# How do you find all points of inflection given y=x^4-3x^2?

Aug 10, 2017

Inflection points at $x = \pm \frac{1}{\sqrt{2}}$.

#### Explanation:

Find the second derivative.

$y ' = 4 {x}^{3} - 6 x$

$y ' ' = 12 {x}^{2} - 6$

Inflection points occur when $y ' ' = 0$.

$0 = 12 {x}^{2} - 6$

$0 = 6 \left(2 {x}^{2} - 1\right)$

$x = \pm \frac{1}{\sqrt{2}}$

If we select test points, we see that the sign of the second derivative does indeed change at $x = - \frac{1}{\sqrt{2}}$ and $x = + \frac{1}{\sqrt{2}}$.

Therefore, $x = - \frac{1}{\sqrt{2}}$ and $x = + \frac{1}{\sqrt{2}}$ are both inflection points.

Hopefully this helps!

Aug 10, 2017

Point of infection

$\left(0.7 , - 1.23\right)$
$\left(- 0.7 , - 1.23\right)$

#### Explanation:

Given -

$y = {x}^{4} - 3 {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} - 6 x$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 12 {x}^{2} - 6$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0 \implies 12 {x}^{2} - 6 = 0$

To find the point of inflection, set the second derivative equal to zero

${x}^{2} = \frac{6}{12} = \frac{1}{2} = 0.5$
$x = \pm \sqrt{0.5} = 0.7$

At $x = 0.7$
At $x = - 0.7$

$y = {0.7}^{4} - 3 {\left(0.7\right)}^{2}$

$y = 0.2401 - 1.47 = - 1.23$
$y = - 1.23$

Point of infection

$\left(0.7 , - 1.23\right)$

$y = 0.2401 - 1.47 = - 1.23$
$y = - 0.5096$
$y = - 1.23$
Point of infection
$\left(- 0.7 , - 1.23\right)$