Question

How do use the first derivative test to determine the local extrema `f(x)=x^3-2x +pi`?

Answer 1

See the explanation.

Explanation:

Find the critical numbers for `f`.

`f(x)=x^3-2x +pi`

`f'(x)=3x^2-2`

`f'` is never undefined and is `f'(x) = 0` at `x= +- sqrt6/3`

On `(-oo, -sqrt6/3)`, we get `f'(x)` is positive
on `(-sqrt6/3, sqrt6/3)`, we get `f'(x)` is negative.

So `f` changes from increasing to decreasing as we move to the left past `x = -sqrt6/3`.
Therefore, `f(-sqrt6/3)` is a local maximum.

Recall: on `(-sqrt6/3, sqrt6/3)`, we get `f'(x)` is negative.
now, on `(sqrt6/3, oo)`, we get `f'(x)` is positive.
So `f` changes from decreasing to increasing as we move to the left past `x = sqrt6/3`.
Therefore, `f(sqrt6/3)` is a local minimum.

Calculating (simplifying) `f(-sqrt6/3)` and `f(sqrt6/3)` is left to the student.