# Inflection point confusion

## Homework Statement

$$f(x) = x^{4} - 2x^{2} + 3$$

Find the intervals of concavity and the inflection points.

## Homework Equations

$$f''(x) = 4(3x^{2}-1)$$

## The Attempt at a Solution

$$f''(x)$$ is zero at $$\pm\frac{1}{\sqrt{3}}$$
I've found the correct intervals of concavity, which are $$(-\infty, -\frac{1}{\sqrt{3}}) \cup (\frac{1}{\sqrt{3}}, \infty)$$
I would expect the inflection points to be $$\pm\frac{1}{\sqrt{3}}$$, which is partly correct, but the answer the book gives is $$\pm\frac{1}{\sqrt{3}}, \frac{22}{9}$$.
I can't see how $$\frac{22}{9}$$ could be an inflection point. It does not equal zero when plugged in to the second derivative.

Thanks

HallsofIvy
22/9 isn't an "inflection point" but then neither are $1/\sqrt{3}$ nor $1/\sqrt{3}$! They are not points! What your book is saying is that $\left(1/\sqrt{3}, 22/9\right)$ and $\left(-1/\sqrt{3}, 22/9\right)$ are the inflection points. That is, when x is $1/\sqrt{3}$ or $-1/\sqrt{3}$, y is equal to 22/9.
22/9 isn't an "inflection point" but then neither are $1/\sqrt{3}$ nor $1/\sqrt{3}$! They are not points! What your book is saying is that $\left(1/\sqrt{3}, 22/9\right)$ and $\left(-1/\sqrt{3}, 22/9\right)$ are the inflection points. That is, when x is $1/\sqrt{3}$ or $-1/\sqrt{3}$, y is equal to 22/9.