3.2 Extreme Value Theorem

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In this section we learn the Extreme Value Theorem, and we find the extremes of a function.

1 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 $f(x)$ is continuous on the closed interval $[a,b]$ , then $f(x)$ has both an absolute maximum and an absolute minimum on the interval.

It is important to note that the theorem contains two hypotheses. The first is that $f(x)$ is continuous, and the second is that the interval is closed. If either of these conditions fails to hold, then $f(x)$ might fail to have either an absolute max or an absolute min (or both).

The theorem tells us that the max and the min exist in the interval, but it does not tell us how to find them. The next theorem, attributed to the $17^{th}$ century polymath and mathematician Pierre de Fermat, gives us a way to find the location of the absolute max and absolute min of a continuous function on a closed interval.

Fermat’s Theorem Suppose $f(x)$ is defined on the open interval $(a,b)$ . If $f(x)$ has an absolute maximum or absolute minimum at a point $x = c$ inside the interval, then $c$ is a critical number for $f(x)$ .

Note that the theorem refers to what are called interior points of the interval, $[a,b]$ , and not the endpoints, $x = a$ and $x = b$ . Fermat’s Theorem tells us that the absolute extremes of a continuous function on $[a,b]$ must occur at either: the endpoints of the interval or critical numbers inside the interval.

2 Closed Interval Method

Finding the absolute extremes of a continuous function, $f(x)$ , on a closed interval $[a,b]$ is a three-step process.

1.: Find the critical numbers of $f(x)$ inside the interval $(a, b)$ .
2.: Compare the values of the function $f(x)$ at these critical numbers and at the endpoints.
3.: The largest of the values from step 2 is the absolute maximum of $f(x)$ on $[a,b]$ , and the smallest is the absolute minimum.

example 1 Find the absolute maximum and the absolute minimum values of $f(x) = x^2 - 4x + 5$ on the closed interval $[0, 3]$ .

Solution. The function $f(x)$ is a polynomial, so it is continuous everywhere. The interval is of the form $[a,b]$ , so it is closed. The Extreme Value Theorem tells us that this function has an absolute maximum and an absolute minimum on the interval $[0,3]$ . We find these extremes by using the Closed Interval Method outlined above.
To find the critical numbers of $f(x)$ in the interval $(0, 3)$ , we need the derivative, $f'(x) = 2x - 4$ which exists for all values of $x$ . Solving the equation $f'(x) =0$ gives $x=2$ as the only critical number of the function. Moreover, this critical number is inside the interval $(0,3)$ . We next compute the function values at the critical numbers inside the interval and the endpoints:

\begin{eqnarray*} f(0) &=& 0^2 - 4(0) + 5 =5\\ f(2) &=& 2^2 - 4(2) + 5 = 1\\ f(3) &=& 3^2 - 4(3) + 5 = 2 \end{eqnarray*}

The largest of these values is $f(0) = 5$ , and the smallest is $f(2) = 1$ . Finally, the absolute maximum of $f(x)$ on $[0,3]$ is $5$ occurring at the left endpoint $x = 0$ and the absolute minimum of $f(x)$ in the interval is $1$ occurring at the critical number $x = 2$ . Below is a graph of the function on the interval $[0,3]$ which confirms our results.

(problem 1) Find the absolute maximum and the absolute minimum of $f(x) = x^2 - 4x + 2$ on the closed interval $[0, 3]$ .

The absolute maximum is $\answer {2}$ and it occurs at $x = \answer {0}.$
The absolute minimum is $\answer {-2}$ and it occurs at $x = \answer {2}.$

example 2 Find the absolute max and the absolute min of $f(x) = x^3 - 6x^2 + 5$ on the closed interval $[\text {-}1, 6]$ .

Solution: First, we find the critical numbers of $f(x)$ in the interval $[\text {-}1, 6]$ . The function is a polynomial, so it is differentiable everywhere. We solve the equation $f'(x) =0$ . This becomes $3x^2 -12x= 0$ which has two solutions $x=0$ and $x = 4$ (verify). Hence $f(x)$ has two critical numbers in the interval. The absolute extremes occur at either the endpoints, $x=\text {-}1, 6$ or the critical numbers $x = 0, 4$ . Plugging these special values into the original function $f(x)$ yields:

\begin{eqnarray*} f(\text{-}1) &=& (\text{-}1)^3 -6(\text{-}1)^2 + 5 = \text{-}2\\ f(0) &=& 0^3 - 6(0)^2 + 5 =5\\ f(4) &=& 4^3 - 6(4)^2 + 5 = \text{-}27\\ f(6) &=& 6^3 - 6(6)^2 + 5 = 5 \end{eqnarray*}

From this data we conclude that the absolute maximum of $f(x)$ on the interval is $5$ occurring at both the critical number $x = 0$ and the right endpoint $x = 6$ and the absolute minimum of $f(x)$ in the interval is $-27$ occurring at the critical number $x = 4$ .

(problem 2a) Find the absolute maximum and the absolute minimum of $f(x) = x^3 - 12x + 1$ on the closed interval $[-3, 3]$ .

The absolute maximum is $\answer {17}$ and it occurs at $x = \answer {-2}.$
The absolute minimum is $\answer {-15}$ and it occurs at $x = \answer {2}.$

(problem 2b) Find the absolute maximum and the absolute minimum of $f(x) = x^3 + 3x^2$ on the closed interval $[-3, 1]$ .

