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## Wednesday, 27 March 2013

### 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) = 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|

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 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)
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 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.