Today we will do one warm-up problem on second derivatives, and then we
will move on to a discussion of asymptotes.
√
x
Problem 0.1. Sketch the graph of f (x) = x+3
. Pay particular attention to
where f (x) is increasing and decreasing, and where f (x) is concave up and
concave down.
Clearly this function is only defined for x ≥ 0
Let’s take the derivative:
f ′ (x) =
(1/2x−1/2 )(x + 3) − x1/2
(x + 3)2
This is zero when the numerator is equal to zero. The numerator is
−1/2x1/2 + 3/2x−1/2
which is x−1/2 (−1/2x + 3/2). This is zero when x = 3. So x = 3 is the only
critical point of this function on (0, ∞).
Let’s make a sign chart for f ′ (x): notice that the denominator is always
negative so we need only check the numerator:
Region 0 < x < 3 x > 3
f ′ (x)
f (x)
Now let’s fill in the sign chart:
If we plug 1 into the numerator, we get the positive value 1. If we plug 4
into the denominator, we get −1/4, which is negative. So we fill in our sign
chart:
Region 0 < x < 3
x>3
f ′ (x)
Positive
Negative
f (x)
Increasing Decreasing
Now, we need to do the same thing for the second derivative of f . This is
(x + 3)2 (−(1/4)x−1/2 − 3/4x−3/2 ) − 2(x + 3)(−1/2x1/2 + 3/2x−1/2 )
(x + 3)4
We can cancel out a factor of (x + 3) to get
(x + 3)(−(1/4)x−1/2 − 3/4x−3/2 ) − 2(−1/2x1/2 + 3/2x−1/2 )
(x + 3)3
The denominator is positive for all x in the domain. Let’s find out where the
numerator is zero. Multiplying in, we get that the numerator is
(3/4)x1/2 − 9/2x−1/2 − 9/4x−3/2 .
this is
(3/4)x−3/2 (x2 − 6x − 3)
1
The roots of x2 − 6x − 3 are
6±
which come out to
√
36 + 12
2
√
3 ± 2 3.
Notice that f ′′ (x) is positive√if and only if x2 − 6x − 3 is positive.
Of these roots, only 3 + 2 3 is in the domain. It’s easy to see that at x = 1,
2
we have
Taking something larger than
√ that x − 6x − 3 is −8, which is negative.
2
−
6x
−
3
is 37, which is positive. So
x
3 + 2 3, such as 10, we can see that
√
there is a point of inflection√at 3 + 2 3. √
region 0 < x < 3 + 2 3 x > 3 + 2 3
f ′′ (x)
negative
positive
f (x)
concave down
concave up
Let’s make one huge table incorporating
we have:
√
√ all of the information
region 0 < x < 3 3 < x < 3 + 2 3 x > 3 + 2 3
f ′ (x)
positive
negative
negative
f ′′ (x)
negative
negative
positive
f (x)
inc, CCD
dec, CCD
dec, CCU
We can see the local maximum and inflection point in the picture in the
geogebra file.
Notice the behavior of this function as x gets large. Although the function
f (x) is decreasing, it seems to be decreasing toward a finite value.
How can we find the value? We need to consider the limit as x → ∞.
√
x
1
lim
= lim 1/2
x→∞ x + 3
x→∞ x
+ 3x−1/2
=0
The limit is seen to be zero, because x1/2 gets large as x approaches infinity,
and x−1/2 goes to zero as x approaches infinity.
Notice the behavior in the graph of f (x): as x gets large, the graph gets
closer to the line y = 0. We say that the line y = 0 is a horizontal asymptote of
f (x).
To find the horizontal asymptotes of a function, we need to take the limits
as x → ∞ and as x → −∞. These limits might be different; so a function can
have up to two horizontal asymptotes.
Definition 0.2. A function f (x) has a horizontal asymptote at y = a if
limx→∞ f (x) = a or limx→−∞ f (x) = −a.
This is one type of “infinite limit”- a limit as x approaches ∞ or −∞.
However, there is another type of “infinite limit” that we are interested in.
Consider the function
1
g(x) =
.
(x − 5)
2
This function is not defined at x = 5. What happens for values of x that are
close to 5? If x is a number that is slightly smaller than 5, then g(x) will be
a very negative number. In fact, g(x) can be made as negative as you want by
taking x close to 5.We say
lim g(x) = −∞.
x→5−
Similarly, if we plug in a value that is very slightly larger than 5, for x, g(x) is
a very large positive number. In fact g(x) can be made as large as you want by
taking x close enough to 5. We say
lim g(x) = ∞.
x→5+
Either one of these conditions, on its own, is sufficient for x = 5 to be a vertical
asymptote of g(x).
Definition 0.3. A function g(x) has a vertical asymptote at x = c if at least
one of the following four things happens:
• limx→c− g(x) = ∞
• limx→c− g(x) = −∞
• limx→c+ g(x) = ∞
• limx→c− g(x) = −∞
Note that we only need one of these things to happen. It may happen that
one of the left-hand and right-hand limits is finite, and the other one is infinite.
Notice that this definition does NOT say anything about whether g(c) is
defined. It is entirely possible for g(x) to be defined at x = c. In other words,
a function may be defined at an x-value at which it has a vertical asymptote.
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