The absolute maximum is and it occurs at
The absolute minimum is and it occurs at
In this section we learn the Extreme Value Theorem and we find the extremes of a function.
The Extreme Value Theorem
In this section we will solve the problem of finding the maximum and minimum values of a continuous function on a closed interval.
Extreme Value Theorem If is continuous on the closed interval , then has both an
absolute maximum and an absolute minimum on the interval.
It is important to note that the theorem contains two hypothesis. The first is that is continuous and the second is that the interval is closed. If either of these conditions fails to hold, then might fail to have either an absolute max or an absolute min (or both). It is also important to note that the theorem tells us that the max and the min occur in the interval, but it does not tell us how to find them.
The main idea is finding the location of the absolute max and absolute min of a continuous function on a closed interval is contained in the following theorem.
Fermat’s Theorem Suppose is defined on the open interval . If has an absolute
maximum or absolute minimum at a point in the interval, then is a critical number
for .
As a result of Fermat’s Theorem, we can conclude that the absolute extremes of a
continuous function on a closed interval must occur at the endpoints of the interval
or at a critical numbers inside the interval.Closed Interval Method
Finding the absolute extremes of a continuous function, , on a closed interval is a three step process.
- 1.
- Find the critical numbers of inside the interval .
- 2.
- Compute the values of at the critical numbers and at the endpoints.
- 3.
- The largest of the values from step 2 is the absolute maximum of on the closed interval , and the smallest of these values is the absolute minimum.
example 1 Find the absolute maximum and the absolute minimum values of on the
closed interval .
The largest of these values is 5, corresponding to , and the smallest is 1,
corresponding to . Thus, we conclude that the absolute maximum of on the closed
interval is occurring at the left endpoint and the absolute minimum of in the
interval is occurring at the critical number . 
Solution: The function is a polynomial, so it is continuous, and the interval is closed, so by the Extreme Value Theorem, we know that this function has an absolute maximum and an absolute minimum on the interval . First, we find the critical numbers of in the interval . The derivative is which exists for all values of . Solving the equation gives as the only critical number of the function. Moreover, this critical number is in the interval . By the closed interval method, we know that the absolute extremes occur at either the endpoints, and , or the critical number . Plugging these values into the original function yields:
(problem 1) Find the absolute maximum and the absolute minimum of on the closed
interval .
example 2 Find the absolute max and the absolute min of on the closed interval
.
From this data we conclude that the absolute maximum of on the interval is
occurring at both the critical number and the right endpoint and the
absolute minimum of in the interval is occurring at the critical number .
Solution: First, we find the critical numbers of in the interval . The function is a polynomial, so it is differentiable everywhere. We solve the equation . This becomes which has two solutions and (verify). Hence has two critical numbers in the interval. The absolute extremes occur at either the endpoints, or the critical numbers . Plugging these special values into the original function yields:
(problem 2a) Find the absolute maximum and the absolute minimum of on the
closed interval .
The absolute maximum is and it occurs at
The absolute minimum is and it occurs at
(problem 2b) Find the absolute maximum and the absolute minimum of on the
closed interval .
(If the max/min occurs in more than one place, list them in ascending order).
The absolute maximum is and it occurs at and
The absolute minimum is and it occurs at and
example 3 Find the absolute max and the absolute min of on the closed interval
.
Solution: First, we find the critical numbers of in the interval . The function is a polynomial, so it is differentiable everywhere. We solve the equation . This becomes and the solutions are and (verify). Noting that is not in the interval we see that has two critical numbers in the interval, namely and . The absolute extremes occur at either the endpoints, or the critical numbers . Plugging these special values into the original function yields:
From this data we conclude that the absolute maximum of on the interval is occurring at the endpoint and the absolute minimum of in the interval is occurring at the critical number .
(problem 3) Find the absolute maximum and the absolute minimum of on the closed
interval .
The absolute maximum is and it occurs at
The absolute minimum is and it occurs at
example 4 Find the absolute max and the absolute min of on the closed interval
.
From this data we conclude that the absolute maximum of on the interval is
occurring at the right endpoint and the absolute minimum of in the interval is
occurring at the critical number . 
Solution: First, we find the critical numbers of in the interval . We solve the equation . Using the product rule and the chain rule, we have which simplifies to . Thus we need to solve (verify) and the only solution is (verify). Hence has one critical number in the interval and it occurs at . The absolute extremes occur at either the endpoints, or the critical number . Plugging these special values into the original function yields:
(problem 4a) Find the absolute maximum and the absolute minimum of on the
closed interval .
The absolute maximum is and it occurs at
The absolute minimum is and it occurs at
(problem 4b) Find the absolute maximum and the absolute minimum of on the
closed interval .
The absolute maximum is and it occurs at
The absolute minimum is and it occurs at
Note that the critical number is not in the interval .
Here is a detailed, lecture style video on the Extreme Value Theorem:
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Proof of Fermat’s Theorem
Suppose that is defined on the open interval and that has an absolute max at . Thus for all in . If is undefined then, is a critical number for . If is defined, then from the definition of the derivative we have The difference quotient in the left hand limit is positive (or zero) since the numerator is negative (or zero) and the denominator is negative. Thus . But the difference quotient in the right hand limit is negative (or zero) and so . Hence and the theorem is proved.