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

MA 3972-MA-Book May 8, 2018 13:52

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

20. Sincef(x) is the antiderivative of

1

x

,

f(x)=

∫

1

x

d=ln|x|+C.

Givenf(1)=5; thus, ln (1)+C= 5

⇒ 0 +C=5orC=5.

Thus,f(x)=ln|x|+5 and

f(e)=ln (e)+ 5 = 1 + 5 =6.

11.7 Solutions to Cumulative Review Problems

21. (a) Att=4, speed is 5 which is the
    greatest on 0≤t≤10.
    (b) The particle is moving to the right
    when 6<t<10.


22. V=

4

3

πr^3 ;

dV

dt

=

(

4

3

)

(3)πr^2

dr

dt

= 4 πr^2

dr

dt

If

dV

dt

= 100

dr

dt

, then 100

dr

dt

= 4 πr^2

dr

dt

⇒ 100.

= 4 πr^2 orr=±

√

25

π

=±

5

√

π

.

Sincer≥0,r=

5

√

π

meters.

23. Letu=lnx;du=

1

x

dx.

Rewrite:

∫

u^3 du=

u^4

4

+C=

(lnx)^4

4

+C

=

ln^4 (x)

4

+C.

24. 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 11.7-1.)

Tangent

0 0.5 1 1.5 2

A B C

D

f E

y

x

Not to scale

Figure 11.7-1

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