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Extreme Value Theorem - Endpoint Extremes

Author: Sophia

what's covered
In this lesson, you will use critical numbers and endpoint analysis to determine the maximum and minimum values of a continuous function on some closed interval. Specifically, this lesson will cover:

Table of Contents

1. The Extreme Value Theorem

If a function f open parentheses x close parentheses is continuous on a closed interval open square brackets a comma space b close square brackets comma then f open parentheses x close parentheses is guaranteed to have global maximum and global minimum values on the interval open square brackets a comma space b close square brackets. This is known as the extreme value theorem.

Here is an illustration of the extreme value theorem:

A graph with an x-axis and a y-axis intersecting at the origin, representing a function that is continuous on a closed interval. A curve starts from a marked point to the right of the y-axis, rises to a marked point labeled ‘max’, and then descends to the marked point on the right labeled ‘min’, which is lower than the starting point.

  • The function is continuous on the interval open square brackets a comma space b close square brackets.
  • The maximum point occurs inside the interval.
  • The minimum occurs at an endpoint.
The following are examples of situations in which one of the criteria is violated.

f   Continuous Open Interval f   Continuous Open Interval f   Not Continuous Closed Interval
A graph with an x-axis and a y-axis intersecting at the origin, representing a function that is continuous on an open interval. A curve starts from an open circle to the right of the y-axis, rises to a marked point labeled ‘max’, and then descends to another open circle on the right, which is lower than the first open circle. The graph has the words ‘no min’ below the curve. A graph with an x-axis and a y-axis intersecting at the origin. The graph has an open curve. The curve begins from an open circle labeled ‘no max’ in the upper left part of the graph and then extends downward to an open circle labeled ‘no min’. A graph with an x-axis and a y-axis intersecting at the origin. The graph has a vertical dashed line and two slanting line segments that run parallel to each other. The line segment to the left of the dashed line starts at a marked point and descends to an open circle, which is labeled 'no min'. The second line segment starts at an open circle well above the previous open circle, this one labeled 'no max', then descends to the right to a marked point.

term to know
Extreme Value Theorem
If f open parentheses x close parentheses is a continuous function on some closed interval open square brackets a comma space b close square brackets comma then f open parentheses x close parentheses has global maximum and global minimum values on the interval open square brackets a comma space b close square brackets.


2. Finding Extreme Values of a Continuous Function on a Closed Interval

As a result of the theorem, here is what we need to do in order to find the global minimum and maximum values of f open parentheses x close parentheses on a closed interval open square brackets a comma space b close square brackets.

  1. Find all critical numbers of f open parentheses x close parentheses that are in the interval open square brackets a comma space b close square brackets.
  2. Evaluate f open parentheses x close parentheses at each endpoint and each critical number. The largest value of f is the global maximum and the smallest value of f is the global minimum.

EXAMPLE

Find the global maximum and minimum points of the function f open parentheses x close parentheses equals x cubed minus 6 x squared plus 5 on the interval open square brackets short dash 1 comma space 3 close square brackets.

First, find the critical numbers.

f open parentheses x close parentheses equals x cubed minus 6 x squared plus 5 Start with the original function.
f apostrophe open parentheses x close parentheses equals 3 x squared minus 12 x Take the derivative.
3 x squared minus 12 x equals 0
3 x open parentheses x minus 4 close parentheses equals 0
x equals 0 comma space x equals 4
Since f apostrophe open parentheses x close parentheses is a polynomial, it is never undefined. Set f apostrophe open parentheses x close parentheses equals 0 and solve for x.

Therefore, the critical numbers are x equals 0 and x equals 4.

However, since only the closed interval open square brackets short dash 1 comma space 3 close square brackets is considered, the critical value x equals 4 is not used.

