0 ⇔  = 2. Applying the Closed Interval Method, we ﬁnd that

 (0) = 5, (2) = −7, and (3) = −4. So (0) = 5 is the absolute maximum value and (2) = −7 is the absolute minimum value.

48. () = ³− 3 + 1, [0 3]. ⁰() = 3²− 3 = 0 ⇔  = ±1, but −1 is not in [0 3]. (0) = 1, (1) = −1, and

 (3) = 19. So (3) = 19 is the absolute maximum value and (1) = −1 is the absolute minimum value.

49. () = 2³− 3²− 12 + 1, [−2 3]. ⁰() = 6²− 6 − 12 = 6(²−  − 2) = 6( − 2)( + 1) = 0 ⇔

 = 2 −1. (−2) = −3, (−1) = 8, (2) = −19, and (3) = −8. So (−1) = 8 is the absolute maximum value and

 (2) = −19 is the absolute minimum value.

50.³− 6²+ 5, [−3 5]. ⁰() = 3²− 12 = 3( − 4) = 0 ⇔  = 0, 4. (−3) = −76, (0) = 5, (4) = −27, and (5) = −20. So (0) = 5 is the absolute maximum value and (−3) = −76 is the absolute minimum value.

51. () = 3⁴− 4³− 12²+ 1, [−2 3]. ⁰() = 12³− 12²− 24 = 12(²−  − 2) = 12( + 1)( − 2) = 0 ⇔

 = −1, 0, 2. (−2) = 33, (−1) = −4, (0) = 1, (2) = −31, and (3) = 28. So (−2) = 33 is the absolute maximum value and (2) = −31 is the absolute minimum value.

52. () = (²− 4)³, [−2 3]. ⁰() = 3(²− 4)²(2) = 6( + 2)²( − 2)²= 0 ⇔  = −2, 0, 2. (±2) = 0,

 (0) = −64, and (3) = 5³ = 125. So (3) = 125 is the absolute maximum value and (0) = −64 is the absolute minimum value.

53. () =  +1

, [02 4]. ⁰() = 1 − 1

² =²− 1

² =( + 1)( − 1)

² = 0 ⇔  = ±1, but  = −1 is not in the given interval, [02 4]. ⁰()does not exist when  = 0, but 0 is not in the given interval, so 1 is the only critical nuumber.

 (02) = 52, (1) = 2, and (4) = 425. So (02) = 52 is the absolute maximum value and (1) = 2 is the absolute minimum value.

54. () = 

²−  + 1, [0 3].

⁰() = (²−  + 1) − (2 − 1)

(²−  + 1)² =²−  + 1 − 2²+ 

(²−  + 1)² = 1 − ²

(²−  + 1)² =(1 + )(1 − )

(²−  + 1)² = 0 ⇔

60. Find the absolute maximum and absolute minimum values of f on the given interval.

f (x) = e^x

1 + x², [0, 3]

Solution:

332 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 60. () = ^

1 + ², [0 3]. ⁰() =(1 + ²)^− ^(2)

(1 + ²)² =^(²− 2 + 1)

(1 + ²)² = ^( − 1)²

(1 + ²)² . ⁰() = 0 ⇒ ( − 1)²= 0 ⇔  = 1. ⁰()exists for all real numbers since 1 + ²is never equal to 0. (0) = 1,

 (1) = 2  1359, and (3) = ³10  2009. So (3) = ³10is the absolute maximum value and (0) = 1 is the absolute minimum value.

61. () = 2 cos  + sin 2, [0, 2].

⁰() = −2 sin  + cos 2 · 2 = −2 sin  + 2(1 − 2 sin²) = −2(2 sin² + sin  − 1) = −2(2 sin  − 1)(sin  + 1).

⁰() = 0 ⇒ sin  = ¹2or sin  = −1 ⇒  = ^6. (0) = 2, (^₆) =√ 3 +¹₂√

3 =³₂√

3  260, and (^2) = 0.

