This is the same problem that I had prepared for the March 09 lecture.
Problem 0.1. Sketch the graph of
f=
p
x2 − x4 .
The first thing to notice about this function is the restriction on the domain.
We need x2 − x4 ≥ 0, which implies that −1 ≤ x ≤ 1.
There is no need to talk about horizontal asymptotes for f (x) because the
domain is bounded. Furthermore, f (x) is continuous on [−1, 1] so there are no
vertical asymptotes either.
Let’s take a derivative:
1
(2x − 4x3 )(x2 − x4 )−1/2 .
2
This is undefined at x = 0 as well as at x = ±1 because of the zero in the
denominator. Let’s compute the left and right limits of f ′ (x) as x → 0. We
could use L’Hôpital’s rule, but it’s easier just to factor:
f ′ (x) =
x − 2x3
lim+ √
x→0
x2 − x4
x(1 − 2x2 )
= lim+ √
x→0
x 1 − x2
1 − 2x2
= lim+ √
x→0
1 − x2
=1
We can also compute the left limit:
x − 2x3
lim− √
x→0
x2 − x4
x(1 − 2x2 )
√
= lim
x→0− −x 1 − x2
1 − 2x2
√
= lim
x→0− − 1 − x2
= −1
Note that these limits are different. If we take the limit as x → 1, we see that the
numerator goes to −1 and the denominator goes to 0 from the right, implying
that the limit is −∞. Similarly, looking in the region where x → −1, we see
that the limit is ∞. So we expect the graph to be near vertical close to x = 1
and x = −1.
Finally, we notice that the derivative
is zero whenever 1 − 2x2 = 0, or
√
whenever x2 = 21 . This happens at ± 22 . The sign chart for f ′ (x) looks like
this:
1
√
√
√
√
region −1 < x < − 2 − 2 < x < 0 0 < x < 2
2<x<1
f ′ (x)
positive
negative
positive
negative
f (x)
increasing
decreasing
increasing
decreasing
The sign chart can be completed using the limit calculations from before
(or
√
2
by plugging in test points). So f (x) has local maxima when x = ± 2 and a
local minimum at x = 0.
Now, let’s look at the second derivative:
1
f ′′ (x) = (1 − 6x2 )(x2 − x4 )−1/2 − (x − 2x3 )(x2 − x4 )−3/2 (2x − 4x3 )
2
which simplifies to
f ′′ (x) = (1 − 6x2 )(x2 − x4 )−1/2 − (x − 2x3 )2 (x2 − x4 )−3/2
This is again undefined at −1, 0, and 1. Let’s factor out a copy of (x2 − x4 )−3/2 :
f ′′ (x) = (x2 − x4 )−3/2 ((x2 − x4 )(1 − 6x2 ) − (x − 2x3 )2 )
Notice that we can also factor an x2 out:
f ′′ (x) = x2 (x2 − x4 )−3/2 ((1 − x2 )(1 − 6x2 ) − (1 − 2x2 )2 )
Now we need to bite the bullet and multiply things out:
f ′′ (x) = x2 (x2 − x4 )−3/2 (1 − 7x2 + 6x4 − 1 + 4x2 − 4x4 )
this is
f ′′ (x) = x2 (x2 − x4 )−3/2 (−3x2 + 2x4 )
Notice that this is always negative which means that f (x) is concave down on
(−1, 0) and (0, 1). Furthermore, notice the behavior near x = 0: The numerator
has essentially 4 powers of x near zero and the denominator looks like (x2 )3/2 ,
which is x3 . So we expect the second derivative to approach 0 as x → 0. This is
indeed what happens. Therefore, the function f ′′ (x) should be almost straight
near x = 0.
Putting this information together gives us the graph.
Let’s do a simpler example now:
Problem 0.2. Graph the function f (x) = x2/3 .
This function is continuous everywhere and defined everywhere. Furthermore, f (x) is nonnegative for all values of x and equal to zero if and only if x = 0.
There are no horizontal asymptotes because limx→∞ f (x) = limx→−∞ f (x) =
∞.
Let’s look at f ′ (x) to determine where f (x0 is increasing and decreasing:
f ′ (x) =
2 −1/3
x
.
3
2
Notice that x−1/3 has the same sign as x, so f ′ (x) is negative for x < 0 and
positive for x > 0. We have that f ′ (x) is undefined when x = 0.
We should compute the limit of f ′ (x) as x → 0 from the left or from the
right. We have
2
lim f ′ (x) = lim
= −∞
−
−
x→0
x→0 3x1/3
because if x is a small negative number, we have that 3x21/3 is a very negative
number. Similarly, we get that limx→0+ f ′ (x) = ∞.
Notice that this is different from the situation from before: this time the
limit from one direction is ∞ and the limit from the other direction is −∞.
This corresponds to a cusp in the graph.
Anyway, let’s create the sign chart for f (x):
region
x<0
x>0
f ′ (x)
negative
positive
f (x)
decreasing increasing
Therefore, f (x) has a local minimum at x = 0.
Finally, we will compute f ′′ (x):
f ′′ (x) =
−2 −4/3
x
9
This function is negative everywhere it is defined: at all values of x except for
x = 0. Therefore, we have from our sign chart:
region
x<0
x>0
f ′′ (x) negative negative
f (x)
CCD
CCD
So f (x) is concave down on both regions x < 0 and x > 0.
This seems a bit odd: f (x) is concave down for all x 6= 0 but f (x) has a local
minimum at x = 0. This kind of bizarre behavior can happen at places where
f (x) has a cusp. Notice that this doesn’t contradict the second derivative test
because f ′′ (0) is not defined.
3