MATH 147
Lab Key #8
4/5/2016
2
(1) Determine the intervals on which f ( x ) = xe−2x is increasing or decreasing.
Solution: First, we find the derivative using the product rule:
2
d
d h −2x2 i
.
e
[ x ] e−2x + x ·
dx
dx
i
2
2
d h
∴ f 0 ( x ) = e−2x + xe−2x ·
−2x2 .
dx
2
0
−2x2
2 −2x2
∴ f (x) = e
− 4x e
= e−2x 1 − 4x2 .
f 0 (x) =
2
Now, to find critical points, we find when f 0 ( x ) = 0. But e−2x > 0, for each
x. Therefore, it remains to solve
1 − 4x2 = 0.
4x2 = 1.
1
x2 = .
4
1
x=± .
2
This gives us our critical values. We now make a sign chart as follows:
1
,
∞
−∞, − 12
− 21 , 12
2
2
e−2x
1 − 4x2
f 0 (x)
+
−
−
+
+
+
+
−
−
Therefore, f is increasing on − 21 , 21 and decreasing on −∞, − 21 ∪ 12 , ∞ .
1
6
4
x
x
− 12
+ x − 1 is concave
(2) Determine ther intervals on which f ( x ) = 30
upward or downward and find the x-coordinates of all inflection points.
Solution: First, we need to find the second derivative:
f 0 (x) =
6x5
4x3
x5
x3
−
+1 =
−
+ 1.
30
12
5
3
3x2
5x4
−
= x4 − x2 .
5
3
Now, we find our possible inflection points. These occur when
∴ f 00 ( x ) =
x4 − x2 = 0.
x2 x2 − 1 = 0.
x2 ( x + 1) ( x − 1) = 0.
x = 0, ±1.
These give us our possible inflection point x-coordinates. We now make a sign
chart as follows:
x2
x+1
x−1
f 00 ( x )
(−∞, −1) (−1, 0) (0, 1) (1, ∞)
+
+
+
+
−
+
+
+
−
−
−
+
+
−
−
+
Hence, f is concave upward on (−∞, −1) ∪ (1, ∞) and concave downward on
(−1, 0) ∪ (0, 1) . Furthermore, f has inflection points when x = ±1 (but not
when x = 0, since concavity does not change there).
2
2
(3) Let f ( x ) = xx+1 , x 6= −1.
(a) Find all asymptotes of f .
(b) Find the intervals on which f is increasing or decreasing and the local
extrema.
(c) Find the intervals on which f is concave upward or downward and the
inflection points.
(d) Sketch the graph of y = f ( x ) and clearly label the asymptotes, local extrema, and inflection points (if any).
Solution: (a) First, to find the vertical asymptotes, we just note that f ( x ) is
in simplest form so it has a vertical asymptote where the denominator is zero.
But this is at x = −1. Hence, f has a vertical asymptote at x = −1. Using long
1
division, we can see that f ( x ) = x − 1 + x+
1 . Therefore, f has no horizontal
asymptotes (since its limits at ±∞ are ±∞), but it has an oblique asymptote at
y = x − 1.
(b) We use the quotient rule to find the derivative:
f 0 (x) =
2x ( x + 1) − x2
( x + 1)
2
=
x2 + 2x
( x + 1)
2
=
x ( x + 2)
( x + 1)2
.
Therefore, we have critical numbers at the solutions of ( x + 1)2 = 0 and x ( x + 2) =
0. Therefore, we have critical values at x = −2, −1, 0. Hence, we make the following sign chart:
x
x+2
( x + 1)2
f 0 (x)
(−∞, −2) (−2, −1) (−1, 0) (0, ∞)
−
−
−
+
−
+
+
+
+
+
+
+
+
−
−
+
Hence, f is increasing on (−∞, −2) ∪ (0, ∞) and decreasing on (−2, −1) ∪
(−1, 0) . Hence, we have a local maximum at x = −2 and a local minimum at
x = 0. Therefore, the local maximum is (−2, f (2)) = (−2, −4) and the local
minimum is (0, f (0)) = (0, 0) .
1
(c) To find the second derivative, we’ll use the formula f ( x ) = x − 1 + x+
1.
f 0 ( x ) = 1 − ( x + 1 ) −2 .
f 00 ( x ) = 2 ( x + 1)−3 =
2
.
( x + 1)3
Therefore, the only value that we have to consider here is x = −1. Since this
is not in the domain of the original function, there are no points of inflection.
However, we use it to make our sign chart as follows:
2
( x + 1)3
f 00 ( x )
(−∞, −1) (−1, ∞)
+
+
−
+
−
+
3
Hence, f is concave upward on (−1, ∞) and concave downward on (−∞, −1) .
(d)
4