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Let $\mu\in\mathbb R$ and $\sigma >0$. Determine the extrema of the Gauß function $\varphi$ and explain whether they are minima or maxima where $$ \varphi:\mathbb R\to \mathbb R,\qquad \varphi(x)=\frac{1}{\sqrt{2\pi}\sigma}\cdot\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right). $$

So far I got $$ \varphi'(x)=\frac{(\mu-x)}{\sqrt{2\pi}\sigma^3}\cdot\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) $$ and $x=\mu$ as the only root of $\varphi(x)$. To find out whether this is a maximum or minimum I computed $$ \varphi''(x)=\frac{-2\mu x+x^2+\mu^2-\sigma^2}{\sqrt{2\pi}\sigma^5}\cdot\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) $$ but $\varphi''(\mu)=0$ which makes this difficult because we learned the following for any two times continuously differentiable function $f:(a,b)\to\mathbb R$ with $x\in(a,b)$: $$\begin{align*} f'(x)=0,\quad f''(x)>0\qquad&\implies\qquad x \text{ is a local minimum}\\ x \text{ is a local minimum}\quad&\implies\quad f'(x)=0,\quad f''(x)\ge 0\\ f'(x)=0,\quad f''(x)<0\qquad&\implies\qquad x \text{ is a local maximum}\\ x \text{ is a local maximum}\quad&\implies\quad f'(x)=0,\quad f''(x)\le 0\\ \end{align*}$$

I know the $\varphi$ function from probability theory and therefore $x=\mu$ must be a maximum, however I do not know what other approaches could be useful to show that this indeed is a maximum.

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up vote 2 down vote accepted

Look for the sign of $\varphi^{'}(x)$ before and after the extrema point. It will tell you about the growth of $\varphi$.

Here, we can see that $\varphi^{'}(x)>0$ for some $x<\mu$ and so $\varphi$ is increasing to the left of $\mu$.

Similarly, $\varphi$ is decreasing near the right of $\mu$ therefore, $\mu$ must be a maximum (local).

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