Download Quiz #2-Joe

Quiz #2-Joe
March 7, 2014
Name:
Math 151
Section
Row:
This quiz is due by 11:30 on March 7. Do you work either at the bottom of this page, the back of this paper or on
another sheet of paper. Clearly label your work and your answers.
1. For y =
x2
find y ′′ (x).
3x + 1
y′ =
3x2 + 2x
(3x + 1)(2x) − x2 (3)
=
2
(3x + 1)
(3x + 1)2
y ′′ =
(3x + 1)2 (6x + 2) − (3x2 + 2x) ∗ 2(3x + 1)(3)
(3x + 1)[(3x + 1)(6x + 2) − 6(3x2 + 2x)]
=
4
(3x + 1)
(3x + 1)4
y ′′ =
2
18x2 + 12x + 2 − 18x2 − 12x
=
3
(3x + 1)
(3x + 1)3
2. For f (x) =
1
find f (3) (x).
(7x + 1)2
f (x) = (7x + 1)−2
f ′ (x) = −2(7x + 1)−3 ∗ 7
f ′′ (x) = 6(7x + 1)−4 ∗ 72
f (3) (x) = −24(7x + 1)−5 ∗ 73
or f (3) (x) =
−24 ∗ 73
(7x + 1)5
3. If x = 3t + 5 and y = t3 + 2t2 − 4. Compute
dy dx (11,12)
solve x = 3t + 5 = 11 for t to find the value of t that give the point. You get t = 2.
3t2 + 4t
dy
=
dx
3
dy dy 20
=
=
dx (11,12)
dx t=2
3
4. Find the values of t where the curve will have a horiozntal tangent line.
x = t2 − 25
y = (t2 + 4t)2
dx
dy
= 2t
= 2(t2 + 4t) ∗ (2t + 4).
dt
dt
you want dy/dt = 0 and dx/dt =
6 0.
dy
= 2(t2 + 4t) ∗ (2t + 4) = 0
dt
2t(t + 4) ∗ 2(t + 2) = 0
This gives t = 0, t = −4, and t = −2. since t = 0 makes dx/dt = 0 it is not one of the answers.
Answer: t = −4, and t = −2