MA 3972-MA-Book April 11, 2018 16:20
AP Calculus AB Practice Exam 2 395
- The correct answer is (A).
0
f decr.
f′
incr.
rel. min.
Onlyfhas a relative minimum on (a,b).
- The correct answer is (D).
3
09
x = 9
x = y^2
y
x
Volume=π
∫ 3
− 3
(
92 −
(
y^2
) 2 )
dy
=
1944 π
5
.
- The correct answer is (D).
y=ex;
dy
dx
=ex
y=x^2 + 5 x;
dy
dx
= 2 x + 5
If the graphs have parallel tangents at a point,
then the slopes of the tangents are equal.
Entery 1 =exandy 2 = 2 x +5. Using the
[Intersection] function on your calculator, you
obtainx=− 2 .45 andx= 2 .25.
[–4, 3] by [–5, 12]
- The correct answer is (D).
Since
dy
dx
=ky⇒y=y 0 ekt
3
4
y 0 =y 0 ek(^6 )⇒
3
4
=e^6 k
⇒ln
(
3
4
)
=ln
(
e^6 k
)
⇒ln
3
4
= 6 kork=
ln
(
3
4
)
6
=− 0. 048.
- The correct answer is (B).
h′(x)=(x−5)^3
[ [
45
8
x
h(x) decr.
h′(x)
rel. min.
incr.
Thus,hhas a relative minimum atx=5.
