Critical Points

So far we have found extreme points by looking at graphs. This is only an approximate procedure. We now present a method which uses the symbolic representation of functions.

The idea behind this procedure is to find a finite number of points at which an extremum could possibly occur. These are called critical points. We then examine each critical point individually to see if it is an extreme value. Finally, all local maxima and all local minima are compared to find the largest or smallest among them.

Our plan is to now list all types of critical points, i.e. all points at which a local extremum can occur.

Points Where the Derivative is Zero

At local extrema the
tangents are horizontal

The first key observation is that, if a function is differentiable at an extreme point, then its derivative must be zero.

Suppose that f is continuous on an interval J with endpoints a and b and that f is differentiable on the open interval (a,b) contained in J. If f has a local maximum or minimum at x = c with a < c < b, then f '(c) = 0.

Why is THAT true?

If f(x) is differentiable and if we are looking on an open interval the only critical points are points where f '(x) = 0.

Endpoints of Intervals

If the domain of f(x) includes endpoints of intervals, then these endpoints are often local extrema. It is thus necessary to add endpoints of intervals to the list of critical points.

If the domain of f(x) is an interval J with endpoints a and b, and if f(x) is differentiable on the open interval (a,b), then the only critical points are a, b, and points where f '(x) = 0.

Points Where f '(x) is Undefined

"Corner" maximum.

"Cusp" maximum.

The next type of critical point is that where f '(x) is undefined. In the context of local extrema, this can happen at a corner or at a "cusp" as shown at the right. (It can also happen that f '(x) is undefined due to worse behavior, but we will not encounter this very much.)

As illustrated by the graphs, corners or cusps can be local extreme points and are hence critical points.

Points Where f(x) is Discontinuous.

"Isolated" maximum.

A function f(x) is continuous at x = a if its behavior at x = a is predictable. If f(x) is NOT continuous at x = a, then its value could be anything - including a local maximum or minimum.

Points of discontinuity are potential local extrema and are therefore critical points.

Summary

At this juncture we have listed all possible types of critical points.

Given a function f(x) defined on an interval J and a point c in J in its domain, then

either f(x) is continuous at x = c or it is not. If f(x) is not continuous at x = c, then c is a critical point. If f(x) IS continuous at x = c, then

either c is an endpoint of J or it is not. If c is an endpoint of J, then c is a critical point. If c is NOT an endpoint of J, then

either f '(c) is defined or it is not. If f '(c) is not defined, then c is a critical point. If f '(c) IS defined, then

either f '(c) = 0 or f '(c) 0. If f '(c) = 0, then c is a critical point. If f '(c) 0, then f '(c) is not a critical point.

This gives us a procedure for finding all critical points of a function on an interval. Note a point at which f(x) is not defined is a point at which f(x) is not continuous, so even though such a point cannot be a local extrema, it is technically a critical point.

Example: Let us find all critical points of the function f(x) = x^2/3 - 2x on the interval [-1,1].

Solution: First, f(x) is continuous at every point of the interval [-1,1]. The endpoints are -1 and 1, so these are critical points. The derivative of f(x) is given by

f'(x) = (2/3)x^-1/3 -2.

Since x^-1/3 is not defined at x = 0, neither is f'(x) and 0 is a critical point. Finally we need to find where f'(x) = 0. Solving (2/3)x^-1/3 -2 = 0 yields that x^1/3 = 1/3, or x = 1/27.

Thus the complete set of critical points is {-1, 0, 1/27, 1}. Notice how we were able to determine the precise critical points by following an algebraic procedure and without the use of a graph. However, it is not a bad a idea to look at the graph to corroborate our calculations.

Equipment Check: Find all critical point of g(t) = t + 4/t on the real line.

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