Section 4.5 Summary of Curve Sketching

(1)

Section 4.5 Summary of Curve Sketching

11. Use the guidelines of this section to sketch the curve.

y = x − x² 2 − 3x + x² Solution:

SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 359 9.  =  () = ²− 4

²− 2= ( + 2)( − 2)

( − 2) = + 2

 = 1 +2

 for  6= 2. There is a hole in the graph at (2 2).

A.  = { |  6= 0 2} = (−∞ 0) ∪ (0 2) ∪ (2 ∞) B. -intercept: none; -intercept: −2 C. No symmetry D. lim

→±∞

 + 2

 = 1, so  = 1 is a HA. lim

→0⁻

 + 2

 = −∞,

lim

→0⁺

 + 2

 = ∞, so  = 0 is a VA. E. ⁰() = −2²  0 [ 6= 0 2]

so  is decreasing on (−∞ 0), (0 2), and (2 ∞). F. No extrema G. ⁰⁰() = 4³ 0 ⇔   0, so  is CU on (0 2) and (2 ∞) and CD on (−∞ 0). No IP

H.

10.  =  () = ²+ 5

25 − ² = ( + 5)

(5 + )(5 − ) = 

5 − for  6= −5. There is a hole in the graph at

−5 −¹2

.

A.  = { |  6= ±5} = (−∞ −5) ∪ (−5 5) ∪ (5 ∞) B. -intercept = 0, -intercept = (0) = 0 C. No symmetry D. lim

→±∞



5 −  = −1, so  = −1 is a HA. lim

→5⁻



5 − = ∞, lim

→5⁺



5 − = −∞, so  = 5 is a VA.

E. ⁰() = (5 − )(1) − (−1)

(5 − )² = 5

(5 − )²  0for all  in , so  is increasing on (−∞ −5), (−5 5), and (5 ∞). F. No extrema

G. ⁰() = 5(5 − )⁻² ⇒

⁰⁰() = −10(5 − )⁻³(−1) = 10

(5 − )³  0 ⇔   5, so  is CU on (−∞ −5) and (−5 5), and  is CD on (5 ∞). No IP

H.

11.  =  () =  − ²

2 − 3 + ² = (1 − )

(1 − )(2 − ) = 

2 − for  6= 1. There is a hole in the graph at (1 1).

A.  = { |  6= 1 2} = (−∞ 1) ∪ (1 2) ∪ (2 ∞) B. -intercept = 0, -intercept = (0) = 0 C. No symmetry D. lim

→±∞



2 −  = −1, so  = −1 is a HA. lim

→2⁻



2 − = ∞, lim

→2⁺



2 − = −∞, so  = 2 is a VA.

E. ⁰() = (2 − )(1) − (−1)

(2 − )² = 2

(2 − )²  0 [ 6= 1 2], so  is increasing on (−∞ 1), (1 2), and (2 ∞). F. No extrema

G. ⁰() = 2(2 − )⁻² ⇒

⁰⁰() = −4(2 − )⁻³(−1) = 4

(2 − )³  0 ⇔   2, so  is CU on (−∞ 1) and (1 2), and  is CD on (2 ∞). No IP

H.

12.  =  () = 1 +1

+ 1

² = ²+  + 1

² A.  = (−∞ 0) ∪ (0 ∞) B. -intercept: none [ 6= 0];

-intercepts: () = 0 ⇔ ²+  + 1 = 0, there is no real solution, and hence, no -intercept C. No symmetry D. lim

→±∞

 1 +1

+ 1

²



= 1, so  = 1 is a HA. lim

→0 () = ∞, so  = 0 is a VA. E. ⁰() = −1

² − 2

³ = − − 2

³ .

