Section 4.3 What Derivatives Tell Us about the Shape of a Graph 55. Let

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Section 4.3 What Derivatives Tell Us about the Shape of a Graph

55. Let C(x) = x^1/3(x + 4).

(a) Find the intervals of increase or decrease.

(b) Find the local maximum and minimum values.

(c) Find the intervals of concavity and the inflection points.

(d) Use the information from parts (a)(c) to sketch the graph. You may want to check your work with a graphing calculator or computer.

Solution:

SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 331

(b)  changes from decreasing to increasing at  = 0, so (0) = 0 is a local minimum value.  changes from increasing to decreasing at  = 1, so (1) = 3 is a local maximum value. Note that the First Derivative Test applies at  = 0 even though ⁰is not deﬁned at  = 0, since  is continuous at 0.

(c) ⁰⁰() = −¹⁰9^−43−²⁰9^−13= −¹⁰9 ^−43(1 + 2). ⁰⁰()  0 ⇔

  −¹2 and ⁰⁰()  0 ⇔ −¹2    0or   0. So  is CU on

−∞ −¹2

and  is CD on

−¹2 0

and (0 ∞). The only change in concavity occurs at  = −¹2, so there is an inﬂection point at

−¹2 6√³ 4.

(d)

45. (a) () = ^13( + 4) = ^43+ 4^13 ⇒ ⁰() = ⁴₃^13+⁴₃^−23= ⁴₃^−23( + 1) = 4( + 1) 3√³

² . ⁰()  0if

−1    0 or   0 and ⁰()  0for   −1, so  is increasing on (−1 ∞) and  is decreasing on (−∞ −1).

(b) (−1) = −3 is a local minimum value. (d)

(c) ⁰⁰() = ⁴₉^−23−⁸9^−53=⁴₉^−53( − 2) = 4( − 2) 9√³

⁵ .

⁰⁰()  0for 0    2 and ⁰⁰()  0for   0 and   2, so  is concave downward on (0 2) and concave upward on (−∞ 0) and (2 ∞).

There are inﬂection points at (0 0) and 2 6√³

2

≈ (2 756).

46. (a) () = ln(²+ 9) ⇒ ⁰() = 1

²+ 9· 2 = 2

²+ 9. ⁰()  0 ⇔ 2  0 ⇔   0 and ⁰()  0 ⇔

  0. So  is increasing on (0 ∞) and  is decreasing on (−∞ 0).

(b)  changes from decreasing to increasing at  = 0, so (0) = ln 9 is a local minimum value. There is no local maximum value.

(d)

(c) ⁰⁰() =(²+ 9) · 2 − 2(2)

(²+ 9)² = 18 − 2²

(²+ 9)² = −2( + 3)( − 3) (²+ 9)² .

⁰⁰() = 0 ⇔  = ±3. ⁰⁰()  0on (−3 3) and ⁰⁰()  0on (−∞ −3) and (3 ∞). So  is CU on (−3 3), and  is CD on (−∞ −3) and (3 ∞). There are inﬂection points at (±3 ln 18).

47. (a) () = 2 cos  + cos², 0 ≤  ≤ 2 ⇒ ⁰() = −2 sin  + 2 cos  (− sin ) = −2 sin  (1 + cos ).

⁰() = 0 ⇔  = 0  and 2. ⁰()  0 ⇔     2 and ⁰()  0 ⇔ 0    . So  is increasing on ( 2) and  is decreasing on (0 ).

(b) () = −1 is a local minimum value.

62. Let f (x) = _1−e^e^xx.

(a) Find the vertical and horizontal asymptotes.

(b) Find the intervals of increase or decrease.

(c) Find the local maximum and minimum values.

(d) Find the intervals of concavity and the inflection points.

(e) Use the information from parts (a)(d) to sketch the graph of f.

Solution:

SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 333 50.  () = ²− 4

²+ 4has domain R.

