# How do you find the critical points for f(x)=x^(8) and the local max and min?

Jan 5, 2018

Check below.

#### Explanation:

$f \left(x\right) = {x}^{8}$

${D}_{f} = \mathbb{R}$

$f$ is continuous in $\mathbb{R}$

$f ' \left(x\right) = 8 {x}^{7}$ and

$f ' \left(x\right) = 0 \iff x = 0$
${x}_{0} = 0$ is a critical point.
There is no other critical point because $f '$ is defined in $\mathbb{R}$.

If $x < 0$ then $f ' \left(x\right) < 0$ so $f$ is decreasing in $\left(- \infty , 0\right]$
If $x > 0$ then $f ' \left(x\right) > 0$ so $f$ is increasing in $\left[0 , + \infty\right)$

• $0$$<$$x$ $\iff f \left(x\right) > f \left(0\right) = 0$

• $x \ge 0$ $\iff f \left(x\right) \ge f \left(0\right) = 0$

So for $x$$\in$$\mathbb{R}$: $f \left(x\right) \ge f \left(0\right)$

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