⁰()  0 ⇔ −2    0 and ⁰()  0 ⇔   −2 or   0, so  is increasing on (−2 0) and decreasing

y = sin x +√

3 cos x, −2π ≤ x ≤ 2π

Solution:

368 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION

34. =  () =  + cos  A.  = R B. -intercept: (0) = 1; the -intercept is about −074 and can be found using Newton’s method C. No symmetry D. No asymptote E. ⁰() = 1 − sin   0 except for  = ^2 + 2, so  is increasing on R. F. No local extrema

G. ⁰⁰() = − cos . ⁰⁰()  0 ⇒ − cos   0 ⇒ cos   0 ⇒

is in_

2 + 2^3₂ + 2

and ⁰⁰()  0 ⇒

is in

−^2 + 2^₂ + 2, so  is CU on_

2 + 2^3₂ + 2and CD on

−^2 + 2^₂ + 2. IP at

2 +   2 + 

=

2 + ^₂ +  [on the line  = ]

H.

35. =  () =  tan , −^2    ^₂ A.  =

−^2^₂

B. Intercepts are 0 C. (−) = (), so the curve is symmetric about the -axis. D. lim

→(2)⁻ tan  = ∞ and lim

→−(2)⁺ tan  = ∞, so  = ^2 and  = −^2 are VA.

E. ⁰() = tan  +  sec²  0 ⇔ 0    ^2, so  increases on 0^₂ and decreases on

−^2 0

. F. Absolute and local minimum value (0) = 0.

G. ⁰⁰= 2 sec² + 2 tan  sec²  0for −^2    ^₂, so  is CU on

−^2^₂ . No IP

H.

36. =  () = 2 − tan , −^2    ^₂ A.  =

−^2^₂

B. -intercept: (0) = 0; -intercepts: () = 0 ⇔ 2 = tan  ⇔  = 0 or  ≈ ±117 C. (−) = −(), so  is odd; the graph is symmetric about the origin.

D. lim

→(−2)⁺(2 − tan ) = ∞ and lim

→(2)⁻(2 − tan ) = −∞, so  = ±^2 are VA. No HA.

E. ⁰() = 2 − sec²  0 ⇔ |sec | √

2and ⁰()  0 ⇔ |sec | √

2, so  is decreasing on

−^2 −^4

,

increasing on

−^4^₄

, and decreasing again on_

4^₂

F. Local maximum value _

4

=^₂ − 1,

local minimum value 

−^4

= −^2 + 1

G. ⁰⁰() = −2 sec  · sec  tan  = −2 tan  sec² = −2 tan (tan² + 1) so ⁰⁰()  0 ⇔ tan   0 ⇔ −^2    0, and ⁰⁰()  0 ⇔ tan   0 ⇔ 0    ^2. Thus,  is CU on

−^2 0

and CD on 0^₂

. IP at (0 0)

H.

37. =  () = sin  +√

3 cos , −2 ≤  ≤ 2 A.  = [−2 2] B. -intercept: (0) =√

3; -intercepts:

 () = 0 ⇔ sin  = −√

3 cos  ⇔ tan  = −√

3 ⇔  = −^43  −^3^2₃or^5₃ C.  is periodic with period 2. D. No asymptote E. ⁰() = cos  −√3 sin . ⁰() = 0 ⇔ cos  =√

3 sin  ⇔ tan  = 1

√3 ⇔

 = −^116 , −^56 ,^₆, or^7₆ . ⁰()  0 ⇔ −^116    −^56 or^₆    ^7₆, so  is decreasing on

−^116  −^56



SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 369 and

6^7₆ , and  is increasing on

−2 −^116

,

−^56^₆, and7

6  2. F. Local maximum value



−^116

=  6

=¹₂ +√ 31

2

√3= 2, local minimum value 

−^56

= 7

6

= −¹2+√ 3

−¹2

√3

= −2 G. ⁰⁰() = − sin  −√3 cos . ⁰⁰() = 0 ⇔ sin  = −√

3 cos  ⇔ tan  = − 1

√3 ⇔  = −^43, −^3, ^2₃ , or^5₃ . ⁰⁰()  0 ⇔

−^43    −^3 or ^2₃   ^5₃, so  is CU on

−^43  −^3

and2

3 ^5₃ , and

is CD on

−2 −^43

,

−^3^2₃  , and_5

3  2

. There are IPs at

−^43 0 ,

−^3 0,2

3 0, and5

3 0.

