Question

What are the global and local extrema of `f(x)=x^3 + 4x^2 - 5x` ?

Answer 1

The function has no global extrema. It has a local maximum of `f((-4-sqrt31)/3) = (308+62sqrt31)/27` and a local minimum of `f((-4+sqrt31)/3) = (308-62sqrt31)/27`

Explanation:

For `f(x)=x^3 + 4x^2 - 5x` ,

`lim_(xrarr-oo)f(x)=-oo` so `f` has no global minimum.

`lim_(xrarroo)f(x)=oo` so `f` has no global maximum.

`f'(x)=3x^2+8x-5` is never undefined and is `0` at

`x=(-4+-sqrt31)/3`

For numbers far from `0` (both positive and negative), `f'(x)` is positive.
For numbers in `((-4-sqrt31)/3,(-4+sqrt31)/3)`, 3f'(x)# is negative.

The sign of `f'(x)` changes from + to - as we move past `x=(-4-sqrt31)/3`, so `f((-4-sqrt31)/3)` is a local maximum.

The sign of `f'(x)` changes from - to + as we move past `x=(-4+sqrt31)/3`, so `f((-4+sqrt31)/3)` is a local minimum.

Finish by doing the arithmetic to get the answer:

`f` has a local maximum of `f((-4-sqrt31)/3) = (308+62sqrt31)/27` and a local minimum of `f((-4+sqrt31)/3) = (308-62sqrt31)/27`