Find all the maxima and inflection points of the following function:
$$f(x)=\frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}$$
Answer is local Max is at $(0,\frac{1}{\sqrt{2\pi}})$
inflection points $(1,\frac{1}{\sqrt{2\pi{e}}})$ & $(1,\frac{-1}{\sqrt{2\pi{e}}})$
If
$$f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}},$$
then
$$f'(x)=-\frac{x}{\sqrt{2\pi}}e^{-\frac{x^2}{2}},$$
and
$$f''(x)=\frac{x^2-1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}.$$
Now as $e^{-\frac{x^2}{2}}>0$ for all $x\in\mathbb{R}$, we get that
$$f'(x)=-\frac{x}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}=0 \iff x=0$$
and
$$f''(x)=\frac{x^2-1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}=0 \iff x^2-1=0 \iff x=\pm 1.$$
It is easy to check that these give a global maximum and two inflection points respectively. Furthermore
$$f(0)=\frac{1}{\sqrt{2\pi}}$$
and
$$f(\pm 1)=\frac{1}{\sqrt{2\pi}}e^{-\frac{(\pm 1)^2}{2}}=\frac{1}{\sqrt{2\pi e}}.$$
Thus the global minimum is at
$$\left(0,\frac{1}{\sqrt{2\pi}}\right)$$
and the inflection points are at
$$\left(-1,\frac{1}{\sqrt{2\pi e}}\right)\text{ and }\left(1,\frac{1}{\sqrt{2\pi e}}\right).$$