# − ( ) = +

```MATH 147
Quiz Key #7
3/31/2016
(1) Consider the function f ( x ) = 13 x3 + 21 x2 − 6x + 2.
(a) Determine whether f is increasing or decreasing.
(b) Determine where f is concave up or down and the x-coordinates of all
inflection points.
Solution: (a) f 0 ( x ) = x2 + x − 6. Therefore,
f 0 ( x ) = ( x + 3) ( x − 2) = 0.
Therefore, the critical values are at x = −3 and x = 2. We now make a sign
chart as follows:
x+3
x−2
f 0 (x)
(−∞, −3) (−3, 2) (2, ∞)
−
+
+
−
−
+
+
−
+
Therefore, f is increasing on (−∞, −3) ∪ (2, ∞) and decreasing on (−3, 2) .
(b) f 00 ( x ) = 2x + 1. Hence, the only possible point of inflection is at x = − 21 .
We make a sign chart:
− 12 , ∞
−∞, − 21
f 00 ( x )
−
+
Therefore, f is concave up on − 21 , ∞ and concave down on −∞, − 21 , and
it has a point of inflection at x = − 12 .
1
2
(2) Consider the function f ( x ) = e− x .
(a) Determine whether f is increasing or decreasing.
(b) Determine where f is concave up or down and the x-coordinates of all
inflection points.
Solution: (a)
f 0 (x) =
2
2
d h 2i
d h − x2 i
= e− x &middot;
e
− x = −2xe− x .
dx
dx
To find the critical values of f , we need to set the derivative equal to zero.
However, since no exponential function is ever zero, f 0 ( x ) = 0 only when
x = 0. Now, we make a sign chart:
−2x
2
e− x
0
f (x)
(−∞, 0) (0, ∞)
+
−
+
+
+
−
Hence, f is increasing on (−∞, 0) and decreasing on (0, ∞) .
(b)
i
2
2
d h
d
d h − x2 i
f 00 ( x ) =
−2xe− x =
e
.
[−2x ] &middot; e− x − 2x &middot;
dx
dx
dx
2
2
2
∴ f 00 ( x ) = −2e− x − 2x (−2x ) e− x = −2e− x 1 − 2x2 .
Now to solve f 00 ( x ) = 0, we only need to find solutions of 1 − 2x2 = 0.
1 − 2x2 = 0
2x2 = 1.
1
x2 = .
2
1
x = &plusmn;√ .
2
Hence, the values at which inflection points can occur are these two values. We
now set up a sign chart as follows:
√1 , ∞
−∞, − √1
− √1 , √1
2
−2e− x
1 − 2x2
f 00 ( x )
2
2
−
−
+
2
2
−
−
+
−
−
+
∪ √1 , ∞ and concave down on
Therefore, f is concave up on −∞, − √1
2
2
− √1 , √1 , as well as inflection points at x = &plusmn; √1 .
2
2
2
2
```