# How do you find the absolute extrema of the function on the indicated interval by using the concept of the Extreme-Value Theorem f(x) = { |x| if -3 ≤ x ≤ 2 , 4-x if 2 < x ≤ 3 ; [ -3, 3]?

Jul 1, 2015

The minimum is $0$ (it occurs at $0$).
The maximum is $3$ (it occurs at $- 3$)

#### Explanation:

$f \left(x\right) = \left\{\begin{matrix}\left\mid x \right\mid & \text{ if " & -3 <= x <= 2 \\ 4-x & " if } & 2 < x \le 3\end{matrix}\right.$

Note first that this function is continuous on $\left[- 3 , 3\right]$.

The only possible "problem point" is $x = 2$ but we see that
${\lim}_{x \rightarrow {2}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {2}^{+}} f \left(x\right) = f \left(2\right)$. So $f$ is continuous at $2$.

Thus, the Extreme Value Theorem guarantees that the function attains both a minimum and a maximum on the interval.

These extrema occur at values of $x$ at which either $f ' \left(x\right) = 0$, or $f ' \left(x\right)$ does not exist, or $x$ is an endpoint of the interval. (That is $x = - 3$ or $x = 3$.)

(Values of $x$ that are in the domain of $f$ and at which either $f ' \left(x\right) = 0$, or $f ' \left(x\right)$ does not exist are called critical numbers for $f$.)

For the function in this question, we get the following piecewise defined derivative:

$f ' \left(x\right) = \left\{\begin{matrix}- 1 & \text{ if " & -3 < x < 0 \\ "DNE" & " if " & x = 0 \\ 1 & " if " & 0 < x < 2 \\ "DNE" & " if " & x = 2 \\ -1 & " if } & 2 < x < 3\end{matrix}\right.$

Note: $D N E$ is used here to indicate that the derivative does not exist. We really could, and perhaps should, leave those lines out of the definition.

The only critical numbers are $0$ and $2$.

The minimum and the maximum must occur at one of the values of $x$:
$x = - 3 , 0 , 2 , 3$

Evaluating $f$ at each of these, we get:

$f \left(- 3\right) = 3$
$f \left(0\right) = 0$
$f \left(2\right) = 2$
$f \left(3\right) = 1$

The minimum is $0$ (it occurs at $0$).
The maximum is $3$ (it occurs at $- 3$).

Some teachers and textbooks express this by writing:

Minimum: $f \left(0\right) = 0$
Maximum: $f \left(- 3\right) = 3$