Now, evaluate f open parentheses x close parentheses at the endpoints, x equals short dash 1 and x equals 3 comma and the remaining critical number, x equals 0.

x bold italic f open parentheses bold x close parentheses Result
-1 open parentheses short dash 1 close parentheses cubed minus 6 open parentheses short dash 1 close parentheses squared plus 5 equals short dash 2 Neither a Global Maximum or Global Minimum
0 open parentheses 0 close parentheses cubed minus 6 open parentheses 0 close parentheses squared plus 5 equals 5 Global Maximum
3 3 cubed minus 6 open parentheses 3 close parentheses squared plus 5 equals short dash 22 Global Minimum

In conclusion, the global maximum occurs at the point (0, 5) and the global minimum occurs at the point (3, -22). In order words, the global maximum value is 5 and occurs when x equals 0 semicolon and the global minimum value is -22 and occurs when x equals 3.

The graph of the function on open square brackets short dash 1 comma space 3 close square brackets is shown below, which confirms the results.

A graph with an x-axis and a y-axis ranging from −20 to 20 in increments of 5. The graph has a parabolic portion that starts from a marked point at (−1, −2), reaches a peak at the marked point at (0, 5), and then ends at another marked point at (3, −22).

watch
In this video, we’ll find the global minimum and maximum values of f open parentheses x close parentheses equals 10 square root of x minus x on the interval open square brackets 16 comma space 64 close square brackets.

When finding critical numbers, it’s important to consider only those that are in the interval open square brackets a comma space b close square brackets. Here is an example that helps illustrate this.

EXAMPLE

Find all global extreme values of f open parentheses x close parentheses equals short dash 1 fourth x to the power of 4 plus 2 over 3 x cubed on the interval open square brackets 1 comma space 4 close square brackets.

To start, we find the critical numbers.

f open parentheses x close parentheses equals short dash 1 fourth x to the power of 4 plus 2 over 3 x cubed Start with the original function.
f apostrophe open parentheses x close parentheses equals short dash x cubed plus 2 x squared Use the power rule to find the derivative.
short dash x cubed plus 2 x squared equals 0 Set f apostrophe open parentheses x close parentheses equals 0.
short dash x squared open parentheses x minus 2 close parentheses equals 0 Factor out short dash x squared.
short dash x squared equals 0 space space or space space x minus 2 equals 0 Set each factor equal to 0.
x equals 0 space space or space space x equals 2 Solve each equation.

Thus, the critical numbers are x equals 0 and x equals 2.

Considering the interval we are interested in, open square brackets 1 comma space 4 close square brackets comma notice that x equals 0 is not contained in this interval. This means that x equals 0 is not considered in any further analysis.

To determine the values of the local minimum and maximum values, we consider x equals 2 along with the endpoints of the interval, x equals 1 and x equals 4.

Here is a table of values for f open parentheses x close parentheses equals short dash 1 fourth x to the power of 4 plus 2 over 3 x cubed. Note the approximations are also provided to make comparisons easier.

x 1 2 4
f  (x  ) 5 over 12 almost equal to 0.433... 4 over 3 almost equal to 1.333... short dash 64 over 3 almost equal to short dash 21.333...

The global minimum value is short dash 64 over 3 when x equals 4. and the global maximum value is 4 over 3 when x equals 2.

summary
In this lesson, you learned that when f open parentheses x close parentheses is continuous on a closed interval, the extreme value theorem guarantees a global minimum value and a global maximum value at some location within the closed interval. Then, you applied this theorem to find extreme values of a continuous function on a closed interval, by first finding all critical numbers of f open parentheses x close parentheses that are in the interval open square brackets a comma space b close square brackets, then evaluating f open parentheses x close parentheses at each endpoint and each critical number. This concept is going to be very useful once we use derivatives to solve optimization problems.

Source: THIS TUTORIAL HAS BEEN ADAPTED FROM CHAPTER 3 OF "CONTEMPORARY CALCULUS" BY DALE HOFFMAN. ACCESS FOR FREE AT WWW.CONTEMPORARYCALCULUS.COM. LICENSE: CREATIVE COMMONS ATTRIBUTION 3.0 UNITED STATES.

Terms to Know
Extreme Value Theorem

If f open parentheses x close parentheses is a continuous function on some closed interval open square brackets a comma space b close square brackets, then f open parentheses x close parentheses has global maximum and global minimum values on the interval open square brackets a comma space b close square brackets.