So (^₆) =³₂√

3is the absolute maximum value and (^₂) = 0is the absolute minimum value.

62. () = 1 + cos², [4 ]. ⁰() = 2 cos (− sin ) = −2 sin  cos  = − sin 2. ⁰() = 0 ⇒ − sin 2 = 0 ⇒

2 =  ⇒  = ^2 . Only  = ^₂ [ = 1] is in the interval (4, ). (4) = 1 +

 √2 2

²

=3 2,

 (2) = 1 + 0² = 1, and () = 1 + (−1)²= 2. So () = 2 is the absolute maximum value and (2) = 1 is the absolute minimum value.

63. () = ⁻²ln , 1

2 4. ⁰() = ⁻²·1

+ (ln )(−2⁻³) = ⁻³− 2⁻³ln  = ⁻³(1 − 2 ln ) = 1 − 2 ln 

³ .

⁰() = 0 ⇔ 1 − 2 ln  = 0 ⇔ 2 ln  = 1 ⇔ ln  = ¹2 ⇔  = ^12≈ 165. ⁰()does not exist when  = 0, which is not in the given interval,₁

2 4. ₁

2

= ln 12

(12)² =ln 1 − ln 2

14 = −4 ln 2 ≈ −2773,

 (^12) = ln ^12 (^12)² =12

 = 1

2 ≈ 0184, and (4) = ln 4 4² = ln 4

16 ≈ 0087. So (^12) = 1

2 is the absolute maximum value and ₁

2

= −4 ln 2 is the absolute minimum value.

64. () = ^2, [−3 1]. ⁰() = ^21 2

+ ^2(1) = ^21

2 + 1. ⁰() = 0 ⇔ ¹2 + 1 = 0 ⇔  = −2.

 (−3) = −3^−32≈ −0669, (−2) = −2⁻¹≈ −0736, and (1) = ^12≈ 1649. So (1) = ^12is the absolute maximum value and (−2) = −2 is the absolute minimum value.

65. () = ln(²+  + 1), [−1 1]. ⁰() = 1

²+  + 1· (2 + 1) = 0 ⇔  = −¹2. Since ²+  + 1  0for all , the domain of  and ⁰is . (−1) = ln 1 = 0, 

−¹2

= ln³₄  −029, and (1) = ln 3  110. So (1) = ln 3  110 is

the absolute maximum value and 

−¹2

= ln³₄  −029 is the absolute minimum value.

66. () =  − 2 tan⁻¹, [0 4]. ⁰() = 1 − 2 · 1

1 + ² = 0 ⇔ 1 = 2

1 + ² ⇔ 1 + ²= 2 ⇔ ²= 1 ⇔

 = ±1. (0) = 0, (1) = 1 −^2 ≈ −057, and (4) = 4 − 2 tan⁻¹4 ≈ 1 35. So (4) = 4 − 2 tan⁻¹4is the absolute maximum value and (1) = 1 −^2 is the absolute minimum value.

f (x) = x⁻²ln x, [1 2, 4]

Solution:

SECTION 4.1 MAXIMUM AND MINIMUM VALUES ¤ 307

 = ±1, but  = −1 is not in the given interval, [0 3]. (0) = 0, (1) = 1, and (3) = ³7. So (1) = 1 is the absolute maximum value and (0) = 0 is the absolute minimum value.

55. () =  −√³

, [−1 4]. ⁰() = 1 − ¹3^−23= 1 − 1

3^23. ⁰() = 0 ⇔ 1 = 1

3^23 ⇔ ^23=1

3 ⇔

 = ±

1 3

32

= ±

1

27 = ± 1 3√

3 = ±

√3

9 . ⁰()does not exist when  = 0. (−1) = 0, (0) = 0,



 −1 3√

3



= −1 3√

3−−1

√3 = −1 + 3 3√

3 =2√ 3

9 ≈ 03849, 

 1 3√

3



= 1

3√ 3− 1

√3 = −2√ 3 9 , and

 (4) = 4 −√³

4 ≈ 2413. So (4) = 4 −√³

4is the absolute maximum value and 

 √3 9



= −2√ 3

9 is the absolute minimum value.

