Prove that the distance from the mean to the inflection points is one standard deviation. Prove that the distance from the mean to the inflection points is one standard deviation. 
So I know in order to find critical points one must find the second derivative and set it equal to 0. However I'm confused how to approach this proof. Step by step explanation please! 
 A: Start by focusing on the standard normal distribution.
$\phi\left(x\right):=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^{2}}\tag1$
$\phi'\left(x\right)=\phi\left(x\right)\left(-x\right)\tag2$ 
$\phi''\left(x\right)=\phi'\left(x\right)\left(-x\right)+\phi\left(x\right)\left(-1\right)=\phi\left(x\right)\left(x^{2}-1\right)\tag3$ 
To achieve (2) we apply the chain rule. To achieve (3) we apply the product rule and former result (2).
If $f\left(x\right):=\frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)$
then $$f''\left(x\right)=0\iff\phi''\left(\frac{x-\mu}{\sigma}\right)=0\iff\left(\frac{x-\mu}{\sigma}\right)^{2}=1\iff x=\mu \pm \sigma$$

Personal note: I don't like parameters and always try to avoid them as much as possible.
A: $$f(x)=\frac 1{\sigma \sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$
To simplify things, let $k=\frac 1{\sigma \sqrt{2\pi}}$
$$f'(x)=k\frac{de^{-\frac{(x-\mu)^2}{2\sigma^2}}}{d(-\frac{(x-\mu)^2}{2\sigma^2})}*\frac{d(-\frac{(x-\mu)^2}{2\sigma^2})}{dx} $$
We performed chain rule in the above differentiation, by firstly treating the power of $e$ as a whole variable, and differentiate $e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ to give itself (because the derivative of $e^{f(x)}$ with respect to $f(x)$ is just itself). Then we multiply it with the derivative of the quadratic function with respect to $x$. So we will get
$$f'(x)= k\frac{-2(x-\mu)}{2\sigma^2}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$
Now the second derivative of the function is,
$$f''(x)=\frac{-1}{\sigma^3 \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} +\frac{(\mu-x)^2}{\sigma^5 \sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$
Can you try differentiating it using chain rule, like what I did above, and product rule? 
After that, it is easy to solve:
$f''(x)=0$ iff
$$(\mu -x)^2 -\sigma^2=0$$
$$x=\mu \pm \sigma$$
