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

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

28 STEP 2. Determine Your Test Readiness

there is a total of eight points of inflection on

[−π, π]. An alternate solution is to enter

y^2 =

d^2

dx^2

(y 1 (x),x, 2). The graph ofy 2 crosses

thex-axis eight times, thus eight zeros on

[−π, π].

18. Sinceg(x)=

∫x

a

f(t)dt, g′(x)= f(x).

See the figure below.

The only graph that satisfies the behavior ofg

is choice (A).

[ [

ab 0

rel. max.

0+ −

incr. decr.

g′(x) = f(x)

g(x)

19. See the figure below.
    A change of concavity occurs atx=0 forq.
    Thus,qhas a point of inflection atx=0.
    None of the other functions has a point of
    inflection.

[ [

a 0 b

incr. decr.

+ −

Concave

upward

Concave

downward

q′

q′′

q

Chapter 9

20. Letzbe the diagonal of a square. Area of a
    squareA=
    z^2
    2
    dA
    dt

=

2 z

2

dz

dt

=z

dz

dt

.

Since

dA

dt

= 4

dz

dt

;4

dz

dt

=z

dz

dt

⇒z= 4.

Letsbe a side of the square. Since the

diagonalz=4,s^2 +s^2 =z^2 or 2s^2 =16. Thus,

s^2 =8ors= 2

√

2.

21. See the figure below.
    The graph ofgindicates that a relative
    maximum occurs atx=2;gis not
    differentiable atx=6, since there is acuspat
    x=6, andgdoes not have a point of
    inflection atx=−2, since there is no tangent
    line atx=−2. Thus, only statement I is true.

Chapter 10

22. y=

√

x− 1 =(x−1)^1 /^2 ;

dy

dx

=

1

2

(x−1)−^1 /^2

=

1

2(x−1)^1 /^2

dy

dx

∣∣

∣∣

x= 5

=

1

2(5−1)^1 /^2

=

1

2(4)^1 /^2

=

1

4

Atx=5, y=

√

x− 1 =

√

5 − 1

=2; (5, 2).

Slope of normal line=negative reciprocal of

(

1

4

)

=−4.

Equation of normal line:

y− 2 =−4(x−5)⇒y=−4(x−5)+2or

y=− 4 x+ 22.

23. y=cos(xy);
    dy
    dx
    =[−sin(xy)]

(

1 y+x

dy

dx

)