# How do you find the relative extrema for f(x)=(x^4)+(4x^3)-12?

Dec 30, 2017

$f$ has global minimum at ${x}_{0} = - 3$ , $f \left(- 3\right) = - 39$

#### Explanation:

$f \left(x\right) = {x}^{4} + 4 {x}^{3} - 12$, ${D}_{f} = \mathbb{R}$

$f ' \left(x\right) = 4 {x}^{3} + 12 {x}^{2} = 4 {x}^{2} \left(x + 3\right)$

$f ' \left(x\right) = 0 \iff$ (x=0 , x=-3)

$f ' \left(x\right) > 0 \iff 4 {x}^{2} \left(x + 3\right) > 0$ $\iff$ $x$$\in$$\left(- 3 , 0\right) \cup \left(0 , + \infty\right)$

and $f$ is continuous at ${x}_{0} = 0$ so $f$ is strictly increasing in $\left[- 3 , + \infty\right)$

$f ' \left(x\right) < 0 \iff 4 {x}^{2} \left(x + 3\right) < 0$ $\iff$ $x$$\in$$\left(- \infty , - 3\right)$

so $f$ is strictly decreasing in $\left(- \infty , - 3\right]$

• $x < - 3$ $\implies$ $f \left(x\right) > f \left(- 3\right) = - 39$ - ($f$ decreasing)
• $x \ge - 3$ $\implies$ $f \left(x\right) \ge f \left(- 3\right) = - 39$ - ($f$ increasing)

As a result $f$ has relative minimum at ${x}_{0} = - 3$ , $f \left(- 3\right) = - 39$ which also is an global minimum.
(you can mention that $f \left(0\right)$ is a saddle point.)