H.

38. =  () = csc  − 2 sin , 0     A.  = (0 ) B. No -intercept; -intercept: () = 0 ⇔ csc  = 2 sin  ⇔ ¹2 = sin² ⇔ sin  = ±¹2

√2 ⇔  =^4 or ^3₄ C. No symmetry D. lim

→0⁺ () = ∞ and lim

→⁻ () = ∞, so  = 0 and  =  are VAs.

E. ⁰() = − csc  cot  − 2 cos  = −cos 

sin²− 2 cos  = − cos 

 1

sin²+ 2



. ⁰()  0when − cos   0 ⇔ cos   0 ⇔ ^2    , so ⁰is increasing on

2 , and  is decreasing on

0^₂

. F. Local minimum value _

2

= −1

G. ⁰⁰() = (− csc )(− csc²) + (cot )(csc  cot ) + 2 sin 

=1 + cos² + 2 sin⁴ sin³

⁰⁰has the same sign as sin , which is positive on (0 ), so  is CU on (0 ).

No IP

H.

39. =  () = sin  1 + cos 





when cos 6= 1

= sin 

1 + cos ·1 − cos 

1 − cos  = sin  (1 − cos )

sin² = 1 − cos 

sin  = csc  − cot 





A. The domain of  is the set of all real numbers except odd integer multiples of ; that is, all reals except (2 + 1), where

is an integer. B. -intercept: (0) = 0; -intercepts:  = 2,  an integer. C. (−) = −(), so  is an odd function; the graph is symmetric about the origin and has period 2. D. When  is an odd integer, lim

→()⁻ () = ∞ and lim

→()⁺ () = −∞, so  =  is a VA for each odd integer . No HA.

E. ⁰() =(1 + cos ) · cos  − sin (− sin )

(1 + cos )² = 1 + cos 

(1 + cos )² = 1

1 + cos . ⁰()  0for all  except odd multiples of

, so  is increasing on ((2 − 1) (2 + 1)) for each integer . F. No extreme values G. ⁰⁰() = sin 

(1 + cos )²  0 ⇒ sin   0 ⇒

 ∈ (2 (2 + 1)) and ⁰⁰()  0on ((2 − 1) 2) for each integer .  is CU on (2 (2 + 1)) and CD on ((2 − 1) 2) for each integer .  has IPs at (2 0) for each integer .

H.

1

(2)

y = tan⁻¹ x − 1 x + 1

Solution:

374 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 52. =  () = ln 

² A.  = (0 ∞) B. -intercept: none; -intercept: () = 0 ⇔ ln  = 0 ⇔  = 1 C. No symmetry D. lim

→0⁺ () = −∞, so  = 0 is a VA; lim

→∞

ln 

²

= limH

→∞

1

2 = 0, so  = 0 is a HA.

E. ⁰() = ²(1) − (ln )(2)

(²)² =(1 − 2 ln )

⁴ =1 − 2 ln 

³ . ⁰()  0 ⇔ 1 − 2 ln   0 ⇔ ln  ¹2 ⇒ 0    ^12and ⁰()  0 ⇒   ^12, so  is increasing on

0√



and decreasing on√

 ∞ . F. Local maximum value (^12) =12

 = 1 2

G. ⁰⁰() = ³(−2) − (1 − 2 ln )(3²) (³)²

=²[−2 − 3(1 − 2 ln )]

⁶ = −5 + 6 ln 

⁴

⁰⁰()  0 ⇔ −5 + 6 ln   0 ⇔ ln   ⁵6 ⇒   ^56 [  is CU]

and ⁰⁰()  0 ⇔ 0    ^56 [  is CD]. IP at (^56 5(6^53))

H.

