# What are the global and local extrema of f(x)=x^3-3x+6 ?

Dec 3, 2015

That function has no global extrema. It has local maximum of $8$ (at ($x = - 1$) and local minimum of $4$ (at $x = 1$)

#### Explanation:

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

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

$f ' \left(x\right) = 3 {x}^{2} - 3$ which is never undefined and is $0$ at $x = - 1$ and at $x = 1$. The domain of $f$ is $\mathbb{R}$.

Therefore, the only critical numbers are $- 1$ and $1$.

The sign of $f '$ changes from + to - as we pass $x = - 1$, so $f \left(- 1\right) = 8$ is a local maximum.

The sign of $f '$ changes from - to + as we pass $x = 1$, so $f \left(1\right) = 4$ is a local minimum.