Question

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]?

Answer 1

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

Explanation:

`f(x) = { ( absx," if ", -3 <= x <= 2), (4-x," if ", 2 < x <= 3) :}`

Note first that this function is continuous on `[-3,3]`.

The only possible "problem point" is `x=2` but we see that
`lim_(xrarr2^-)f(x) = lim_(xrarr2^+)f(x) = f(2)`. 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'(x) =0`, or `f'(x)` 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'(x) =0`, or `f'(x)` does not exist are called critical numbers for `f`.)

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

`f'(x) = { ( -1," if ", -3 < x < 0), ("DNE"," if ",x = 0), ( 1," if ", 0 < x < 2), ("DNE"," if ",x = 2), (-1," if ", 2 < x < 3) :}`

Note: `DNE` 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(-3) =3`
`f(0)=0`
`f(2)=2`
`f(3) = 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(0) =0`
Maximum: `f(-3)=3`