How do you find the local max and min for x^3-2x^2-8xx32x28x?

1 Answer
Jan 17, 2016

Take a look at the derivative.

Explanation:

Here, f(x) = x^3 - 2x^2 - 8xf(x)=x32x28x 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.