5 Steps to a 5 AP Calculus BC 2019

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

28. Letu=lnx;du=

1

x

dx.

Rewrite:

∫

u^3 du=

u^4

4

+C=

(lnx)^4

4

+C

=

ln^4 (x)

4

+C.

29. Label given points as A, B, C, D, and E.
    Sincef′′(x)> 0 ⇒ f is concave upward
    for allxin the interval [0, 2].
    Thus,mBC<f′(x)<mCD
    mBC= 1 .5 andmCD= 2 .5.
    Therefore, 1. 5 <f′(1)< 2 .5, choice (c).
    (See Figure 10.8-1.)

Tangent

0 0.5 1 1.5 2

A

B

C

D

f E

y

x

Not to Scale

Figure 10.8-1

30. (a) f′′is decreasing on [1, 6)⇒ f′′′<
    0 ⇒ f′is concave downward on
    [1, 6) and f′′is increasing on (6, 8]
    ⇒ f′is concave upward on (6, 8].
    Thus, atx=6,f′has a change of
    concavity. Since f′′exists atx= 6
    (which implies there is a tangent to the
    curve of f′atx=6),f′has a point of
    inflection atx=6.

(b) f′′>0 on [1, 4]⇒f′is increasing

and f′′<0 on (4, 8]⇒ f′is

decreasing. Thus atx=4,f′has a

relative maximum atx=4. There is no

relative minimum.

(c) f′′is increasing on [6, 8]⇒ f′> 0

⇒f′is concave upward on [6, 8].

31. xlim→− 2
    x^2 −x− 6
    x^2 − 4

→

0

0

=xlim→− 2

2 x− 1

2 x

=

5

4

32. Att=2,x=4 sin(2π)=0, and
    y= 22 − 3 · 2 + 1 =−1, so the position of
    the object att=2 is (0,−1).


33. To find the slope of the tangent line to the
    curver=3 cosθwhenθ=
    π
    4
    , begin with

x=rcosθandy=rsinθ, and find

dx

dθ

and

dy

dθ

.x=3 cos^2 θso

dx

dθ

=−6 cosθsinθ.

y=3 cosθsinθso

dy

dθ

=3 cos^2 θ−3 sin^2 θ.

Then the slope of the tangent line is

dy

dx

=

dy

dθ

·

dθ

dx

=

3 cos^2 θ−3 sin^2 θ

−6 cosθsinθ

=

cos 2θ

−sin 2θ

.

Evaluate atθ=

π

4

to get

cos

(

π/ 2

)

−sin

(

π/ 2

)=^0

− 1

=0. The slope of the

tangent line is zero, indicating that the

tangent is horizontal.