# How do I find concavity and points of inflection for y = 3x^5 - 5x^3?

Mar 2, 2016

At $\left(0 , 0\right)$ There is point of inflection
At $\left(1 , - 2\right)$ The curve is concave upwards.
At $\left(- 1 , 2\right)$ The curve is concave downwards.

#### Explanation:

Given -

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

Find the first derivative -

$\frac{\mathrm{dy}}{\mathrm{dx}} = 15 {x}^{4} - 15 {x}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 15 {x}^{4} - 15 {x}^{2} = 0$

Then -

$15 {x}^{2} \left({x}^{2} - 1\right) = 0$
$15 {x}^{2} = 0$
$x = 0$

${x}^{2} - 1 = 0$
${x}^{2} = 1$
$x = \pm 1$
$x = 1$
$x = - 1$

Find the second derivative -

$\frac{{d}^{2} x}{{\mathrm{dx}}^{2}} = 60 {x}^{3} - 30 x$

At $x = 0$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 60 {\left(0\right)}^{3} - 30 \left(0\right) = 0$

The value of the function -

$y = 3 {\left(0\right)}^{5} - 5 {\left(0\right)}^{3} = 0$

At $\left(0 , 0\right)$

dy/dx=0; (d^2y)/(dx^2)=0

Hence there is a point of inflection at $\left(0 , 0\right)$

At $x = 1$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 60 {\left(1\right)}^{3} - 30 \left(1\right) = 60 - 30 = 30 > 0$

The value of the function -

$y = 3 {\left(1\right)}^{5} - 5 {\left(1\right)}^{3} = 03 - 5 = - 2$

At $\left(1 , - 2\right)$

dy/dx=0; (d^2y)/(dx^2)>0

Hence there is a minimum at $\left(1 , - 2\right)$
The curve is concave upwards

At $x = - 1$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 60 {\left(- 1\right)}^{3} - 30 \left(- 1\right) = - 60 + 30 = - 30 < 0$

The value of the function -

$y = 3 {\left(- 1\right)}^{5} - 5 {\left(- 1\right)}^{3} = - 3 + 5 = 2$

At $\left(1 , 2\right)$

dy/dx=0; (d^2y)/(dx^2)<0

Hence there is a Maximum at $\left(1 , 2\right)$
The curve is concave downwards.