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

Nov 18, 2015

The function has no global extrema. It has a local maximum of $f \left(\frac{- 4 - \sqrt{31}}{3}\right) = \frac{308 + 62 \sqrt{31}}{27}$ and a local minimum of $f \left(\frac{- 4 + \sqrt{31}}{3}\right) = \frac{308 - 62 \sqrt{31}}{27}$

#### Explanation:

For $f \left(x\right) = {x}^{3} + 4 {x}^{2} - 5 x$ ,

${\lim}_{x \rightarrow - \infty} f \left(x\right) = - \infty$ so $f$ has no global minimum.

${\lim}_{x \rightarrow \infty} f \left(x\right) = \infty$ so $f$ has no global maximum.

$f ' \left(x\right) = 3 {x}^{2} + 8 x - 5$ is never undefined and is $0$ at

$x = \frac{- 4 \pm \sqrt{31}}{3}$

For numbers far from $0$ (both positive and negative), $f ' \left(x\right)$ is positive.
For numbers in $\left(\frac{- 4 - \sqrt{31}}{3} , \frac{- 4 + \sqrt{31}}{3}\right)$, 3f'(x)# is negative.

The sign of $f ' \left(x\right)$ changes from + to - as we move past $x = \frac{- 4 - \sqrt{31}}{3}$, so $f \left(\frac{- 4 - \sqrt{31}}{3}\right)$ is a local maximum.

The sign of $f ' \left(x\right)$ changes from - to + as we move past $x = \frac{- 4 + \sqrt{31}}{3}$, so $f \left(\frac{- 4 + \sqrt{31}}{3}\right)$ is a local minimum.

Finish by doing the arithmetic to get the answer:

$f$ has a local maximum of $f \left(\frac{- 4 - \sqrt{31}}{3}\right) = \frac{308 + 62 \sqrt{31}}{27}$ and a local minimum of $f \left(\frac{- 4 + \sqrt{31}}{3}\right) = \frac{308 - 62 \sqrt{31}}{27}$