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

MA 3972-MA-Book May 9, 2018 10:9

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

=lim

h→ 0

sin

(

π

2

)

[cosh−1]+cos

(

π

2

)

sinh

h

=limh→ 0 sin

(

π

2

)(

cosh− 1

h

)

−lim

h→ 0

cos

(

π

2

)(

sinh

h

)

=sin

(

π

2

)

hlim→ 0

(

cosh− 1

h

)

−cos

(

π

2

)

lim

h→ 0

(

sinh

h

)

=

[

sin

(

π

2

)]

0 +cos

(

π

2

)

(1)

=cos

(

π

2

)

= 0.

27. Using the chain rule, letu=(π−x).
    Then, f′(x)=2 cos(π−x)−sin(π−x)
    =2 cos(π−x) sin(π−x)
    f′(0)=2 cosπsinπ= 0.


28. Since the degree of the polynomial in the
    denominator is greater than the degree of
    the polynomial in the numerator, the limit
    is 0.


29. You can use the difference quotient
    f(a+h)−f(a)
    h
    to approximate f′(a).

Leth=1; f′(2)≈

f(3)−f(2)

3 − 2

≈

6. 5 − 4. 8

1

≈ 1. 7.

Leth=2; f′(2)≈

f(4)− f(2)

4 − 2

≈

8. 9 − 4. 8

2

≈ 2. 05.

Or, you can use the symmetric difference

quotient

f(a+h)−f(a−h)

2 h

to

approximate f′(a).

Leth=1; f′(2)≈

f(3)− f(1)

3 − 1

≈

6. 5 − 4

2

≈ 1. 25.

Leth=2; f′(2)≈

f(4)− f(0)

4 − 0

≈

8. 9 − 3. 9

4

≈ 1. 25.

Thus,f′(2)= 1 .7, 2.05, or 1.25

depending on your method.

30. (See Figure 7.12-1.) Checking the three
    conditions of continuity:

[–10, 10] by [–10, 10]

Figure 7.12-1

(1) f(3)= 3

(2) lim

x→ 3

x^2 − 9

x− 3

=lim

x→ 3

(

(x+3)(x−3)

(x−3)

)

=lim

x→ 3

(x+3)=(3)+ 3 = 6

(3) Since f(3)=lim

x→ 3

f(x), f(x)is

discontinuous atx=3.