Calculus

Applications of the Derivative

Applications of Derivative Logo

Local Extrema of Functions

Definition of Local Maximum and Local Minimum

Let a function y = f (x) be defined in a δ-neighborhood of a point x0, where δ > 0. The function f (x) is said to have a local (or relative) maximum at the point x0, if for all points xx0 belonging to the neighborhood (x0δ, x0 + δ) the following inequality holds:

If the strict inequality holds for all points xx0 in some neighborhood of x0:

then the point x0 is a strict local maximum point.

Similarly, we define a local (or relative) minimum of the function In this case, the following inequality is valid for all points of the -neighborhood of the point

Accordingly, a strict local minimum is described by the inequality

The concepts of local maximum and local minimum are united under the general term local extremum. The word "local" is often ommitted for brevity, so it is said simply about maxima and minima of functions.

Points of local extrema
Figure 1.

Figure schematically shows the different extrema points. The point is a strict local minimum point, since there exists a -neighborhood in which the following inequality holds:

Similarly, the point is a strict local maximum point. At this point, we have the inequality

(Of course, the number at each point may be different.)

The subsequent points are classified as follows:

Necessary Condition for an Extremum

We introduce some more concepts.

The points at which the derivative of the function is equal to zero are called the stationary points.

The points at which the derivative of the function is equal to zero or does not exist are called the critical points of the function. Consequently, the stationary points are a subset of the set of critical points.

A necessary condition for an extremum is formulated as follows:

If the point is an extremum point of the function then the derivative at this point either is zero or does not exist. In other words, the extrema of a function are contained among its critical points.

The proof of the necessary condition follows from Fermat's theorem.

Note that the necessary condition does not guarantee the existence of an extremum. A classic illustration here is the cubic function Despite the fact that the derivative of the function at the point is zero: this point is not an extremum.

Local extrema of differentiable functions exist when the sufficient conditions are satisfied. These conditions are based on the use of the first-, second-, or higher-order derivative. Respectively, sufficient conditions for local extrema are considered. Now we turn to their formulation and proof.

First Derivative Test

Let the function be differentiable in a neighborhood of the point except perhaps at the point itself, in which, however, the function is continuous. Then:

Proof.

We confine ourselves to the case of the minimum. Suppose that the derivative changes sign from minus to plus when passing through the point To the left from the point the following condition is satisfied:

By Lagrange's theorem, the difference of the values of the function at the points and is written as

where the point belongs to the interval in which the derivative is negative, i.e. Since to the left of the point then

Likewise, it is established that

(to the right of the point ).

Based on the definition, we conclude that is a strict minimum point of the function

Similarly, we can prove the first derivative test for a strict maximum.

Note that the first derivative test does not require the function to be differentiable at the point If the derivative at this point is infinite or does not exist (i.e. the point is critical, but not stationary), the first derivative test can still be used to investigate the local extrema of the function.

Second Derivative Test

Let the first derivative of a function at the point be equal to zero: that is is a stationary point of Suppose also that there exists the second derivative at this point. Then

Proof.

In the case of a strict minimum Then the first derivative is an increasing function at the point Consequently, there exists a number such that

Since (because is a stationary point), therefore the first derivative is negative in the -neighborhood to the left of the point , and is positive to the right, i.e. the derivative changes sign from minus to plus when passing through the point By the first derivative test, this means that is a strict minimum point.

The case of the maximum can be considered in a similar way.

The second derivative test is convenient to use when calculation of the first derivatives in the neighborhood of a stationary point is difficult. On the other hand, the second test may be used only for stationary points (where the first derivative is zero) − in contrast to the first derivative test, which is applicable to any critical points.

Third Derivative Test

Let the function have derivatives at the point up to the th order inclusively. Then if

the point for even is

For odd the extremum at does not exist.

It is clear that for , we obtain as a special case the second derivative test for local extrema considered above. To avoid such a transition, the third derivative test implies that

Proof.

Expand the function at the point in a Taylor series:

Since, by assumption, all of the first derivatives up to the th order are equal to zero, we obtain:

where the remainder term has a higher order of smallness than As a result, the sign of the difference in the -neighborhood of the point will be determined by the sign of the th term in the Taylor series:

or

If is an even number (), then

Consequently, in this case

If in the -neighborhood of the point , then the following inequality holds:

By definition, this means that is a strict minimum point of the function

Similarly, if in the -neighborhood of the point , we have the inequality

that corresponds to a strict maximum point.

If is an odd number the degree of will change sign when passing through the point Then it follows from the formula

that the difference also changes sign when passing through In this case, the extremum at the point does not exist.

See solved problems on Pages 2,3.

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