53. =  () = ^{arctan } A.  = R B. -intercept: (0) = ⁰= 1; no -intercept since ^{arctan }is positive for all .

C. No symmetry D. lim

→−∞ () = ^−2[≈ 021], so  = ^−2is a HA. lim

→∞ () = ^2[≈ 481], so  = ^2is a HA. E. ⁰() = ^{arctan }

 1

1 + ²



. ⁰()  0for all , so  is increasing on R. F. No local extrema

G. ⁰⁰() =

(1 + ²)^{arctan }

 1

1 + ²



− ^{arctan }(2) (1 + ²)²

=^{arctan }(1 − 2) (1 + ²)²

⁰⁰()  0for  ¹₂, so  is CU on

−∞¹2

 and  is CD on₁

2 ∞ . IP at

1

2 ^{arctan 12}

≈ (05 159)

H.

54. =  () = tan⁻¹

 − 1

 + 1



A.  = { |  6= −1} B. -intercept = 1, -intercept = (0) = tan⁻¹(−1) = −^4

→±∞tan⁻¹

 − 1

 + 1



= lim

→±∞tan⁻¹

1 − 1

1 + 1



= tan⁻¹1 =^₄, so  = ^₄ is a HA.

Also lim

→−1⁺tan⁻¹

 − 1

 + 1



= −

2 and lim

→−1⁻tan⁻¹

 − 1

 + 1



= 2.

E. ⁰() = 1

1 + [( − 1)( + 1)]²

( + 1) − ( − 1)

( + 1)² = 2

( + 1)²+ ( − 1)² = 1

²+ 1 0, so  is increasing on (−∞ −1) and (−1 ∞)  F. No extreme values

G. ⁰⁰() = −2

²+ 12

 0 ⇔   0, so  is CU on (−∞ −1) and (−1 0), and CD on (0 ∞). IP at

0 −^4



H.

56. The graph of a function f is shown. (The dashed lines indicate horizontal asymptotes.) Find each of the following for the given function g(x) =pf(x).³

(a) The domains of g and g⁰. (b) The critical numbers of g.

(c) The approximate value of g⁰(6).

(d) All vertical and horizontal asymptotes of g.

328ζHAPTER 4 Applications of Differentiation

(d) AII vertical and horizontal asymptotes of 9

55. g(x) = J.河三;

57. g(x) = I j(x) I

56. g(x) =

刃而

58. g(x) ⁼ l/f(x)

59. ln the theory of relativity, the mass of a particle is

m= n10

J l -

^V²^{/C 2}

where mo is the rest mass of the particle, m is the mass when the particle moves with speed v relative to the observer, and c is the speed of light. Sketch the graph of m as a function of v.

60. ln the theory of relativity, the energy of a particle is E =

J

^m0²^C4+

h

²c²/λ2

where mo is the rest mass of the particle，λis its wave length, and h is Planck's constant. Sketch the graph of E as a func- tion of 入.What does the graph say about the energy?

61. A model for the spread of a rumor is given by the equation

p(t)= l h

1 + ae-'I

where p(t) is the proportion of the population that knows the rumor at time t and a and k are positive constants.

(a) When will half the population have heard the rumor?

(b) When is the rate of spread of the rumor greatest?

(c) Sketch the graph of p.

62. A model for the concentration at time t of a drug i 吋 ectedinto the bloodstream is

C(t) = K(e 叫 -e 仙

where a, b, and K are positive constants and b > α. Sketch the graph of the concentration function. What does the graph tell us about how the concentration varies as time passes?

