Let’s suppose that we want to sketch the graph of the following function
√
3
x
f (x) = 2
x −1
. This looks kind of like the function from last time. Notice that this function is
undefined at x = ±1. A quick check reveals that plugging ±1 into the numerator
gives a nonzero value, so f (x) has a vertical asymptote at x = 1 and at x = −1.
Let’s take the derivative: using the power rule and the chain rule, we get:
1/3(x2 − 1)x−2/3 − 2x4/3
(x2 − 1)2
4/3
1/3x − 1/3x−2/3 − 2x4/3
=
(x2 − 1)2
4/3
−5/3x − 1/3x−2/3
=
(x2 − 1)2
4/3
−5x − x−2/3
=
3(x2 − 1)2
f ′ (x) =
Notice that this function is equal to zero if and only if −x4/3 = x−2/3 which is
impossible because x2/3 and x4/3 are always positive.
In particular, the numerator of this function is negative and the denominator
is positive, so f (x) is decreasing everywhere it is defined.
The other big thing to notice is that x−2/3 is undefined at x = 0. Taking
both the left and right limits as x → 0 gives that x−2/3 goes to ∞ from both
directions. Therefore, we have that f ′ (x) approaches −∞ as x → 0. This
means that we have a vertical tangent line: the function is going straight down
at x = 0.
Let’s take the second derivative now:
(x2 − 1)2 (−20/3x1/3 + 2/3x−5/3 ) − 4x(x2 − 1)(−5x4/3 − x−2/3 )
3(x2 − 1)4
(x2 − 1)(−20/3x1/3 + 2/3x−5/3 ) − 4x(−5x4/3 − x−2/3 )
=
3(x2 − 1)3
(−20/3x7/3 + 22/3x1/3 − 2/3x−5/3 ) + 20x7/3 + 4x1/3
=
3(x2 − 1)3
(40/3x7/3 + 34/3x1/3 − 2/3x−5/3
=
9(x2 − 1)3
f ′ (x) =
Let’s factor 2/3x−5/3 out of the numerator to get:
(2/3x−5/3 )(20x4 + 17x2 − 1)
9(x2 − 1)3
1
Note that this is, of course, undefined when x = ±1 as well as when x = 0. The
second derivative is equal to zero when 20x4 + 17x2 − 1 = 0. This expression is
quadratic in x2 . So we can solve for the roots:
√
−17 ± 369
x2 =
40
√
x2 must be positive, so we only have the −9+4 89 root. Taking the square roots
gives:
s
√
−17 369
±
40
as the possible inflection points of f .
Notice that these roots are
in the interval
(−1, 1). So we haverour sign chart:
r
r
r
region
f ′′ (x)
f (x)
x < −1
√
−17 369
40
negative
CCU
−1 < x < −
positive
CCD
−
√
−17 369 < x < 0
40
negative
CCD
√
−17 369
40
positive
CCU
0 < x<
√
−17 369 < x < 1
40
positive
CCD
Finally, we notice that
x1/3
x→∞ x2 + 1
1
= lim 5/3
x→∞ x
+ x−1/3
=0
lim f (x) = lim
x→∞
and the same thing is true as x → −∞.
So we put our sketch together to get something similar to the geogebra file.
Let’s do another (hopefully easier!) curve sketching problem:
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
This appears to come out to 0 × ∞ which doesn’t mean anything.
2
x > 1
negative
CCU
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.
3