(a) lim

→±∞

²− 4

²+ 4= lim

→±∞

1 − 4² 1 + 4² = 1

1= 1, so  = 1 is a HA. There is no vertical asymptote.

(b) ⁰() = (²+ 4)(2) − (²− 4)(2)

(²+ 4)² = 2[(²+ 4) − (²− 4)]

(²+ 4)² = 16

(²+ 4)². ⁰()  0 ⇔   0 and

⁰()  0 ⇔   0. So  is increasing on (0 ∞) and  is decreasing on (−∞ 0).

(c)  changes from decreasing to increasing at  = 0, so (0) = −1 is a local minimum value.

(d) ⁰⁰() = (²+ 4)²(16) − 16 · 2(²+ 4)(2)

[(²+ 4)²]² =16(²+ 4)[(²+ 4) − 4²]

(²+ 4)⁴ =16(4 − 3²) (²+ 4)³ .

⁰⁰() = 0 ⇔  = ±2√

3. ⁰⁰()  0 ⇔ −2√

3    2√ 3 and ⁰⁰()  0 ⇔   −2√

3 or   2√3. So  is CU on

−2√ 3 2√

3

and  is CD on

−∞ −2√ 3

and 2√

3 ∞ . There are inﬂection points at

±2√ 3 −¹2

.

(e)

51. (a) lim

→−∞

√²+ 1 − 

= ∞ and

lim→∞

√²+ 1 − 

= lim

→∞

√²+ 1 − √

²+ 1 + 

√²+ 1 + = lim

→∞

√ 1

²+ 1 + = 0, so  = 0 is a HA.

(b) () =√

²+ 1 −  ⇒ ⁰() = 

√²+ 1− 1. Since 

√²+ 1 1for all , ⁰()  0, so  is decreasing on R.

(c) No minimum or maximum

(d) ⁰⁰() = (²+ 1)^12(1) −  ·¹2(²+ 1)^−12(2)

√²+ 12

=

(²+ 1)^12− ² (²+ 1)^12

²+ 1 = (²+ 1) − ²

(²+ 1)^32 = 1

(²+ 1)^32  0, so  is CU on R. No IP

(e)

52.  () = ^

1 − ^ has domain { | 1 − ^6= 0} = { | ^6= 1} = { |  6= 0}.

(a) lim

→∞

^

1 − ^ = lim

→∞

^^

(1 − ^)^ = lim

→∞

1

1^− 1 = 1

0 − 1 = −1, so  = −1 is a HA.

→−∞lim

^

1 − ^ = 0

1 − 0 = 0, so  = 0 is a HA. lim

→0⁺

^

1 − ^ = −∞ and lim

→0⁻

^

1 − ^ = ∞, so  = 0 is a VA.

(b) ⁰() = (1 − ^)^− ^(−^)

(1 − ^)² =^[(1 − ^) + ^]

(1 − ^)² = ^

(1 − ^)². ⁰()  0for  6= 0, so  is increasing on (−∞ 0) and (0 ∞).

(c) There is no local maximum or minimum.

334 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (d) ⁰⁰() = (1 − ^)²^− ^· 2(1 − ^)(−^)

[(1 − ^)²]²

= (1 − ^)^[(1 − ^) + 2^]

(1 − ^)⁴ =^(^+ 1) (1 − ^)³

⁰⁰()  0 ⇔ (1 − ^)³ 0 ⇔ ^ 1 ⇔   0 and

⁰⁰()  0 ⇔   0. So  is CU on (−∞ 0) and  is CD on (0 ∞).

There is no inﬂection point.

(e)

53. (a) lim

→±∞^−² = lim

→±∞

1

^² = 0, so  = 0 is a HA. There is no VA.

(b) () = ^−² ⇒ ⁰() = ^−²(−2). ⁰() = 0 ⇔  = 0. ⁰()  0 ⇔   0 and ⁰()  0 ⇔

  0. So  is increasing on (−∞ 0) and  is decreasing on (0 ∞).

(c)  changes from increasing to decreasing at  = 0, so (0) = 1 is a local maximum value. There is no local minimum value.

