# How do you find the points of inflection of the curve y=e^(x^2)?

Mar 19, 2018

None, it's concave up for $\left(- \infty , \infty\right)$.

#### Explanation:

Compute the first derivative.

$y ' = 2 x {e}^{{x}^{2}}$

Second derivative is given by the product rule.

$y ' ' = 2 \left({e}^{{x}^{2}}\right) + 2 x \left(2 x\right) {e}^{{x}^{2}}$

$y ' ' = 2 {e}^{{x}^{2}} + 4 {x}^{2} {e}^{{x}^{2}}$

We need to set this to $0$ and solve to determine inflection points.

$0 = 2 {e}^{{x}^{2}} + 4 {x}^{2} {e}^{{x}^{2}}$

$0 = 2 {e}^{{x}^{2}} \left(1 + 2 {x}^{2}\right)$

We see this has no solution because $2 {e}^{{x}^{2}} \ne 0$ for all values of $x$. Furthermore, the second equation states that $1 + 2 {x}^{2} = 0 \to 2 {x}^{2} = - 1 \to x = \sqrt{- \frac{1}{2}}$

This doesn't have a real value so no real solution to this equation. This simply means the function $y = {e}^{{x}^{2}}$ will have no inflection points ($y ' '$ is positive on all it's domain, therefore concave up on $\left(- \infty , \infty\right)$). We can even confirm graphically. Hopefully this helps!

Mar 19, 2018

In principle, by differentiating twice, setting the result to zero, and checking whether the result is a genuine point of inflexion. However, this function has no points of inflexion.

#### Explanation:

Differentiating twice gives $f ' ' \left(x\right) = 2 x \cdot 2 x {e}^{{x}^{2}} + 2 \cdot {e}^{{x}^{2}}$
$= 2 {e}^{{x}^{2}} \left(2 {x}^{2} + 1\right)$
This can never be zero because all three terms of the product are always strictly positive

Did you mean ${e}^{- {x}^{2}}$? This has points of inflection when $2 {x}^{2} - 1 = 0$, that is, x=±1/sqrt2

You should then test that the second derivative changes sign at these points, which it clearly does as $2 {x}^{2} - 1$ is a parabola passing through through (±1/sqrt2,0), and the exponential term is always positive.