The document discusses local maxima and minima of functions. It defines a local maximum/minimum as a point where the function values in an interval around that point are less than/greater than the function value at that point. Critical points, where the derivative is zero, are important for determining local extrema. Two tests are described - the first derivative test examines the sign of the derivative on each side of the critical point, and the second derivative test uses the concavity at the critical point to determine if it is a maximum or minimum. Examples are provided to illustrate applying these tests.

1. TARUN GEHLOT (B.E, CIVIL, HONOURS)
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.

2. TARUN GEHLOT (B.E, CIVIL, HONOURS)
Example. Consider the function f(x) = x3
. 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) = |x2
-x|
Answer. We have
Clearly we have
Clearly we have

3. TARUN GEHLOT (B.E, CIVIL, HONOURS)
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 toc),
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) = |x2
-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)

4. TARUN GEHLOT (B.E, CIVIL, HONOURS)
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.

5. TARUN GEHLOT (B.E, CIVIL, HONOURS)
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 Second-Derivative Test.
Example. Find the local extrema of
f(x) = x5
- 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 5x4
- 5 = 0, or equivalently x4
- 1 =0. Since x4
- 1 = (x-
1)(x+1)(x2
+1), then the critical points are 1 and -1. Since f''(x) = 20 x3
, then
The second-derivative test implies that x=1 is a local minimum and x= -1 is a local
maximum.