(d) ⁰⁰() = ^−²(−2) + (−2)^−²(−2) = −2^−²(1 − 2²).

⁰⁰() = 0 ⇔ ²=¹₂ ⇔  = ±1√

2. ⁰⁰()  0 ⇔

  −1√

2or   1√

2and ⁰⁰()  0 ⇔ −1√

2    1√ 2. So

is CU on

−∞ −1√ 2and

1√

2 ∞, and  is CD on

−1√ 2 1√

2. There are inﬂection points at

±1√

2 ^−12 .

(e)

54.  () =  −¹6²−²3ln has domain (0 ∞).

(a) lim

→0⁺

 −¹6²−²3ln 

= ∞ since ln  → −∞ as  → 0⁺, so  = 0 is a VA. There is no HA.

(b) ⁰() = 1 −1 3 − 2

3 =3 − ²− 2

3 = −(²− 3 + 2)

3 = −( − 1)( − 2)

3 . ⁰()  0 ⇔

( − 1)( − 2)  0 ⇔ 1    2 and ⁰()  0 ⇔ 0    1 or   2. So  is increasing on (1 2) and

is decreasing on (0 1) and (2 ∞).

(c)  changes from decreasing to increasing at  = 1, so (1) = ⁵₆is a local minimum value.  changes from increasing to decreasing at  = 2, so (2) =⁴₃−²3ln 2 ≈ 087 is a local maximum value.

(d) ⁰⁰() = −1 3+ 2

3² =2 − ²

3² . ⁰⁰()  0 ⇔ 0   √ 2 and

⁰⁰()  0 ⇔  √

2. So  is CU on 0√

2and  is CD on

√2 ∞. There is an inﬂection point at√

2√

2 −¹3−¹3ln 2.

(e)

64. Let f (x) = x −¹₆x²−²₃ln x.

1

(2)

(a) Find the vertical and horizontal asymptotes.

(b) Find the intervals of increase or decrease.

(c) Find the local maximum and minimum values.

(d) Find the intervals of concavity and the inflection points.

(e) Use the information from parts (a)(d) to sketch the graph of f.

Solution:

334 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (d) ⁰⁰() = (1 − ^)²^− ^· 2(1 − ^)(−^)

[(1 − ^)²]²

=(1 − ^)^[(1 − ^) + 2^]

(1 − ^)⁴ =^(^+ 1) (1 − ^)³

⁰⁰()  0 ⇔ (1 − ^)³ 0 ⇔ ^ 1 ⇔   0 and

⁰⁰()  0 ⇔   0. So  is CU on (−∞ 0) and  is CD on (0 ∞).

There is no inﬂection point.

(e)

53. (a) lim

→±∞^−² = lim

→±∞

1

^² = 0, so  = 0 is a HA. There is no VA.

(b) () = ^−² ⇒ ⁰() = ^−²(−2). ⁰() = 0 ⇔  = 0. ⁰()  0 ⇔   0 and ⁰()  0 ⇔

  0. So  is increasing on (−∞ 0) and  is decreasing on (0 ∞).

(c)  changes from increasing to decreasing at  = 0, so (0) = 1 is a local maximum value. There is no local minimum value.

(d) ⁰⁰() = ^−²(−2) + (−2)^−²(−2) = −2^−²(1 − 2²).

⁰⁰() = 0 ⇔ ²=¹₂ ⇔  = ±1√

2. ⁰⁰()  0 ⇔

  −1√

2or   1√

2and ⁰⁰()  0 ⇔ −1√

2    1√ 2. So

is CU on

−∞ −1√ 2and

1√

2 ∞, and  is CD on

−1√ 2 1√

2. There are inﬂection points at

±1√

2 ^−12 .

(e)

54.  () =  −¹6²−²3ln has domain (0 ∞).

