# What are the points of inflection, if any, of f(x)=x^4 ?

Jan 1, 2016

The function has no points of inflection.

#### Explanation:

A point of inflection occurs when $f ' ' \left(x\right)$ switches from positive to negative, or vice versa. This correlates to $f \left(x\right)$ switching concavity.

Possible points of inflection occur when $f ' ' \left(x\right) = 0$. So, to find the possible points of inflection, differentiate $f \left(x\right)$ twice.

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

Set $f ' ' \left(x\right) = 0$ to find possible points of inflection.

$12 {x}^{2} = 0$
$x = 0$

Create a sign chart to determine if the second derivative switches signs when $x = 0$.

$\textcolor{w h i t e}{- - - - - - - - - - -} 0$
$f ' ' \left(x\right) \textcolor{w h i t e}{- - -} \leftarrow - - - - - - - - - - \rightarrow$
$\textcolor{w h i t e}{- - - - - -} \text{POSITIVE"color(white)(---)"POSITIVE}$

$f ' ' \left(x\right)$ doesn't change signs at $x = 0$, so there's not a point of inflection when $x = 0$. Since $x = 0$ was the only possible point, we know that ${x}^{4}$ has no points of inflection.

graph{x^4 [-10.04, 9.96, -2.2, 7.8]}