# What are the local extrema, if any, of f(x)= 120x^5 - 200x^3?

Mar 11, 2016

Local maximum of $80$ (at $x = - 1$) and local minimum of $- 80$ (at $x = 1$.

#### Explanation:

$f \left(x\right) = 120 {x}^{5} - 200 {x}^{3}$

$f ' \left(x\right) = 600 {x}^{4} - 600 {x}^{2} = 600 {x}^{2} \left({x}^{2} - 1\right)$

Critical numbers are: $- 1$, $0$, and $1$

The sign of $f '$ changes from + to - as we pass $x = - 1$, so $f \left(- 1\right) = 80$ is a local maximum.

(Since $f$ is odd, we can immediately conclude that $f \left(1\right) = - 80$ is a relative minimum and $f \left(0\right)$ is not a local extremum.)

The sign of $f '$ does not change as we pass $x = 0$, so $f \left(0\right)$ is not a local extremum.

The sign of $f '$ changes from - to + as we pass $x = 1$, so $f \left(1\right) = - 80$ is a local minimum.