5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 15:57

268 STEP 4. Review the Knowledge You Need to Score High

12.8 Solutions to Cumulative Review Problems

21. Asx→−∞, x=−

√

x^2.

lim

x→−∞

√

x^2 − 4

3 x− 9

=lim

x→−∞

√

x^2 − 4 /−

√

x^2

(3x−9)/x

=lim

x→−∞

−

√

(x^2 −4)/x^2

3 −(9/x)

=lim

x→−∞

−

√

1 −(4/x)^2

3 − 9 /x

=

−

√

1 − 0

3 − 0

=−

1

3

22. y=ln

∣∣

x^2 − 4

∣∣

,

dy

dx

=

1

(x^2 −4)

(2x)

dy

dx

∣

∣∣

∣x= 3 =

2(3)

(3^2 −4)

=

6

5

23. (a) (See Figure 12.8-1.)

x

f′

f decr. incr. decr. incr. decr.

• 68 –^5 –^137

rel.

min.

rel.

max.

rel.

min.

rel.

max.

• 
  
  
  
  • – + –
  
  
  
  

Figure 12.8-1

The function fhas a relative

minimum atx=−5 andx=3, andf

has a relative maximum atx=−1 and

x=7.

(b) (See Figure 12.8-2.)

x

f′

f′′

f

incr.

• 6
  
  
  • – + –
    concave
    downward
  
  
  
  

concave

upward

concave

downward

– (^3158)
decr. incr. decr.
concave
upward
Figure 12.8-2
The functionfis concave upward on
intervals (−6,−3) and (1, 5).
(c) A change of concavity occurs at
x=−3,x=1, andx=5.

24. (Calculator) Differentiate both sides of
    9 x^2 + 4 y^2 − 18 x+ 16 y=11.

18 x+ 8 y

dy

dx

− 18 + 16

dy

dx

= 0

8 y

dy

dx

+ 16

dy

dx

=− 18 x+ 18

dy

dx

(8y+16)=− 18 x+ 18

dy

dx

=

− 18 x+ 18

8 y+ 16

Horizontal tangent⇒

dy

dx

=0.

Set

dy

dx

= 0 ⇒− 18 x+ 18 =0orx= 1.

Atx=1, 9+ 4 y^2 − 18 + 16 y= 11

⇒ 4 y^2 + 16 y− 20 = 0.

Using a calculator, enter [Solve]

(4y∧ 2 + 16 y− 20 =0,y); obtainingy=− 5

ory=1.

Thus, at each of the points at (1, 1) and

(1,−5), the graph has a horizontal tangent.

Vertical tangent⇒

dy

dx

is undefined.

Set 8y+ 16 = 0 ⇒y=−2.

Aty=−2, 9x^2 + 16 − 18 x− 32 = 11

⇒ 9 x^2 − 18 x− 27 = 0.

Enter [Solve](9x^2 − 18 x− 27 =0,x) and

obtainx=3orx=−1.

Thus, at each of the points (3,−2) and

(−1,−2), the graph has a vertical tangent.

(See Figure 12.8-3.)