MA 3972-MA-Book April 11, 2018 16:20
AP Calculus AB Practice Exam 2 399
f′(x)=− 12 xe−^2 x^2 =
− 12 x
e^2 x^2
f(0)=3, since fhas only one critical
point (atx=0), thus atx=0,fhas an
absolute maximum. The absolute
maximum value is 3.
(E) f(x)=ae−bx
2
, a>0,b> 0
f′(x)=ae−bx^2 (− 2 bx)=− 2 abxe−bx^2
Setting
f′(x)=0,− 2 abxe−bx^2 = 0 ⇒x= 0
f′(x)=
− 2 abx
ebx^2
.
f′(x)>0ifx<0 andf′(x)<0if
x>0. Thus, f has a relative maximum at
x=0, and since it is the only critical point,
fhas an absolute maximum atx=0. Since
f(0)=a, the absolute maximum for fisa.
- (A) f(−3)=
∫− 3
0
g(t)dt=−
∫ 0
− 3
g(t)dt
=−
∫− 1
− 3
g(t)dt−
∫ 0
− 1
g(t)dt
=−
(
−
1
2
(2)(2)
)
−
(
1
2
(1)(2)
)
= 2 − 1 = 1
f(3)=
∫ 3
0
g(t)dt
=
∫ 1
0
g(t)dt+
∫ 3
1
g(t)dt
=
1
2
(1)(2)+
(
−
1
2
(1)(2)
)
= 1 − 1 = 0
(B) Note thatf′(x)=g(x), andg(x)<0on
(−3,−1) and (1, 3) and thatg(x)>0on
(−1, 1). The functionf increases on
(−1, 1) and decreases on (1, 3). Thusf
has a relative maximum atx=1. Also,f
decreases on (−3,−1) and increases on
(−1, 1). Thus,f has a relative minimum
atx=−1.
(C) f′(x)=g(x) andf′′(x)=g′(x)
f′′(x) = g′(x)+ – +
x
g(x) incr. decr. incr.
f Concave
upward
Concave
upward
Concave
downward
Change of
concavity
Change of
concavity
The functionfhas a change of concavity
atx=0 andx=2.
(D) f(1)=
∫ 1
0
g(t)dt=
1
2
(1)(2)= 1
f′(1)=g(1)= 0
Thus,m=0, point (1, 1); the equation of
the tangent line to f(x)atx=1isy=1.
- (A)
dy
dx
=
y
2 x^2
; (2,1)
dy
dx
∣∣
∣
∣x=2,y= 1 =
1
2(2)^2
=
1
8
Equation of tangent:
y− 1 =
1
8
(x−2) or
y=
1
8
(x−2)+ 1.
(B) f(2.5)≈
1
8
(2. 5 −2)+ 1 = 1. 0625
≈ 1. 063
(C)
dy
dx
=
y
2 x^2
dy
y
=
dx
2 x^2
and
∫
dy
y
=
∫
dx
2 x^2
