AP Calculus BC Practice Exam 2 429
Section II Part B
- (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) Think in terms of area above and below
the curve. The functionfincreases on
(0, 1) and decreases on (1, 3). Thus,f has
a relative maximum atx=1. Also,f
decreases on (−3,−1) and increases on
(−1, 0). Thus,f has a relative minimum
atx=−1. Another approach is as follows:
Note thatf′(x)=g(x) and the behavior
of the graph off(x)on[−3, 3] is
summarized below.
g
f
[ [
–3 –1 13
0
decr. decr.incr.
rel. min.
0
rel. max.
––– +++ ––– 0
(C) f′(x)=g(x) andf′′(x)=g′(x)
f′′(x) = g′(x)+ – +
x
–3 0 2 3
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); atx=1, the
equation of the tangent line tof(x)is
y=1.