63. The fìgure shows a beam of length L embedded in concrete walls. If a constant load W is distributed evenly along its length, the beam takes the shape of the deflection curve

W WL 司 WL² 月

y= 一一一一-x崎 +一一一-x^J 一一一一-x^L

24EI 12EI 24EI

where E and I are positive constants. (E is Young's modu- lus of elasticity and I is the moment of inertia of a cross-

section of the beam.) Sketch the graph of the deflection curve.

64. Coulomb's Law states that the force of attraction between two charged particles is directly proportional to the product of the charges and inversely proportional to the square of the dis tance between them. The fìgure shows particles with charge 1 located at positions 0 and 2 on a coordinate line and a particle with charge -1 at a position x between them. lt follows from Coulomb's Law that the net force acting on the middle particle is

k k

F(x) = 一 + 一一一一一

x² (x - 2)2 O<x< 2

where k is a positive constant. Sketch the graph of the net force function. What does the graph say about the force?

65-68 Find an equation of the slant asymptote. Do not sketch the curve.

+一+

x-x 2

一一

J V F3

AU 4x³- 10x² 一 llx+ 1

66. y = ? 而

x- - .)x

2X3 - 5x²

+

3x -6x⁴

+

2X3

+

3

67. Y = 勻 68. Y = 冉、

x~ - x - L 乙 x~ - x

69-74 Use the guidelines of this section to sketch the curve. In guideline D, fìnd an equation of the slant asymptote.

x 2 1+ 5x - 2X2

69. y = 一一 70. Y = 用

x 一 I X - L

x³+ 4

71. Y = 一一τ一一

x-

一、‘'，'，

X一十

-x

一一

、、pd

、4勻，

73. Y = 1 +

tx

+ e-' 74. y= 1 -x+e^l+^r/3

75. Show that the curve y = x - tan -IX has two slant asymptotes: y = x + ^7f/2 and y = x - 7f /2. Use this fact to help sketch the curve.

76. Show that the curve y =

Jx訂立了 h as

two slant 的 mptotes:

y = x + 2 and y = - x - 2. Use this fact to help sketch the curve.

77. Show that the lines y = (b/α)x and y = 一 (b/α)xare slant asymptotes of the hyperbola (x2/ a 2) 一 (y2/b²) = 1.

Solution:

SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 403

G. ⁰⁰() =

(1 + ²)^{arctan }

 1

1 + ²



− ^{arctan }(2) (1 + ²)²

= ^{arctan }(1 − 2) (1 + ²)²

⁰⁰()  0for  ¹₂, so  is CU on

−∞¹2

 and  is CD on₁

2 ∞ . IP at

1

2 ^{arctan 12}

≈ (05 159)

H.

54. =  () = tan⁻¹

 − 1

 + 1



A.  = { |  6= −1} B. intercept = 1, intercept = (0) = tan⁻¹(−1) = −^4

→±∞tan⁻¹

 − 1

 + 1



= lim

→±∞tan⁻¹

1 − 1

1 + 1



= tan⁻¹1 =^₄, so  = ^₄ is a HA.

Also lim

→−1⁺tan⁻¹

 − 1

 + 1



= −

2 and lim

→−1⁻tan⁻¹

 − 1

 + 1



= 2.

E. ⁰() = 1

1 + [( − 1)( + 1)]²

( + 1) − ( − 1)

( + 1)² = 2

( + 1)²+ ( − 1)² = 1

²+ 1  0, so  is increasing on (−∞ −1) and (−1 ∞)  F. No extreme values

G. ⁰⁰() = −2

²+ 12

 0 ⇔   0, so  is CU on (−∞ −1) and (−1 0), and CD on (0 ∞). IP at

0 −^4



H.

55.() =

 ()

(a) The domain of  consists of all  such that () ≥ 0, so  has domain (−∞ 7]. ⁰() = 1 2

 ()· ⁰() = ⁰() 2

 (). Since ⁰(3)does not exist, ⁰(3)does not exist. (Note that (7) = 0, but 7 is an endpoint of the domain of .) The domain of ⁰is (−∞ 3) ∪ (3 7).

