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

MA 3972-MA-Book May 7, 2018 9:52

Limits and Continuity 107

18. The graph indicates that (a)f(3)=4,
    (b) limx→ 3 +f(x)=0, (c) limx→ 3 −f(x)=0,
    (d) limx→ 3 f(x)=0, and (e) therefore,f(x)is
    not continuous atx=3 since
    f(3)=limx→ 3 f(x).


19. (See Figure 6.7-3.) Ifb=0, thenr=0, but
    rcannot be 0. Ifb=−3,−2, or−1,
    fwould have more than one root. Thus
    b=1. Choice (D).

(–2, 3)

1

• 2


• 22


• 1

(2, 4)

y

x

0

Figure 6.7-3

20. Substitutingx=0 would lead to 0/0.
    Substitute (1−cos^2 x) in place of sin^2 x
    and obtain

limx→ 0

1 −cosx

sin^2 x

=limx→ 0

1 −cosx

(1−cos^2 x)

=limx→ 0

1 −cosx

(1−cosx)(1+cosx)

=limx→ 0

1

(1+cosx)

=

1

1 + 1

=

1

2

.

Verify your result with a calculator. (See

Figure 6.7-4)

[–10, 10] by [–4, 4]

Figure 6.7-4

6.8 Solutions to Cumulative Review Problems

21. Rewrite 3x− 2 y=6iny=mx+bform,
    which isy=

3

2

x− 3 .The slope of this line

whose equation isy=

3

2

x−3ism=

3

2

.

Thus, the slope of a line perpendicular to

this line ism=−

2

3

. Since the
perpendicular line passes through the
point (2,−4), therefore, an equation of
the perpendicular line is

y−(−4)=−

2

3

(x−2), which is equivalent

toy+ 4 =−

2

3

(x−2).

22. The graph indicates that lim
    x→ 4 −
    f(x)=3,
    f(4)=1, and limx→ 4 f(x) does not exist.
    Therefore, only statements I and III are
    true.