# What are the global and local extrema of f(x) = x^3-9x+3  ?

Nov 20, 2015

There are no global extrema. $3 + 3 \sqrt{3}$ is a local maximum (Which occurs at $- \sqrt{3}$) and $3 - 6 \sqrt{3}$ is a local minimum. (It occurs at $\sqrt{3}$.)

#### Explanation:

The domain of $f$ is $\left(- \infty , \infty\right)$.

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

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

$f ' \left(x\right) = 3 {x}^{2} - 9$ is never undefined and is $0$ at $x = \pm \sqrt{3}$.

We look at the sign of $f '$ on each interval.

{: (bb "Interval", bb"Sign of "f',bb" Incr/Decr"), ((-oo,-sqrt3)," " +" ", " "" Incr"), ((-sqrt3,sqrt3), " " -, " " " Decr"), ((sqrt3 ,oo), " " +, " "" Incr") :}

$f$ has a local maximum at $- \sqrt{3}$, which is $f \left(- \sqrt{3}\right) = 3 + 3 \sqrt{3}$

and a local minimum at $\sqrt{3}$, hich is $f \left(\sqrt{3}\right) = 3 - 6 \sqrt{3}$