# How do you find all points of inflection given y=x^5-2x^3?

Nov 10, 2016

There is just one point of inflection at the origin

#### Explanation:

$y = {x}^{5} - 2 {x}^{3}$

Differentiating wrt $x$ we have'

$\frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{4} - 6 {x}^{2}$

At a critical point , $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 5 {x}^{4} - 6 {x}^{2} = 0$
$\therefore {x}^{2} \left(5 {x}^{2} - 6\right) = 0$
So, Either ${x}^{2} = 0 \implies x = 0$, or $5 {x}^{2} - 6 = 0 \implies x = \pm \sqrt{\frac{6}{5}}$

So critical points occurs when $x = 0 , x = \pm \sqrt{\frac{6}{5}}$

Next we can find the nature of th critical points by looking at the second derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{4} - 6 {x}^{2}$
$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 20 {x}^{3} - 12 x$
$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 4 x \left(5 {x}^{2} - 3\right)$

When  { (x=-sqrt(6/5), => (d^2y)/(dx^2)<0,=>,"maximum"), (x=0, => (d^2y)/(dx^2)=0,=>,"inflection"), (x =sqrt(6/5), => (d^2y)/(dx^2)>0,=>,"minimum") :}

So therei is one point of inflection when $x = 0 \implies y = 0$