56. () =

√

1 + ², [0 2]. ⁰() = (1 + ²)(1(2√

 )) −√

 (2)

(1 + ²)² =(1 + ²) − 2√

√

 (2) 2√

 (1 + ²)² = 1 − 3² 2√

 (1 + ²)².

⁰() = 0 ⇔ 1 − 3²= 0 ⇔ ² =1

3 ⇔  = ± 1

√3, but  = − 1

√3is not in the given interval, [0 2]. ⁰()does

not exist when  = 0, which is an endpoint. (0) = 0, 

 1

√3



= 1√⁴ 3

1 + 13= 3^−14 43 =3^34

4 ≈ 0570, and

 (2) =

√2

5 ≈ 0283. So 

 1

√3



=3^34

4 is the absolute maximum value and (0) = 0 is the absolute minimum value.

57. () = 2 cos  + sin 2, [0, 2].

⁰() = −2 sin  + cos 2 · 2 = −2 sin  + 2(1 − 2 sin²) = −2(2 sin² + sin  − 1) = −2(2 sin  − 1)(sin  + 1).

⁰() = 0 ⇒ sin  = ¹2 or sin  = −1 ⇒  =^6. (0) = 2, (^₆) =√ 3 +¹₂√

3 =³₂√

3 ≈ 260, and (^2) = 0.

So (^₆) =³₂√

3is the absolute maximum value and (^₂) = 0is the absolute minimum value.

58. () =  + cot(2), [4 74]. ⁰() = 1 − csc²(2) ·¹2.

⁰() = 0 ⇒ ¹2csc²(2) = 1 ⇒ csc²(2) = 2 ⇒ csc(2) = ±√

2 ⇒ ¹2 =^₄ or¹₂ = ^3₄

_

4 ≤  ≤ ^74 ⇒ ^8 ≤¹2 ≤^78 and csc(2) 6= −√

2in the last interval

⇒  = ^2 or  = ^3₂ .

_

4

=^₄ + cot^₈ ≈ 320, _

2

=^₂ + cot^₄ =^₂ + 1 ≈ 257, _3

2

= ^3₂ + cot^3₂ =^3₂ − 1 ≈ 371, and

_7

4

=^7₄ + cot^7₈ ≈ 308. So _3

2

= ^3₂ − 1 is the absolute maximum value and _

2

= ^₂ + 1is the absolute minimum value.

59. () = ⁻²ln , 1

2 4. ⁰() = ⁻²·1

+ (ln )(−2⁻³) = ⁻³− 2⁻³ln  = ⁻³(1 − 2 ln ) = 1 − 2 ln 

³ .

⁰() = 0 ⇔ 1 − 2 ln  = 0 ⇔ 2 ln  = 1 ⇔ ln  = ¹2 ⇔  = ^12≈ 165. ⁰()does not exist when  = 0, which is not in the given interval,1

2 4. 1 2

= ln 12

(12)² = ln 1 − ln 2

14 = −4 ln 2 ≈ −2773,



^12

= ln ^12 (^12)² =12

 = 1

2 ≈ 0184, and (4) = ln 4 4² = ln 4

16 ≈ 0087. So (^12) = 1

2

60.  () = ^2, [−3 1]. ⁰() = ^21 2

+ ^2(1) = ^21

2 + 1. ⁰() = 0 ⇔ ¹2 + 1 = 0 ⇔  = −2.

61.  () = ln(²+  + 1), [−1 1]. ⁰() = 1

²+  + 1 · (2 + 1) = 0 ⇔  = −¹2. Since ²+  + 1  0for all , the domain of  and ⁰is R. (−1) = ln 1 = 0, 

−¹2

= ln³₄ ≈ −029, and (1) = ln 3 ≈ 110. So (1) = ln 3 ≈ 110 is the absolute maximum value and 

−¹2

= ln³₄ ≈ −029 is the absolute minimum value.

