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.

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 hypothesis. 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). 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.

(Closed Interval Method) If $f(x)$ is continuous on the closed interval $[a,b]$ , then the absolute extremes occur at either the endpoints or at critical numbers inside the interval.

By the Closed Interval Method, finding the absolute extremes of a continuous function, $f(x)$ , on a closed interval $[a,b]$ is a three step process.

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

The largest of these function values is the absolute max and the smallest is the absolute min of $f(x)$ on $[a,b]$ .

Examples of the Closed Interval Method

CI1.

Find the absolute max and the absolute min of $f(x) = x^2 - 4x + 7$ on the closed interval $[0, 3]$ .

Solution: First, we find the critical numbers of $f(x)$ in the interval $[0, 3]$ . The function is a polynomial, so it is differentiable everywhere. We solve the equation $f'(x) =0$ . This becomes $2x-4 = 0$ and the solution is $x=2$ . Hence $f(x)$ has one critical number in the interval and it occurs at $x = 2$ . The absolute extremes occur at either the endpoints, $x=0, 3$ or the critical number $x = 2$ . Plugging these special values into the original function $f(x)$ yields:

$\begin{eqnarray*} f(0) &=& 0^2 - 4(0) + 7 =7\\ f(2) &=& 2^2 - 4(2) + 7 = 3\\ f(3) &=& 3^2 - 4(3) + 7 = 4. \end{eqnarray*}$

From this data we conclude that the absolute maximum of $f(x)$ on the interval is $7$ occurring at the left endpoint $x = 0$ and the absolute minimum of $f(x)$ in the interval is $3$ occurring at the critical number $x = 2$ .

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 $x = \answer {2}.$

CI2.

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 $x = 4$ .

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 $x = \answer {2}.$

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 $x=\answer {0}.$

CI3.

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:

$\begin{eqnarray*} f(\text{-}1) &=& \frac14(\text{-}1)^4 -2(\text{-}1)^3 + 4(\text{-}1)^2 - 3 = \frac14 + 2 + 4 - 3 = \frac{13}{4}\\ f(0) &=& \frac14 (0)^4 -2(0)^3 + 4(0)^2 - 3 = \text{-}3\\ f(2) &=& \frac14(2)^4 -2(2)^3 + 4(2)^2 - 3 = 4 -8 + 16 -3 = 9\\ f(3) &=& \frac14(3)^4 -2(3)^3 + 4(3)^2 - 3 = \frac{81}{4} - 54 + 36 - 3 = \text{-}\frac34. \end{eqnarray*}$

From this data we conclude that the absolute maximum of $f(x)$ on the interval is $9$ occurring at the critical number $x = 2$ and the absolute minimum of $f(x)$ in the interval is $\text {-}3$ occurring at the critical number $x = 0$ .

CI4.

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 $x = -1/3$ .

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 $x = \answer {-1/2}.$

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 detailed, lecture style video on the Extreme Value Theorem: