Question

How do you find the local max and min for `x^3-2x^2-8x`?

Answer 1

Take a look at the derivative.

Explanation:

Here, `f(x) = x^3 - 2x^2 - 8x` so `f'(x) = 3x^2 - 4x - 8`.

The local max and min of `f` will be given by the roots of `f'`.

We first calculate `Delta = 16 - 4*3*(-8) = 112` so `f'` has 2 real roots.

By the quadratic formula, the zeros of `f'` are given by `(-b +- sqrtDelta)/2a`, which are here `(4 +-sqrt112)/6`.

So the critical points are `f((4 + sqrt112)/6)` and `f((4 -sqrt112)/6)`. By evaluating them, you will see which one is a local max and which one is the local min.