62.  () =  − 2 tan⁻¹, [0 4]. ⁰() = 1 − 2 · 1

1 + ² = 0 ⇔ 1 = 2

1 + ² ⇔ 1 + ²= 2 ⇔ ²= 1 ⇔

63.  () = ^(1 − )^, 0 ≤  ≤ 1,   0,   0.

⁰() = ^· (1 − )^⁻¹(−1) + (1 − )^· ^⁻¹= ^⁻¹(1 − )^⁻¹[ · (−1) + (1 − ) · ]

= ^⁻¹(1 − )^⁻¹( −  − )

At the endpoints, we have (0) = (1) = 0 [the minimum value of  ]. In the interval (0 1), ⁰() = 0 ⇔  = 

 + 



 

 + 



=

 

 + 

 1 − 

 + 



= ^ ( + )^

 +  − 

 + 



= ^

( + )^· ^

( + )^ = ^^ ( + )^+. So 

 

 + 



= ^^

( + )^+ is the absolute maximum value.

64. The graph of () =

1 + 5 − ³ indicates that ⁰() = 0at  ≈ ±13 and that ⁰()does not exist at  ≈ −21, −02, and 23. Those ﬁve values of  are the critical numbers of .

65. (a) From the graph, it appears that the absolute maximum value is about

 (−077) = 219, and the absolute minimum value is about (077) = 181.

1

(2)

f (x) = x − 2 tan⁻¹x, [0, 4]

Solution:



^12

= ln ^12 (^12)² =12

 = 1

2 ≈ 0184, and (4) = ln 4 4² = ln 4

16 ≈ 0087. So (^12) = 1

2

60. () = ^2, [−3 1]. ⁰() = ^21 2

+ ^2(1) = ^21

2 + 1. ⁰() = 0 ⇔ ¹2 + 1 = 0 ⇔  = −2.

61. () = ln(²+  + 1), [−1 1]. ⁰() = 1

²+  + 1 · (2 + 1) = 0 ⇔  = −¹2. Since ²+  + 1  0for all , the domain of  and ⁰is R. (−1) = ln 1 = 0, 

−¹2

= ln³₄ ≈ −029, and (1) = ln 3 ≈ 110. So (1) = ln 3 ≈ 110 is the absolute maximum value and 

−¹2

= ln³₄ ≈ −029 is the absolute minimum value.

62. () =  − 2 tan⁻¹, [0 4]. ⁰() = 1 − 2 · 1

1 + ² = 0 ⇔ 1 = 2

1 + ² ⇔ 1 + ² = 2 ⇔ ²= 1 ⇔

63. () = ^(1 − )^, 0 ≤  ≤ 1,   0,   0.

⁰() = ^· (1 − )^⁻¹(−1) + (1 − )^· ^⁻¹= ^⁻¹(1 − )^⁻¹[ · (−1) + (1 − ) · ]

= ^⁻¹(1 − )^⁻¹( −  − )

At the endpoints, we have (0) = (1) = 0 [the minimum value of  ]. In the interval (0 1), ⁰() = 0 ⇔  = 

 + 



 

 + 



=

 

 + 

 1 − 

 + 



= ^ ( + )^

 +  − 

 + 



= ^

( + )^· ^

( + )^ = ^^ ( + )^+. So 

 

 + 



= ^^

( + )^+ is the absolute maximum value.

64. The graph of () =

1 + 5 − ³ indicates that ⁰() = 0at  ≈ ±13 and that ⁰()does not exist at  ≈ −21, −02, and 23. Those ﬁve values of  are the critical numbers of .

65. (a) From the graph, it appears that the absolute maximum value is about

 (−077) = 219, and the absolute minimum value is about (077) = 181.

2