(b) ⁰() = 0 ⇒ ⁰() = 0on the domain of  ⇒  = 5 [there is a horizontal tangent line there]. From part (a), ⁰(3) does not exist. So the critical numbers of  are 3 and 5.

(c) From part (a), ⁰() = ⁰() 2

 (). ⁰(6) = ⁰(6) 2

 (6) −2 2√

3 = − 1

√3  −058.

(d) lim

→−∞() = lim

→−∞

 () =√

2, so  =√

2is a horizontal asymptote. No VA

56.() = ³

 ()

(a) Since the cuberoot function is defined for all reals, the domain of  equals the domain of , (−∞ ∞).

⁰() = 1 3

3

 ()2 · ⁰() = ⁰() 3

3

 ()2. Since ⁰(3)does not exist and (7) = 0, ⁰(3)and ⁰(7)do not exist.

The domain of ⁰is (−∞ 3) ∪ (3 7) ∪ (7 ∞).

2

(3)

(b) ⁰() = 0 ⇒ ⁰() = 0 ⇒  = 5 or  = 9 [there are horizontal tangent lines there]. From part (a), ⁰()does not exist at  = 3 and  = 7. So the critical numbers of  are 3, 5, 7, and 9.

(c) From part (a), ⁰() = ⁰() 3

3

 ()2. ⁰(6) = ⁰(6) 3

3

 (6)2  −2 3√3

3² = − 2

3^53  −032.

(d) lim

→−∞() = lim

→−∞

3

 () =√³

2and lim

→∞() = lim

→∞

3

 () =√³

−1 = −1, so  = √³

2and  = −1 are horizontal asymptotes. No VA

57.() = |()|

(a) Since the absolutevalue function is defined for all reals, the domain of  equals the domain of , (−∞ ∞). The domain of ⁰equals the domain of ⁰except for any values of  such that both () = 0 and ⁰() 6= 0. ⁰(3)does not exist,

 (7) = 0, and ⁰(7) 6= 0. Thus, the domain of ⁰is (−∞ 3) ∪ (3 7) ∪ (7 ∞).

(b) ⁰() = 0 ⇒ ⁰() = 0 ⇒  = 5 or  = 9 [there are horizontal tangent lines there]. From part (a), ⁰does not exist at  = 3 and  = 7. So the critical numbers of  are 3, 5, 7, and 9.

(c) Since  is positive near  = 6, () = |()| = () near 6, so ⁰(6) = ⁰(6)  −2.

(d) lim

→−∞() = lim

→−∞|()| = |2| = 2 and lim

→∞() = lim

→∞|()| = |−1| = 1, so  = 2 and  = 1 are horizontal asymptotes. No VA

58.() = 1 ()

(a) The domain of  consists of all  such that () 6= 0, so  has domain (−∞ 7) ∪ (7 ∞). () = 1

 () = [ ()]⁻¹ ⇒

⁰() = −1[()]⁻²· ⁰() = − ⁰()

[ ()]². The domain of ⁰will equal the domain of  except for any values of  such that

 () = 0. ⁰(3)does not exist, and (7) = 0. Thus, the domain of ⁰is (−∞ 3) ∪ (3 7) ∪ (7 ∞).

(b) ⁰() = 0 ⇒ ⁰() = 0 ⇒  = 5 or  = 9 [there are horizontal tangent lines there]. From part (a), ⁰does not exist at  = 3 and  = 7. So the critical numbers of  are 3, 5, 7, and 9.

(c) From part (a), ⁰() = − ⁰()

[ ()]². ⁰(6) = − ⁰(6) [ (6)]²  −

−2 3²



= 2 9.

