Question

If `f '(x) = 8x^3 - 16x`, how do you use the second derivative test to find maximum and minimum?

Answer 1

Without knowing the function `f(x)` we cannot find the values, but we can find where the extrema occur.

Explanation:

Given `f'(x) = 8x^3-16x` and assuming that domain of `f` includes the zeros of `f'`, we proceed:

Finding Critical Numbers for `f`

A critical number for `f` is a number in the domain of `f` at which either `f(x)=0` or `f'(x)` does not exist.

In this problem, `f'` is a polynomial, so it exists for all `x`. The critical numbers for this `f` are the zeros of `f'`.

`f'(x) = 8x^3-16x =8x(x^2-2) = 0` at `-sqrt2`, `0`, and `sqrt2`

These are our critical numbers.

Note To use the first derivative test for local extrema, we check the sign of `f'` on both sides of each critical number.

Testing the Critical Numbers

To use the second derivative test for local extrema, we check the sign of `f''` at each critical number.

`f''(x) = 24x^2-16`

At `-sqrt2`, we get `f''(-sqrt2) = 24(-sqrt2)^2-6` which is clearly positive.
`f(-sqrt2)` is a local minimum.

At `0`, we get `f''(0) = 24(0)^2-16` which is negative.
`f(0)` is a local maximum.

At `sqrt2`, we get `f''(sqrt2) = 24(sqrt2)^2-6` which is clearly positive.
`f(sqrt2)` is a local minimum.