# What are the points of inflection, if any, of f(x) =x^3 - 12x^2 ?

May 22, 2017

$x = 4$

#### Explanation:

Points of inflection are where $f ' ' \left(x\right) = 0$ or when $f ' ' \left(x\right)$ instantaneously changes from positive to negative or vice versa.

$f \left(x\right) = {x}^{3} - 12 {x}^{2}$ is a polynomial -- it is a smooth curve so we don't have to worry about instantaneous jumps (e.g. asymptotes).

All we have to do is differentiate twice and set that equal to $0$.

$f ' \left(x\right) = 3 {x}^{2} - 24 x$

$f ' ' \left(x\right) = 6 x - 24$

Now set $f ' ' \left(x\right)$ equal to $0$.

$6 x - 24 = 0$
$6 x = 24$
$x = 4$

This means that there is one point of inflection on the graph of $f \left(x\right)$ and it is $x = 4$.