Question

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

Answer 1

They occur at `x=\pm\frac{1}{\sqrt{3}}`. On the graph of `f`, they are at the points `(x,y)=(\pm\frac{1}{\sqrt{3}},f(\pm\frac{1}{\sqrt{3}}))=(\pm\frac{1}{\sqrt{3}},31/9)`.

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