5 Steps to a 5 AP Calculus BC 2019

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

Thus, limh→ 0

sin(π/ 2 +h)−sin(π/2)

h

=

limh→ 0

sin(π/2) cosh+cos(π/2) sinh−sin(π/2)

h

=limh→ 0

sin(π/2)[cosh−1]+cos(π/2) sinh

h

=limh→ 0 sin

(

π

2

)(

cosh− 1

h

)

−limh→ 0 cos

(

π

2

)(

sinh

h

)

=sin

(

π

2

)

limh→ 0

(

cosh− 1

h

)

−cos

(

π

2

)

hlim→ 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 6.11-1.) Checking the three
    conditions of continuity:

[−10,10] by [−10,10]

Figure 6.11-1

(1) f(3)= 3

(2) lim

x→ 3

x^2 − 9

x− 3

=limx→ 3

(

(x+3)(x−3)

(x−3)

)

=limx→ 3 (x+3)=(3)+ 3 = 6

(3) Sincef(3)/=limx→ 3 f(x), f(x)is

discontinuous atx=3.