$$\begin{align} f(x):= \underbrace{{1 \over \sqrt{2\pi}\sigma }\exp \left(- {(x-\mu)^2 \over 2\sigma^2 } \right)}_{~~\text{pdf of normal distribution}~~ } ~~\text{for}~~(-\infty <x < \infty) \end{align}$$
I want to prove that the inflection points of x-corrdinatess are $~ \mu\pm\sigma ~$
$$\begin{align} {\mathrm{d}f \over \mathrm{d}x }&=- {(x-\mu) \over \sigma^2 }f(x)\\ {\mathrm{d^2}f \over \mathrm{d}x^2 }&=- {1 \over \sigma^2 } \left\{ 1- \left({x-\mu \over \sigma } \right)^2 \right\} f(x)\\ \end{align}$$
$~ \mu\pm\sigma ~$ are the only values each of which makes $~ f''(x) ~$ zero since $~ f(x)\in\mathbb{R_{>0}} ~$ is held for any $~ x ~$
Intuitively, it can be strongly asserted that $~ f(x)|_{x\in[\mu-\sigma,~\mu+\sigma]} ~$ is a concave function from the graphs below.
Currently I've been struggling to understand that $~ f(x)|_{x<\mu-\sigma,~x>\mu+\sigma}~$is a convex function,since slopes of it seem too shallow.
Can anyone give me what I've been missing or some advice about it?
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I've derived that $~f''\big|_{x>\mu+\sigma}>0~$ is held and this suggests that $~f''\big|_{x<\mu-\sigma}>0~$ is also satisfied because of a symmetry.
And the remaining problem is to derive $f''\big|_{x\in[\mu-\sigma,~\mu+\sigma]}<0~$is held.
Apply $~x=\mu+\sigma+c~~\text{where}~~c\in\mathbb R_{>0} $ to the right side of $~ f(x) ~$
This can easily prove $~ f''\big|_{x>\mu+\sigma}>0 ~$
And about the concave section, apply $~x=\mu+\sigma-c~~\text{where}~~c\in\mathbb R_{>0}~\bigcap ~\mathbb R_{\leq \mu+\sigma} ~$
This can prove $~ f''\big|_{0\leq x<\mu+\sigma}<0 ~$
Since $~ f(x) ~$is an even function, finding out the inflection point of positive x-axis side is enough to achieve the given task.