Critical Points
We will discuss the occurrence of local maxima and local minima of
a function. In fact, these points are crucial to many questions
related to optimization problems. We will discuss these problems
in later pages.
Definition. A function f(x) is said to have a local
maximum at c iff there exists an interval I around c such
that
Analogously, f(x) is said to have a local minimum at c iff
there exists an interval I around c such that
A local extremum is a local maximum or a local minimum.
Using the definition of the derivative, we can easily show that:
If f(x) has a local extremum at c, then either
These points are called critical points.

Example. Consider the function
f(x) = x^{3}. Then f'(0) =
0 but 0 is not a local extremum. Indeed, if x < 0, then
f(x)
< f(0) and if x > 0, then
f(x) > f(0).
Therefore the conditions
do not imply in general that c is a local extremum. So a
local extremum must occur at a critical point, but the converse
may not be true.
Example. Let us find the critical points of
f(x) = x^{2}x
Answer. We have
Clearly we have
Clearly we have
Also one may easily show that f'(0) and f'(1) do not exist.
Therefore the critical points are
Let c be a critical point for f(x). Assume that there exists
an interval I around c, that is c is an interior point of
I, such that f(x) is increasing to the left of c and
decreasing to the right, then c is a local maximum. This
implies that if
for
(x close to c),
and
for
(x close to c), then c is
a local maximum. Note that similarly if
for
(x close to c), and
for (x close to c), then c is a local minimum.
So we have the following result:
First Derivative Test. If c is a critical point for
f(x), such that f '(x) changes its sign as x crosses from
the left to the right of c, then c is a local extremum.

Example. Find the local extrema of
f(x) = x^{2}x
Answer. Since the local extrema are critical points, then
from the above discussion, the local extrema, if they exist, are
among the points
Recall that
 (1)
 For x = 1/2, we have
So the critical point
is a local maximum.
 (2)
 For x = 0, we have
So the critical point 0 is a local minimum.
 (3)
 For x = 1, we have
So the critical point 1 is a local minimum.
Let c be a critical point for f(x) such that f'(c) =0.
 (i)
 If
f''(c) > 0, then f'(x) is increasing in an interval around c. Since f'(c) =0, then f'(x) must be negative to the left of c and positive to the right of c. Therefore, c is a local minimum.
 (ii)
 If
f''(c) < 0, then f'(x) is decreasing in an interval around c. Since f'(c) =0, then f'(x) must be positive to the left of c and negative to the right of c. Therefore, c is a local maximum.
This test is known as the SecondDerivative Test.

Example. Find the local extrema of
f(x) = x^{5}  5 x.
Answer. First let us find the critical points. Since
f(x) is a polynomial function, then f(x) is continuous and
differentiable everywhere. So the critical points are the roots
of the equation f'(x) = 0, that is
5x^{4}  5 = 0, or
equivalently
x^{4}  1 =0. Since
x^{4}  1 = (x1)(x+1)(x^{2}+1),
then the critical points are 1 and 1. Since
f''(x) = 20
x^{3}, then
The secondderivative test implies that x=1 is a local minimum
and x= 1 is a local maximum.
Exercise 1. Find the local extrema of
Answer.
Exercise 2. Find the local extrema of
Answer.
[Back]
[Next]
[Trigonometry]
[Calculus]
[Geometry]
[Algebra]
[Differential Equations]
[Complex Variables]
[Matrix Algebra]
S.O.S MATHematics home page
Do you need more help? Please post your question on our
S.O.S. Mathematics CyberBoard.
Mohamed A. Khamsi
Helmut Knaust
Copyright © 19992018 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC.  P.O. Box 12395  El Paso TX 79913  USA
users online during the last hour