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

MA 3972-MA-Book April 11, 2018 16:32

AP Calculus AB Practice Exam 1 371

ln| 2 +y|=

x^2

2

+c. Applying the

exponential function to both sides, you

haveeln|^2 +y|=e(x

22 +c)

or| 2 +y|=

(

ex

22 )

(ec).

Sinceecis an arbitrary constant, letk=ec

and rewrite:| 2 +y|=

(

ex

22 )

(ec)as

| 2 +y|=(k)

(

e

x 22 )

. Substituting (0, 1) into
the equation, you have| 2 + 1 |=(k)(e^0 ),
which yieldsk=3. Since the point (0, 1) is
on the curve ofy= f(x),| 2 +y|becomes
2 +y, and you have 2+y= 3 e(
x 22 )
,or
y= 3 e(
x 22 )
−2.
4. (A)

• 20 –1

I

II

(–2, 1)

(–1, 4)

(0, 6)

(2, 7)

(6, 8)

III IV

2 6

y

x

Area of a Trapezoid=

1

2

(b 1 +b 2 )h

AreaTrapI=

1

2

(1+4)[− 1 −(−2)]=

5

2

= 2. 5

AreaTrapII=

1

2

(4+6)[0−(−1)]= 5

AreaTrapIII=

1

2

(6+7)(2−0)= 13

AreaTrapIV=

1

2

(7+8)(4)= 30

Therefore,∫

6

− 2

f(x)dx≈ 2. 5 + 5 + 13 + 30 ≈ 50 .5.

Note that you cannot use the formula∫

b

a

f(x)dx≈

(

b−a

n

)[

y 0 + 2 y 1 +··· 2 yn− 1 +yn

]

,

since the four given intervals have

different lengths.

(B) The average value of an integrable

function on [a,b]is

faverage=

1

b−a

∫b

a

f(x)dx. In this

case, faverage=

1

6 −(−2)

∫ 6

− 2

f(x)dx≈

1

8

(50.5)≈ 6 .313.

(C) The average rate of change of a continuous

function on [a,b]is

f(b)−f(a)

b−a

. In this
case, the average rate of change offis
f(6)−f(−2)
6 −(−2)

=

8 − 1

8

=

7

8

,or0.875.

(D) Sincef′′(x)<0 for allx∈(−2, 6),

f′(x) is decreasing on the interval. Thus,

f′(0)>f′(2).

5. (A) Begin by summarizingf′(x)on a
   number line.

–4

f ́ 00

f decreasing Increasing

local min local max

decreasing

decr

––+

–2 12

The graph off has a local minimum at

x=−2 on the interval (−4, 2).

(B) Summarizef′′(x) on the number line.

–4

f ́ Increasing deceasing Increasing

Concave

up

f

f ́ ́

Concave

down

pt. of

inflection

pt. of

inflection

Concave

up

+–+

–1 1.5 2

Since the graph of fchanges concavity at

x=−1 andx= 1 .5, the graph off has a

point of inflection atx=−1 andx= 1 .5.