# 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
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
```