(a) lim

→0⁺

 −¹6²−²3ln 

= ∞ since ln  → −∞ as  → 0⁺, so  = 0 is a VA. There is no HA.

(b) ⁰() = 1 −1 3 − 2

3 = 3 − ²− 2

3 = −(²− 3 + 2)

3 = −( − 1)( − 2)

3 . ⁰()  0 ⇔

( − 1)( − 2)  0 ⇔ 1    2 and ⁰()  0 ⇔ 0    1 or   2. So  is increasing on (1 2) and

is decreasing on (0 1) and (2 ∞).

(c)  changes from decreasing to increasing at  = 1, so (1) =⁵₆is a local minimum value.  changes from increasing to decreasing at  = 2, so (2) =⁴₃ −²3ln 2 ≈ 087 is a local maximum value.

(d) ⁰⁰() = −1 3+ 2

3² =2 − ²

3² . ⁰⁰()  0 ⇔ 0   √ 2 and

⁰⁰()  0 ⇔  √

2. So  is CU on 0√

2and  is CD on

√2 ∞. There is an inﬂection point at√

2√

2 −¹3 −¹3ln 2.

(e)

84. For what values of the numbers a and b does the function

f (x) = axe^bx²

have the maximum value f (2) = 1?

Solution:

340 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 74.  () = ^² ⇒ ⁰() = 

^²· 2 + ^²· 1

= ^²(2²+ 1). For (2) = 1 to be a maximum value, we must have ⁰(2) = 0. (2) = 1 ⇒ 1 = 2^4and ⁰(2) = 0 ⇒ 0 = (8 + 1)^4. So 8 + 1 = 0 [ 6= 0] ⇒

 = −¹8and now 1 = 2^−12 ⇒  = √/2.

75. (a) () = ³+ ²+  ⇒ ⁰() = 3²+ 2 + .  has the local minimum value −²9

√3at  = 1√3, so

⁰(√¹

3) = 0 ⇒ 1 +^√²₃ +  = 0 (1) and  (√¹ 3) = −²9

√3 ⇒ ¹9

√3 +¹₃ +¹₃√ 3 = −²9

√3 (2).

Rewrite the system of equations as

2 3

√3 +  = −1 (3)

1

3 + ¹₃√

3 = −¹3

√3 (4) and then multiplying (4) by −2√

3gives us the system

2 3

√3 +  = −1

−²3

√3 − 2 = 2

Adding the equations gives us − = 1 ⇒  = −1. Substituting −1 for  into (3) gives us

2 3

√3 − 1 = −1 ⇒ ²3

√3 = 0 ⇒  = 0. Thus, () = ³− .

(b) To find the smallest slope, we want to find the minimum of the slope function, ⁰, so we’ll find the critical numbers of ⁰. () = ³−  ⇒ ⁰() = 3²− 1 ⇒ ⁰⁰() = 6. ⁰⁰() = 0 ⇔  = 0.

At  = 0,  = 0, ⁰() = −1, and ⁰⁰changes from negative to positive. Thus, we have a minimum for ⁰and

 − 0 = −1( − 0), or  = −, is the tangent line that has the smallest slope.

76. The original equation can be written as (²+ ) +  = 0. Call this (1). Since (2 25) is on this curve, we have (4 + )₅

2

+ 2 = 0, or 20 + 5 + 4 = 0. Let’s rewrite that as 4 + 5 = −20 and call it (A). Differentiating (1) gives (after regrouping) (²+ )⁰= −(2 + ). Call this (2). Differentiating again gives (²+ )⁰⁰+ (2)⁰= −2⁰− 2, or (²+ )⁰⁰+ 4⁰+ 2 = 0. Call this (3). At (2 25), equations (2) and (3) say that (4 + )⁰= −(10 + ) and (4 + )⁰⁰+ 8⁰+ 5 = 0. If (2 25) is an inflection point, then ⁰⁰= 0there, so the second condition becomes 8⁰+ 5 = 0, or ⁰= −⁵8. Substituting in the first condition, we get −(4 + )⁵8 = −(10 + ), or 20 + 5 = 80 + 8, which simplifies to

−8 + 5 = 60. Call this (B). Subtracting (B) from (A) yields 12 = −80, so  = −²⁰3. Substituting that value in (A) gives

−⁸⁰3 + 5 = −20 = −⁶⁰3, so 5 =²⁰₃ and  = ⁴₃. Thus far we’ve shown that IF the curve has an inﬂection point at (2 25), then  = −²⁰3 and  = ⁴₃.

