We will continue our discussion of curve sketching. First, we will complete
the example from class.
I’d like to remind everyone that we showed that f ′ (x) is negative everywhere
it is defined, and that limx→0 f ′ (x) = −∞.
Furthermore, we computed:
f ′′ (x) =
(2/3x−5/3 )(20x4 + 17x2 − 1)
9(x2 − 1)3
So we are interested in determining when 20x4 + 17x2 − 1 = 0. To do this, make
the substitution u = x2 to get: 20u2 + 17u − 1 = 0. Then the solution is
√
−17 ± 369
u=
40
√
by the quadratic formula. Notice that only −17+40 369 is positive, which is important because u = x2 must be nonnegative.
Therefore, we get that f ′′ (x) = 0 when x is equal to
s
√
−17 + 369
±
.
40
As usual, the next stepr is to construct
a sign chart for thersecond derivative:
r
r
region
f ′′ (x)
f (x)
x < −1
√
−17 369
40
positive
CCU
−1 < x < −
negative
CCD
−
√
−17 369 < x < 0
40
negative
CCD
√
−17 369
40
positive
CCU
0 < x <
√
−17 369 < x < 1
40
negative
CCD
This tells us everything we need to know to complete the sketch. Notice that
the concavity changes at x = 0 so we have an inflection point. However, it’s not
a typical inflection point- there is a vertical tangent line at x = 0.
Let’s move on to the next example I had planned for Friday.
Problem 0.1. Sketch the graph of g(x) = x ln x.
First, we note the domain: the expression is defined for x > 0. Furthermore,
it is clear that limx→∞ g(x) = ∞. Let’s consider the first derivative:
g ′ (x) = ln x + x ·
1
x
which is equal to
ln x + 1.
This is equal to zero when ln x = −1, which happens when x = 1e . Note that
g ′ (x) < 0 for 0 < x < 1e and g ′ (x) > 0 for x > 1e . Now, we take a second
derivative:
1
g ′′ (x) =
x
which is positive on (0, ∞). Therefore, there are no inflection points.
The only question that remains is the behavior as x → 0. We need to
evaluate:
lim x ln x
x→0
1
x > 1
positive
CCU
This appears to come out to 0 × ∞ which doesn’t mean anything.
It would be nice if we could find a way to use L’Hôpital’s rule to evaluate
this limit.
1
The trick is to rewrite x as 1/x
.
So we get
ln x
lim
x→0 1/x
Which is, by L’Hôpital’s rule:
1/x
x→0 −1/x2
lim
which is
−x2
x→0 x
lim
which comes out to
lim −x = 0
x→0
so the limit is zero.
Problem 0.2. 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
2
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
√
1
2
2
whenever x = 2 . This happens at ± 2 . The sign chart for f ′ (x) looks like
this:
√
√
√
√
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 )
3
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.
4