Written Homework 7 Solutions

Now set f ′ (x) = 0
1 − 3x 2 = 0
1 = 3x 2
1
3 = x2 , hence
1
x = ±
3
So we need to test the following intervals: (−∞, −
1
3
1 ), (− 3 ,
1
3 ) and (
1
3
, ∞).
By plugging in points from these intervals into f ′ (x), we can figure out the behavior
of f(x). These calculations have been omitted, but f ′ (x) > 0 for x ∈ (−
thus f(x) is increasing on this interval.
1
3 ,
1
3 ),
(b) Where is the graph of y = x − x3 rising most steeply?
Solution: We want to know when y = x − x3 is rising most steeply. We know it has
1
to be in the interval (− 3 ,
1
3 ) since this is the only time when f is increasing. One
way to figure out when f is rising most steeply is to look at the second derivative. If
f ′ (x) = 1 − 3x2 , then f ′′ = −6x. Now we set f ′′ (x) = 0 to find its critical points and
we see that f ′′ (x) = 0 when x = 0. Notice f ′ (0) = 1 > 0 so x = 0 is a maximum for
f ′ (x). So to answer our original question, f is rising most steeply when x = 0.
(c) At what points is the graph of y = x − x 3 horizontal?
Solution: We have already seen that the critical points of f are at x = ±
1
3
. Recall
that f obtains its critical points when f ′ (x) = 0. So f(x) is horizontal, or of zero
slope, or f ′ (x) = 0, when x = ±
1
(d) Make a sketch of the graph of y = x − x 3 that reflects all these results.
Solution:
3 .
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
2
1.6
1.2
0.8
0.4
-0.4
-0.8
-1.2
-1.6