# How do you find inflection points f(x) = x^4 - 2x^2 + 4?

They occur at $x = \setminus \pm \setminus \frac{1}{\setminus \sqrt{3}}$. On the graph of $f$, they are at the points $\left(x , y\right) = \left(\setminus \pm \setminus \frac{1}{\setminus \sqrt{3}} , f \left(\setminus \pm \setminus \frac{1}{\setminus \sqrt{3}}\right)\right) = \left(\setminus \pm \setminus \frac{1}{\setminus \sqrt{3}} , \frac{31}{9}\right)$.
The first derivative is $f ' \left(x\right) = 4 {x}^{3} - 4 x$ and the second derivative is $f ' ' \left(x\right) = 12 {x}^{2} - 4 = 4 \left(3 {x}^{2} - 1\right)$. Setting $f ' ' \left(x\right) = 0$ gives possible inflection points at $x = \setminus \pm \setminus \frac{1}{\setminus \sqrt{3}}$. That these are actual inflection points follows from the fact that $f ' ' \left(x\right)$ changes sign at these values of $x$. The corresponding $y$-coordinate of both of the inflection point is $f \left(\setminus \pm \setminus \frac{1}{\setminus \sqrt{3}}\right) = \frac{31}{9}$.