Today, we will discuss a method for classifying critical points. It’s important
to keep in mind that our methods will only work if f is continuous at, and
differentiable near, the critical point in question.
The key fact about derivatives that you need to know for this section is this:
f ′ (x) can only change sign at critical points. Therefore, between critical points,
we know that f ′ (x) will always have the same sign. This is incredibly useful
information for classifying critical points. Suppose first that f is differentiable
on some interval (a, b). Let’s consider a critical point c in (a, b) with f ′ (c) = 0.
How can we determine whether f (x) has a local maximum or minimum at a c?
Well, in order to find the zeros of f ′ , we typically have access to f ′ . Suppose
that f (x) has a minimum at x = c. If this is the case, then we have that
f (c) ≤ f (x) for all x close to c. This means that for x < c; x close to c, we have
that f (x) > f (c). Thus, we would expect that f (x) is decreasing to the left of
c. Similarly, for x > c; x close to c, we have that f (x) > f (c), so the function
f (x) should be increasing to the right of c. See the geogebra file for a picture.
So what this means is that we expect that, to the left of a local minimum,
the function f (x) should be decreasing, and to the right of a critical point, the
function f (x) should be increasing. Therefore, we expect that f ′ (x) is negative
for x to the left of a local minimum and f ′ (x) is positive for x to the right of a
local minimum.
By the same reasoning, f ′ (x) should be positive to the left of a local maximum, and negative to the right of a local maximum.
As we will see later, it’s important to check both sides of a critical point.
It’s possible to have critical points c for which f ′ (x) has the same sign both
immediately to the left and immediately to the right of c.
Example 0.1. Find and classify the critical points of the function f (x) =
1 3
3 x − 4x. (When we say “classify,” we’re asking you to determine if the critical
point is a local maximum, local minimum, or neither).
To solve this problem, we need to differentiate f (x) to find the critical points.
The derivative is
f ′ (x) = x2 − 4
which is zero when x = −2 and x = 2. But what kinds of critical points are
these? To do this, we plug in “test points” to the left and right of our critical
points:
f ′ (−3) = (−3)2 − 4 = 5
f ′ (0) = (0)2 − 4 = −4
f ′ (3) = e2 − 4 = 5
So f (x) is increasing for x < −2, decreasing for −2 < x < 2, and increasing
for x > 2. (It helps to draw a little table to keep this information straight).
Therefore, f (x) has a local maximum at x = −2 and a local minimum at x = 2.
1
What about this example?
f (x) = (x − 1)3 (x − 2).
We can differentiate this using the product rule:
f ′ (x) = 3(x − 1)2 (x − 2) + (x − 1)3
now, we factor out (x − 1)2 :
f ′ (x) = (x − 1)2 (3(x − 2) + (x − 1))
And we get
f ′ (x) = (x − 1)2 (4x − 7)
So f ′ (x) = 0 when x = 1 and when x = 47 . Let’s plug stuff into the derivative:
f ′ (0) = (−1)2 (0 − 7) = −7
2
−1
1
3
(6 − 7) =
<0
f ′( ) =
2
2
4
f ′ (2) = (1)2 (8 − 1) = 1
So we see that 1 is neither a local minimum nor a local maximum, and 47 is a
local minimum.
Let’s look to see what this function is doing at x = 1. What you can see
from the picture is that f (x) is flat at x = 1, but decreasing on both sides. We
call x = 1 an inflection point (or a point of inflection) of f (x).
You need to check on both sides of a critical point in order to classify it.
Never forget that you can have critical points that are neither maxima nor
minima.
Let’s do one more for good measure:
Find and classify the critical points of
f (x) =
x2 − 1
.
x+3
We can differentiate this using the quotient rule:
f ′ (x) =
(x + 3)(2x) − (x2 − 1)
(x + 3)2
This can be rewritten as
f ′ (x) =
x2 + 6x + 1
(x + 3)2
Therefore f is differentiable everywhere except for x = −3, where it has a
vertical asymptote (we will discuss asymptotes in more detail after the midterm).
So we only need to check to see when the numerator is zero:
x2 + 6x + 1 = 0
2
We can solve for x using the quadratic formula:
√
−6 ± 32
x=
2
which is equal to
√
8
√
We know 8 is between 2 and 3. Let’s quickly test f ′ at places other than 2 or
3.
Notice that the denominator is always positive, so it’s only necessary to
check the sign of the numerator: If we plug in −6, we get that the numerator is
36 − 36 + 1 = 1, which is positive. If we plug in −4, which is between the critical
points, then we see that the numerator is 16 − 24 + 1, which is −7. If we plug
in −2, which is between the critical points but to the right of the asymptote,
we get −7, which is negative. If we plug in x = 0, which is to the right of both
critical points, we get 1, which is positive. A graph of this function is available
in the geogebra file.
x = −3 ±
3