# What are the points of inflection of f(x)=x^{2}e^{1 -x^2} ?

May 2, 2018

Points of inflection:
$\frac{1}{2} \sqrt{\sqrt{17} + 5}$
$- \frac{1}{2} \sqrt{\sqrt{17} + 5}$
1/2 sqrt(5 - sqrt(17)
-1/2 sqrt(5 - sqrt(17)

#### Explanation:

We find that we have possible points of inflection wherever $f ' ' \left(x\right) = 0$. So we need to find the second derivative. Note that we need to use both the power rule and the chain rule to evaluate this derivative.

Before we begin, note that the derivative of ${e}^{1 - {x}^{2}} = - 2 x {e}^{1 - {x}^{2}}$. This is calculated using the chain rule on the term ${e}^{f \left(x\right)}$, the derivative of which is $f ' \left(x\right) \cdot {e}^{f \left(x\right)}$.

$f \left(x\right) = {x}^{2} {e}^{1 - {x}^{2}}$
$f ' \left(x\right) = {x}^{2} \left(- 2 x\right) {e}^{1 - {x}^{2}} + 2 x {e}^{1 - {x}^{2}} = {e}^{1 - {x}^{2}} \left(2 x - 2 {x}^{3}\right)$
$f ' ' \left(x\right) = {e}^{1 - {x}^{2}} \left(2 - 6 {x}^{2}\right) + \left(- 2 x\right) {e}^{1 - {x}^{2}} \left(2 x - 2 {x}^{3}\right)$
$= {e}^{1 - {x}^{2}} \left(2 - 6 {x}^{2} - 4 {x}^{2} + 4 {x}^{4}\right) = {e}^{1 - {x}^{2}} \left(4 {x}^{4} - 10 {x}^{2} + 2\right)$

We now wish to find what values of $x$ makes $f ' ' \left(x\right) = {e}^{1 - {x}^{2}} \left(4 {x}^{4} - 10 {x}^{2} + 2\right) = 0$.

We note that ${e}^{1 - {x}^{2}}$ is never zero, so we really only need to find where $4 {x}^{4} - 10 {x}^{2} + 2 = 0$.

$4 {x}^{4} - 10 {x}^{2} + 2 = 0 \implies 2 {x}^{4} - 5 {x}^{2} + 1 = 0$
$\implies {x}^{4} - \frac{5}{2} {x}^{2} + \frac{1}{2} = 0$

This form is similar enough to a quadratic that we can use the quadratic formula, with a slight modification.

$x = \frac{\frac{5}{2} \pm \sqrt{\frac{25}{4} - \frac{8}{4}}}{2} = \frac{5 \pm \sqrt{17}}{4}$

Instead of writing with the $\left(x - {r}_{1}\right) \left(x - {r}_{2}\right)$ form (as we would usually do after applying the quadratic formula), we instead write it in the form $\left({x}^{2} - {r}_{1}\right) \left({x}^{2} - {r}_{2}\right)$.

This gives $\left({x}^{2} - \frac{5}{4} - \frac{\sqrt{17}}{4}\right) \left({x}^{2} - \frac{5}{4} + \frac{\sqrt{17}}{4}\right) = 0$. Since we now have a product, we compute when each term equals zero.

${x}^{2} - \frac{5}{4} - \frac{\sqrt{17}}{4} = 0 \implies {x}^{2} = \frac{\sqrt{17}}{4} + \frac{5}{4} \implies x = \pm \sqrt{\frac{\sqrt{17}}{4} + \frac{5}{4}}$

${x}^{2} - \frac{5}{4} + \frac{\sqrt{17}}{4} = 0 \implies {x}^{2} = \frac{5}{4} - \frac{\sqrt{17}}{4} \implies x = \pm \sqrt{\frac{5}{4} - \frac{\sqrt{17}}{4}}$

This yields four values of $x$ such that $f ' ' \left(x\right) = 0$. These values are $x = \sqrt{\frac{\sqrt{17}}{4} + \frac{5}{4}} , - \sqrt{\frac{\sqrt{17}}{4} + \frac{5}{4}} , \sqrt{\frac{5}{4} - \frac{\sqrt{17}}{4}} , - \sqrt{\frac{5}{4} - \frac{\sqrt{17}}{4}}$.

What remains is testing whether these are points of inflection by making sure $f ' ' \left(x\right)$ changes sign as we transition from each $x$. I leave this exercise to the reader. (Typically helps to draw a number line.) These answers can be slightly simplified, as seen above in the answer.