Question

How do you find the exact relative maximum and minimum of the polynomial function of `f(x)=x^3-3x+6`?

Answer 1

Relative maximum: `8` (at `-1`) and relative minimum: `4` (at `1`)

Explanation:

`f(x) = x^3-3x+6`

Find critical numbers for `f` .

`f'(x) = 3x^2-3` never fails to exist and is `0` at the solutions of

`3x^2-3=0`.

The solution are `+-1`, so the critical numbers are `+-1`.

Test the critical numbers

Use either the first or second derivative test to see that

`f(-1)` is a relative maximum.

(Second derivative test: `f''(x) = 6x`, so `f''(-1) < 0` and we conclude `f(-10` is a relative maximum.)

And `f(1)` is a relative minimum.

(f''(1) > 0`, so`f(1) is a relative minimum.)

Find the minimum and maximum values

Rel max: `f(-1)=(-1)^3-3(-1)+6 = 8`

Rel min: `f(1) = (1)^3-3(1)+6 = 4`