# How do you find all critical point and determine the min, max and inflection given f(x)=x^4?

Oct 26, 2016

$f \left(x\right) = {x}^{4}$

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

A critical number for $f$ is a number $c$ in the domain of $f$ at which $f ' \left(c\right)$ does not exist or $f ' \left(c\right) = 0$

$f ' \left(x\right) = 4 {x}^{3}$ is defined for all $x$ and is $0$ at $x = 0$.

The only critical number for $f$ is $0$.

(If your treatment of calculus says that a critical point is a point on the graph, then the critical point is $\left(0 , 0\right)$)

Because $f ' \left(x\right)$ is negative for $x < 0$ and positive for $x > 0$, we know that $f \left(0\right) = 0$ is a local minimum.

$f ' ' \left(x\right) = 12 {x}^{2}$ which is always positive.

Since the sign of $f ' '$ never changes, the concavity of $f$ never changes, so there are no inflection points.