To prove the converse, suppose that  = −²⁰3 and  = ⁴₃. Then by (1), (2), and (3), our curve satisﬁes

and

²+⁴₃

 = ²⁰₃ (4)

²+⁴₃

⁰= −2 +²⁰3 (5)

²+⁴₃

⁰⁰+ 4⁰+ 2 = 0. (6) Multiply (6) by

²+⁴₃and substitute from (4) and (5) to obtain

²+⁴₃2

⁰⁰+ 4

−2 +²⁰3

+ 220

3= 0, or

94. Show that if f (x) = x⁴, then f⁰⁰(0) = 0, but (0, 0) is not an inflection point of the graph of f . Solution:

SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 343

85.Let the cubic function be () = ³+ ²+  +  ⇒ ⁰() = 3²+ 2 +  ⇒ ⁰⁰() = 6 + 2.

So  is CU when 6 + 2  0 ⇔   −(3), CD when   −(3), and so the only point of inﬂection occurs when  = −(3). If the graph has three -intercepts ¹, 2and 3, then the expression for () must factor as

 () = ( − ¹)( − ²)( − ³). Multiplying these factors together gives us

 () = [³− (¹+ 2+ 3)²+ (12+ 13+ 23) − ¹23]

Equating the coefﬁcients of the ²-terms for the two forms of  gives us  = −(¹+ 2+ 3). Hence, the -coordinate of the point of inﬂection is − 

3 = −−(¹+ 2+ 3)

3 = 1+ 2+ 3

3 .

86. () = ⁴+ ³+ ² ⇒ ⁰() = 4³+ 3²+ 2 ⇒ ⁰⁰() = 12²+ 6 + 2. The graph of ⁰⁰()is a parabola. If ⁰⁰()has two roots, then it changes sign twice and so has two inflection points. This happens when the discriminant of ⁰⁰()is positive, that is, (6)²− 4 · 12 · 2  0 ⇔ 36²− 96  0 ⇔ ||  ²^√3⁶ ≈ 163 If 36²− 96 = 0 ⇔  = ±²^√3⁶, ⁰⁰()is 0 at one point, but there is still no inflection point since ⁰⁰()never changes sign, and if 36²− 96  0 ⇔ ||  ²^√3⁶, then ⁰⁰()never changes sign, and so there is no inflection point.

 = 6  = 3  = 18

 =²^√₃⁶  = 0  = −2

For large positive , the graph of  has two inflection points and a large dip to the left of the -axis. As  decreases, the graph of  becomes flatter for   0, and eventually the dip rises above the -axis, and then disappears entirely, along with the inflection points. As  continues to decrease, the dip and the inflection points reappear, to the right of the origin.

87.By hypothesis  = ⁰is differentiable on an open interval containing . Since ( ()) is a point of inﬂection, the concavity changes at  = , so ⁰⁰()changes signs at  = . Hence, by the First Derivative Test, ⁰has a local extremum at  = .

Thus, by Fermat’s Theorem ⁰⁰() = 0.

88. () = ⁴ ⇒ ⁰() = 4³ ⇒ ⁰⁰() = 12² ⇒ ⁰⁰(0) = 0. For   0, ⁰⁰()  0, so  is CU on (−∞ 0);

for   0, ⁰⁰()  0, so  is also CU on (0 ∞). Since  does not change concavity at 0, (0 0) is not an inﬂection point.

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