# How do you find the x and y coordinates of all inflection points f(x) = x^4 - 12x^2?

Sep 3, 2015

The coordinates of the two inflection points are $\left(x , y\right) = \left(\pm \sqrt{2} , - 20\right)$

#### Explanation:

The first derivative is $f ' \left(x\right) = 4 {x}^{3} - 24 x$ and the second derivative is $f ' ' \left(x\right) = 12 {x}^{2} - 24 = 12 \left({x}^{2} - 2\right)$.

The second derivative is zero only at $x = \pm \sqrt{2}$ and, in fact, changes sign as $x$ increases through these two values. Therefore the $x$-coordinates of the two inflection points are $x = \pm \sqrt{2}$.

Since $f \left(\pm \sqrt{2}\right) = 4 - 12 \cdot 2 = 4 - 24 = - 20$, it follows that the coordinates of the two inflection points are $\left(x , y\right) = \left(\pm \sqrt{2} , - 20\right)$.

Here's the graph. See if you can find the inflection points in the graph:

graph{x^4-12x^2 [-10, 10, -40,40]}