(d) lim

→−∞() = lim

→−∞

1

 ()= 1

2and lim

→∞() = lim

→∞

1

 ()= 1

−1= −1, so  = 1

2and  = −1 are horizontal asymptotes. lim

→7⁻

1

 ()= 1 lim

→7⁻ () = ∞ and lim

→7⁺() = lim

→7⁺

1

 ()= 1 lim

→7⁺ () = −∞, so  = 7 is a vertical asymptote.

75. Show that the curve y = x − tan⁻¹x has two slant asymptotes: y = x + ^π₂ and y = x − ^π₂. Use this fact to help sketch the curve.

Solution:

70. =  () = 1 −  + ^1+3 A.  = R B. -intercept = (0) = 1 + , no -intercept [see part F]

C. No symmetry D. No VA or HA lim

→−∞[ () − (1 − )] = lim_

→−∞^1+3= 0, so  = 1 −  is a SA.

E. ⁰() = −1 +¹3^1+3 0 ⇔ ¹3^1+3 1 ⇔ ^1+3 3 ⇔ 1 +

3  ln 3 ⇔ 

3 ln 3 − 1 ⇔

  3(ln 3 − 1) ≈ 03, so  isincreasing on (3 ln 3 − 3 ∞) and decreasing on (−∞ 3 ln 3 − 3). F. Local and absolute minimum value

 (3 ln 3 − 3) = 1 − (3 ln 3 − 3) + ^{1+ln 3}⁻¹= 4 − 3 ln 3 + 3 = 7 − 3 ln 3 ≈ 37, no local maximum value G. ⁰⁰() = ¹₉^1+3 0for all , so  is CU on (−∞ ∞). No IP

H.

71. =  () =  − tan⁻¹, ⁰() = 1 − 1

1 + ² = 1 + ²− 1 1 + ² = ²

1 + ²,

⁰⁰() = (1 + ²)(2) − ²(2)

(1 + ²)² =2(1 + ²− ²)

(1 + ²)² = 2

(1 + ²)².

lim→∞

 () −

 −^2

= lim

→∞

_

2 − tan⁻¹

=^₂ −^2 = 0, so  =  −^2 is a SA.

Also, lim

→−∞

 () −

 +^₂

= lim

→−∞

−^2 − tan⁻¹

= −^2 −

−^2

= 0,

so  =  +^₂ is also a SA. ⁰() ≥ 0 for all , with equality ⇔  = 0, so  is increasing on R. ⁰⁰()has the same sign as , so  is CD on (−∞ 0) and CU on (0 ∞). (−) = −(), so  is an odd function; its graph is symmetric about the origin.  has no local extreme values. Its only IP is at (0 0).

72. =  () =√

²+ 4 =

( + 4). ( + 4) ≥ 0 ⇔  ≤ −4 or  ≥ 0, so  = (−∞ −4] ∪ [0 ∞).

y-intercept: (0) = 0; x-intercepts: () = 0 ⇒  = −4, 0.

√²+ 4 ∓ ( + 2) =

√²+ 4 ∓ ( + 2)

1 ·

√²+ 4 ± ( + 2)

√²+ 4 ± ( + 2)=(²+ 4) − (²+ 4 + 4)

√²+ 4 ± ( + 2)

= √ −4

²+ 4 ± ( + 2) so lim

→±∞[ () ∓ ( + 2)] = 0. Thus, the graph of  approaches the slant asymptote  =  + 2 as  → ∞ and it approaches the slant asymptote  = −( + 2) as  → −∞. ⁰() =  + 2

√²+ 4, so ⁰()  0for   −4 and ⁰()  0for   0;

that is,  is decreasing on (−∞ −4) and increasing on (0 ∞). There are no local extreme values. ⁰() = ( + 2)(²+ 4)^−12 ⇒

⁰⁰() = ( + 2) ·

−¹2

(²+ 4)^−32· (2 + 4) + (²+ 4)^−12

= (²+ 4)^−32

−( + 2)²+ (²+ 4)

= −4(²+ 4)^−32 0on  so  is CD on (−∞ −4) and (0 ∞). No IP

3