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

MA 3972-MA-Book April 11, 2018 12:11

Take a Diagnostic Exam 27

9. f(x)=

∣∣

2 xex

∣∣

=

{

2 xex ifx≥ 0

− 2 xex ifx< 0

Ifx≥0,f′(x)= 2 ex+ex( 2 x)= 2 ex+ 2 xex

xlim→ 0 +f′(x)=xlim→ 0 +(^2 ex+^2 xex)=^2 e^0 +^0 =^2

Chapter 7

10. ByL’Hôpital’sRule, limx→π
    ex−eπ
    xe−πe
    =xlim→π
    ex
    exe−^1
    =xlim→π

ex−^1

xe−^1

=

eπ−^1

πe−^1

.

11. y=(x+1)(x−3)^2 ;
    dy
    dx
    =(1)(x−3)^2 +2(x−3)(x+1)

=(x−3)^2 +2(x−3)(x+1)

dy

dx

∣∣

∣

∣x=− 1 =(−^1 −3)

(^2) +2(− 1 −3)(− 1 +1)

=(−4)^2 + 0 = 16

12. f′(x 1 )= lim
    Δx→ 0

f(x 1 +Δx)−f(x 1 )

Δx

Thus, lim

Δx→ 0

tan

(

π

4

+Δx

)

−tan

(

π

4

)

Δx

=

d

dx

(tanx)atx=

π

4

=sec^2

(

π

4

)

=(

√

2)^2 =2.

Chapter 8

13. See the graph shown at next column.


14. I. Since the graph ofgis decreasing and
    then increasing, it is not monotonic.
    II. Since the graph ofgis a smooth curve,
    g′is continuous.
    III. Since the graph ofgis concave upward,
    g′′>0.
    Thus, only statements II and III are true.


15. The graph indicates that (1)f(10)=0,
    (2)f′(10)<0, sincef is decreasing; and
    (3) f′′(10)>0, sincef is concave upward.
    Thus,f′(10)< f(10)< f′′(10), choice (C).

Graph for Question 13.

[

[

[

[

a

adb

0 e b

f′ +

f′′

f′

0 − +0

− +

f incr.

f

decr. incr.

decr. incr.

Concave downward

Concave

upward

aex

y

0 d b

A possible graph of f′

Based on the graph of f:

16. See the figure below.
    The graph off is concave upward forx<x 2.

x 2

f′ incr.

f′′

f

decr.

+ −

Concave

upward

Concave

downward

17. See the figure below.
    Entery^1 =sin(x^2 ). Using the [Inflection]
    function of your calculator, you obtain four
    points of inflection on [0, π]. The points of
    inflection occur atx= 0 .81, 1.81, 2.52, and
    3.07. Sincey 1 =sin (x^2 ) is an even function,