The extreme value theorem states that a continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. Let’s look at Figure 2 again.
Figure 2. Graphs (a), (b), and (c) show several possibilities for absolute extrema for functions with a domain of [latex](−\infty ,\infty )[/latex]. Graphs (d), (e), and (f) show several possibilities for absolute extrema for functions with a domain that is a bounded interval.
One or both of these absolute extrema could occur at an endpoint. If an absolute extremum does not occur at an endpoint, however, it must occur at an interior point, in which case the absolute extremum is a local extremum. Therefore, by Fermat’s Theorem, the point [latex]c[/latex] at which the local extremum occurs must be a critical point. We summarize this result in the following theorem.
location of absolute extrema
Let [latex]f[/latex] be a continuous function over a closed, bounded interval [latex]I[/latex]. The absolute maximum of [latex]f[/latex] over [latex]I[/latex] and the absolute minimum of [latex]f[/latex] over [latex]I[/latex] must occur at endpoints of [latex]I[/latex] or at critical points of [latex]f[/latex] in [latex]I[/latex].
With this idea in mind, let’s examine a procedure for locating absolute extrema.
How to: Locate Absolute Extrema over a Closed Interval
Consider a continuous function [latex]f[/latex] defined over the closed interval [latex][a,b][/latex].
Evaluate the Function at Endpoints: Calculate [latex]f(a)[/latex] and [latex]f(b)[/latex], where [latex]f[/latex] is defined on the closed interval [latex][a,b][/latex].
Identify Critical Points: Find all critical points of [latex]f[/latex] within the interval [latex][a,b][/latex] and evaluate [latex]f[/latex] at these points.
Determine Extrema: Compare the values from steps 1 and 2. The largest value is the absolute maximum, and the smallest is the absolute minimum of [latex]f[/latex] on [latex][a,b][/latex].
Now let’s look at how to use this strategy to find the absolute maximum and absolute minimum values for continuous functions.
For each of the following functions, find the absolute maximum and absolute minimum over the specified interval and state where those values occur.
[latex]f(x)=−x^2+3x-2[/latex] over [latex][1,3][/latex].
[latex]f(x)=x^2-3x^{\frac{2}{3}}[/latex] over [latex][0,2][/latex].
Step 1. Evaluate [latex]f[/latex] at the endpoints [latex]x=1[/latex] and [latex]x=3[/latex].
[latex]f(1)=0[/latex] and [latex]f(3)=-2[/latex]
Step 2. Since [latex]f^{\prime}(x)=-2x+3[/latex], [latex]f^{\prime}[/latex] is defined for all real numbers [latex]x[/latex]. Therefore, there are no critical points where the derivative is undefined. It remains to check where [latex]f^{\prime}(x)=0[/latex]. Since [latex]f^{\prime}(x)=-2x+3=0[/latex] at [latex]x=\frac{3}{2}[/latex] and [latex]\frac{3}{2}[/latex] is in the interval [latex][1,3][/latex], [latex]f(\frac{3}{2})[/latex] is a candidate for an absolute extremum of [latex]f[/latex] over [latex][1,3][/latex]. We evaluate [latex]f(\frac{3}{2})[/latex] and find
Step 3. We set up the following table to compare the values found in steps 1 and 2.
[latex]x[/latex]
[latex]f(x)[/latex]
Conclusion
0
0
[latex]\frac{3}{2}[/latex]
[latex]\frac{1}{4}[/latex]
Absolute maximum
3
-2
Absolute minimum
From the table, we find that the absolute maximum of [latex]f[/latex] over the interval [latex][1,3][/latex] is [latex]\frac{1}{4}[/latex], and it occurs at [latex]x=\frac{3}{2}[/latex]. The absolute minimum of [latex]f[/latex] over the interval [latex][1,3][/latex] is -2, and it occurs at [latex]x=3[/latex] as shown in the following graph.
Figure 8. This function has both an absolute maximum and an absolute minimum.
Step 1. Evaluate [latex]f[/latex] at the endpoints [latex]x=0[/latex] and [latex]x=2[/latex].
[latex]f(0)=0[/latex] and [latex]f(2)=4-3\sqrt[3]{4}\approx -0.762[/latex]
Step 2. The derivative of [latex]f[/latex] is given by
for [latex]x\ne 0[/latex]. The derivative is zero when [latex]2x^{\frac{4}{3}}-2=0[/latex], which implies [latex]x=\pm 1[/latex]. The derivative is undefined at [latex]x=0[/latex]. Therefore, the critical points of [latex]f[/latex] are [latex]x=0,1,-1[/latex]. The point [latex]x=0[/latex] is an endpoint, so we already evaluated [latex]f(0)[/latex] in step 1. The point [latex]x=-1[/latex] is not in the interval of interest, so we need only evaluate [latex]f(1)[/latex]. We find that
[latex]f(1)=-2[/latex]
Step 3. We compare the values found in steps 1 and 2, in the following table.
[latex]x[/latex]
[latex]f(x)[/latex]
Conclusion
0
0
Absolute maximum
1
-2
Absolute minimum
2
-0.762
We conclude that the absolute maximum of [latex]f[/latex] over the interval [latex][0,2][/latex] is zero, and it occurs at [latex]x=0[/latex]. The absolute minimum is −2, and it occurs at [latex]x=1[/latex] as shown in the following graph.
Figure 9. This function has an absolute maximum at an endpoint of the interval.
Watch the following video to see the worked solution to the first part of 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.
At this point, we know how to locate absolute extrema for continuous functions over closed intervals. We have also defined local extrema and determined that if a function [latex]f[/latex] has a local extremum at a point [latex]c[/latex], then [latex]c[/latex] must be a critical point of [latex]f[/latex]. However, [latex]c[/latex] being a critical point is not a sufficient condition for [latex]f[/latex] to have a local extremum at [latex]c[/latex]. Later in this module, we show how to determine whether a function actually has a local extremum at a critical point. First, however, we need to introduce the Mean Value Theorem, which will help as we analyze the behavior of the graph of a function.
![This figure has six parts a, b, c, d, e, and f. In figure a, the line f(x) = x3 is shown, and it is noted that it has no absolute minimum and no absolute maximum. In figure b, the line f(x) = 1/(x2 + 1) is shown, which is near 0 for most of its length and rises to a bump at (0, 1); it has no absolute minimum, but does have an absolute maximum of 1 at x = 0. In figure c, the line f(x) = cos x is shown, which has absolute minimums of −1 at ±π, ±3π, … and absolute maximums of 1 at 0, ±2π, ±4π, …. In figure d, the piecewise function f(x) = 2 – x2 for 0 ≤ x < 2 and x – 3 for 2 ≤ x ≤ 4 is shown, with absolute maximum of 2 at x = 0 and no absolute minimum. In figure e, the function f(x) = (x – 2)2 is shown on [1, 4], which has absolute maximum of 4 at x = 4 and absolute minimum of 0 at x = 2. In figure f, the function f(x) = x/(2 − x) is shown on [0, 2), with absolute minimum of 0 at x = 0 and no absolute maximum.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/2332/2018/01/11210805/CNX_Calc_Figure_04_03_010.jpg)
