# What are the points of inflection, if any, of f(x)= x^4-6x^3 ?

Oct 20, 2016

There exists a point of inflection at $\left(0 , 0\right)$

#### Explanation:

Points of inflections exist when $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 0$ or when $f ' ' \left(x\right) = 0$
We start by differentiating the function
$f \left(x\right) = {x}^{4} - 6 {x}^{3}$ we use $\frac{d \left({x}^{n}\right)}{\mathrm{dx}} = n {x}^{n - 1}$
So $f ' \left(x\right) = 4 {x}^{3} - 18 {x}^{2}$
$f ' \left(x\right) = 0$ when $4 {x}^{3} - 18 {x}^{2} = 0$
$2 {x}^{2} \left(2 x - 9 = 0\right)$ => $x = 0$ and $x = \frac{9}{2}$
Then we calculate $f ' ' \left(x\right) = 12 {x}^{2} - 36 x$
we continue with the values of x obtained above
$f ' ' \left(0\right) = 0$ which is a point of inflection
$f ' ' \left(\frac{9}{2}\right) = 12 \cdot {\left(\frac{9}{2}\right)}^{2} - 36 \cdot \left(\frac{9}{2}\right) = 12 \cdot \frac{9}{4} - 18 \cdot 9 = 27 - 162 = - 135$ which is $< 0$ and is a minimum