Question

How do use the first derivative test to determine the local extrema `f(x)= x^3 - x^2 - 40x + 8`?

Answer 1

This function has a local maximum value of `2516/27` at `x=-10/3` and a local minimum value of `-104` at `x=4`.

Explanation:

The first derivative is `f'(x)=3x^2-2x-40`. This can be factored as `f'(x)=3x^2-2x-40=(3x+10)(x-4)`. Setting this equal to zero and solving for `x` gives two critical points for `f` at `x=-10/3` and `x=4`.

You can check that `f'` changes sign from positive to negative as `x` increases through `x=-10/3` and negative to positive as `x` increases through `x=4` (note that `f'` is a continuous function and that, for instance, `f'(-4)=48+8-40=16>0`, `f'(0)=-40<0`, and `f'(5)=75-10-40=25>0`).

Therefore, the First Derivative Test implies that `f` has a local maximum at `x=-10/3` and a local minimum at `x=4`.

The local maximum value (output) is `f(-10/3)=-1000/27-100/9+400/3+8=2516/27` and the local minimum value (output) is `f(4)=64-16-160+8=-104`.