Optimization and Applications: Background You’ll Need 1

Giselle Miller

Optimization and Applications: Background You’ll Need 1

Use the first derivative to determine where a function is increasing or decreasing and identify local extrema

The First Derivative and Function Behavior

The first derivative tells you whether a function is increasing or decreasing. This comes directly from the Mean Value Theorem:

If [latex]f'(x) > 0[/latex] on an interval [latex]I[/latex], then [latex]f[/latex] is increasing on [latex]I[/latex]
If [latex]f'(x) < 0[/latex] on an interval [latex]I[/latex], then [latex]f[/latex] is decreasing on [latex]I[/latex]

A continuous function [latex]f[/latex] has a local maximum at [latex]c[/latex] when [latex]f[/latex] switches from increasing to decreasing at that point. Similarly, [latex]f[/latex] has a local minimum at [latex]c[/latex] when it switches from decreasing to increasing.

For a continuous function [latex]f[/latex] to switch from increasing to decreasing (or vice versa) at point [latex]c[/latex], the derivative must change sign. This can only happen when [latex]f'(c) = 0[/latex] or [latex]f'(c)[/latex] is undefined—these are your critical points.

A critical point of [latex]f[/latex] occurs at [latex]x = c[/latex] when [latex]f'(c) = 0[/latex] or [latex]f'(c)[/latex] is undefined. Critical points are candidates for local extrema, but not every critical point is a local extremum.

first derivative test

Let [latex]f[/latex] be continuous over an interval [latex]I[/latex] containing a critical point [latex]c[/latex]:

If [latex]f'[/latex] changes from positive to negative at [latex]c[/latex], then [latex]f(c)[/latex] is a local maximum
If [latex]f'[/latex] changes from negative to positive at [latex]c[/latex], then [latex]f(c)[/latex] is a local minimum
If [latex]f'[/latex] does not change sign at [latex]c[/latex], then [latex]f(c)[/latex] is neither a maximum nor minimum

How to: Apply the First Derivative Test

Find all critical points of [latex]f(x)[/latex] on the interval (where [latex]f'(x) = 0[/latex] or [latex]f'(x)[/latex] is undefined)
Divide the interval into subintervals using the critical points as boundaries
Test the sign of [latex]f'(x)[/latex] in each subinterval by choosing a test point
Determine extrema by checking if [latex]f'[/latex] changes sign at each critical point

Use the first derivative test to find the location of all local extrema for [latex]f(x)=x^3-3x^2-9x-1[/latex]. Use a graphing utility to confirm your results.

Step 1. The derivative is [latex]f^{\prime}(x)=3x^2-6x-9[/latex]. To find the critical points, we need to find where [latex]f^{\prime}(x)=0[/latex]. Factoring the polynomial, we conclude that the critical points must satisfy:

[latex]3(x^2-2x-3)=3(x-3)(x+1)=0[/latex]

Therefore, the critical points are [latex]x=3,-1[/latex]. Now divide the interval [latex](−\infty ,\infty)[/latex] into the smaller intervals [latex](−\infty ,-1), \, (-1,3)[/latex], and [latex](3,\infty)[/latex].

Step 2. Since [latex]f^{\prime}[/latex] is a continuous function, to determine the sign of [latex]f^{\prime}(x)[/latex] over each subinterval, it suffices to choose a point over each of the intervals [latex](−\infty ,-1), \, (-1,3)[/latex], and [latex](3,\infty)[/latex] and determine the sign of [latex]f^{\prime}[/latex] at each of these points.

For example, let’s choose [latex]x=-2, \, x=0[/latex], and [latex]x=4[/latex] as test points.

Interval	Test Point	Sign of [latex]f^{\prime}(x)=3(x-3)(x+1)[/latex] at Test Point	Conclusion
[latex](−\infty ,-1)[/latex]	[latex]x=-2[/latex]	[latex](+)(−)(−)=+[/latex]	[latex]f[/latex] is increasing.
[latex](-1,3)[/latex]	[latex]x=0[/latex]	[latex](+)(−)(+)=−[/latex]	[latex]f[/latex] is decreasing.
[latex](3,\infty)[/latex]	[latex]x=4[/latex]	[latex](+)(+)(+)=+[/latex]	[latex]f[/latex] is increasing.

Step 3. Since [latex]f^{\prime}[/latex] switches sign from positive to negative as [latex]x[/latex] increases through [latex]1, \, f[/latex] has a local maximum at [latex]x=-1[/latex].

Since [latex]f^{\prime}[/latex] switches sign from negative to positive as [latex]x[/latex] increases through [latex]3, \, f[/latex] has a local minimum at [latex]x=3[/latex].

These analytical results agree with the following graph.

The function f(x) = x3 – 3x2 – 9x – 1 is graphed. It has a maximum at x = −1 and a minimum at x = 3. The function is increasing before x = −1, decreasing until x = 3, and then increasing after that. — Figure 3. The function [latex]f[/latex] has a maximum at [latex]x=-1[/latex] and a minimum at [latex]x=3[/latex]

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “4.5 Derivatives and the Shape of a Graph” here (opens in new window).