(If the max/min occurs in more than one place, list them in ascending order).
The absolute maximum is $\answer {4}$ and it occurs at $x = \answer {-2}$ and $x=\answer {1}.$
The absolute minimum is $\answer {0}$ and it occurs at $x = \answer {-3}$ and $x=\answer {0}.$

example 3 Find the absolute max and the absolute min of $f(x) = \frac 14 x^4 - 2x^3 + 4x^2 - 3$ on the closed interval $[\text {-}1, 3]$ .

Solution: First, we find the critical numbers of $f(x)$ in the interval $[\text {-}1, 3]$ . The function is a polynomial, so it is differentiable everywhere. We solve the equation $f'(x) =0$ . This becomes $x^3 -6x^2 + 8x = 0$ and the solutions are $x=0, x=2$ and $x=4$ (verify). Noting that $x = 4$ is not in the interval $[\text {-}1,3]$ we see that $f(x)$ has two critical numbers in the interval, namely $x = 0$ and $x = 2$ . The absolute extremes occur at either the endpoints, $x=\text {-}1, 3$ or the critical numbers $x = 0,2$ . Plugging these special values into the original function $f(x)$ yields:

\begin{eqnarray*} f(\text{-}1) &=& \frac 14(\text{-}1)^4 -2(\text{-}1)^3 + 4(\text{-}1)^2 - 3 = \frac 14 + 2 + 4 - 3 = \frac{13}{4}\\ f(0) &=& \frac 14 (0)^4 -2(0)^3 + 4(0)^2 - 3 = \text{-}3\\ f(2) &=& \frac 14(2)^4 -2(2)^3 + 4(2)^2 - 3 = 4 -16 + 16 -3 = 1\\ f(3) &=& \frac 14(3)^4 -2(3)^3 + 4(3)^2 - 3 = \frac{81}{4} - 54 + 36 - 3 = \text{-}\frac 34 \end{eqnarray*}

From this data we conclude that the absolute maximum of $f(x)$ on the interval is $3.25$ occurring at the endpoint $x = -1$ and the absolute minimum of $f(x)$ in the interval is $-3$ occurring at the critical number $x = 0$ .

(problem 3) Find the absolute maximum and the absolute minimum of $f(x) = x^4 - 4x^3 - 8x^2 - 16$ on the closed interval $[-2, 1]$ .

The absolute maximum is $\answer {0}$ and it occurs at $x = \answer {-2}.$
The absolute minimum is $\answer {-27}$ and it occurs at $x = \answer {1}.$

example 4 Find the absolute max and the absolute min of $f(x) = xe^{3x}$ on the closed interval $[-1, 0]$ .

Solution: First, we find the critical numbers of $f(x)$ in the interval $[-1, 0]$ . We solve the equation $f'(x) =0$ . Using the product rule and the chain rule, we have $f'(x) = 1\cdot e^{3x} + x\cdot e^{3x} \cdot 3$ which simplifies to $(3x+1)e^{3x}$ . Thus, we need to solve $(3x+1)e^{3x} = 0$ (verify) and the only solution is $x=\text {-}1/3$ (verify). Hence, $f(x)$ has one critical number in the interval and it occurs at $x = -1/3$ . The absolute extremes occur at either the endpoints, $x=-1, 2$ or the critical number $x = -1/3$ . Plugging these special values into the original function $f(x)$ yields:

\begin{eqnarray*} f(-1) &=& (-1)e^{-3} = -\frac{1}{e^3} \approx -0.0498\\ f(-1/3) &=& (-1/3)e^{-1} = -\frac{1}{3e} \approx -0.1226\\ f(0) &=& 0 \end{eqnarray*}

From this data we conclude that the absolute maximum of $f(x)$ on the interval is $0$ occurring at the right endpoint $x = 0$ and the absolute minimum of $f(x)$ in the interval is $-\frac {1}{3e}$ occurring at the critical number $x = -1/3$ .

(problem 4a) Find the absolute maximum and the absolute minimum of $f(x) = xe^{2x}$ on the closed interval $[-3, 2]$ .

$f'(x) = (2x+1)e^{2x}$

The absolute maximum is $\answer {2e^4}$ and it occurs at $x = \answer {2}.$
The absolute minimum is $\answer {-1/(2e)}$ and it occurs at $x = \answer {-1/2}.$

Here is a video solution of problem 4a:

(problem 4b) Find the absolute maximum and the absolute minimum of $f(x) = \frac {3x}{x^2+4}$ on the closed interval $[0, 4]$ .

Use the quotient rule to find $f'(x)$

$f'(x) = \frac {12-3x^2}{(x^2+4)^2}$

The absolute maximum is $\answer {3/4}$ and it occurs at $x = \answer {2}.$
The absolute minimum is $\answer {0}$ and it occurs at $x = \answer {0}.$
Note that the critical number $x= -2$ is not in the interval $[0, 4]$ .

Here is a video solution of problem 4b:

Here is a detailed, lecture style video on the Extreme Value Theorem:

3 Proof of Fermat’s Theorem

Suppose that $f(x)$ is defined on the open interval $(a,b)$ and that $f(x)$ has an absolute max at $x=c$ . Thus, $f(c) \geq f(x)$ for all $x$ in $(a,b)$ . If $f'(c)$ is undefined then, $x=c$ is a critical number for $f(x)$ . If $f'(c)$ is defined, then from the definition of the derivative we have $f'(c) = \lim _{h \to 0^-} \frac {f(c+h) -f(c)}{h} = \lim _{h \to 0^+} \frac {f(c+h) -f(c)}{h}$ 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, $f'(c) \geq 0$ . But the difference quotient in the right-hand limit is negative (or zero) and so $f'(c) \leq 0$ . Hence, $f'(c) = 0$ and the theorem is proved.

2026-03-19 15:00:34