if the mode of a normal distribution is 0, then what's the value of the mean so if the PDF attains its maximum at 0 it means that $f'(0) = 0$: 
$$f'(x) = -\frac{x}{\sigma^2 \sqrt{\pi}}\exp({-\frac{(x - \mu)^2}{2\sigma^2}}) $$
$$f'(0) = 0 \iff 0 =0$$
yep, no valuable information whatsoever,
now onto the second derivative criterion : $f''(0) < 0$ 
$$f''(0) = 0 -\frac{1}{\sigma^2 \sqrt{\pi}}\exp({-\frac{\mu^2}{2\sigma^2}}) < 0$$
again, no valuable information whatsoever.
maybe if we try using the fact that $f(0) \geq f(y), \, \forall y \in \mathbb{R}$ for $y = \mu$
then : 
$$\frac{1}{\sigma\sqrt{2 \pi}}\exp(-\frac{\mu^2}{2\sigma^2}) \geq\frac{1}{\sigma\sqrt{2 \pi}}\iff \exp(-\frac{\mu^2}{2\sigma^2}) \geq1 \iff -\frac{\mu^2}{2\sigma^2} \geq 0 \iff \mu^2 \leq 0$$
therefore $\mu = 0$ since $0$ is the only 'negative' positive number.
I'm pretty confident that my work is correct but I'd like a confirmation, thanks !
 A: Assuming $\sigma>0$, the derivative of $$f(x)=\frac{1}{\sqrt{2\pi}\sigma} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$$ is not quite what you wrote but 
$$f'(x)=-(x-\mu)\frac{1}{\sqrt{2\pi}\sigma^3} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$$
so $f'(x)=0$ when and only when $x-\mu=0$, i.e. $x=\mu$, as the other terms are all positive.
Since you have a probability density function which is differentiable on the whole of $\mathbb R$, this unique zero means that the mode of the distribution must be at $x=\mu$.  You could check the second derivative if you wish, but that is not necessary  
A: We have
$$\begin{align*}
f(x\mid\mu,\sigma^2)
&=\frac{1}{\sqrt{2\pi\sigma^2}}\text{exp}\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\\\\
\end{align*}$$
Taking the log we get
$$\text{log}(1)-\text{log}\left(\sqrt{2\pi\sigma^2}\right)-\frac{(x-\mu)^2}{2\sigma^2}$$
Setting the derivative equal to zero we get
$$\begin{align*}
\frac{d}{dx}\text{log }f(x\mid\mu,\sigma^2)
&=-\frac{(x-\mu)}{\sigma^2}\\\\
&=0\\\\
&\Rightarrow x=\mu\\\\
\end{align*}$$
so the mean and mode are the same for a normal distribution. Hence $\mu=0$. 
For completeness, since 
$$\frac{d^2}{dx^2}\text{log }f(x\mid\mu,\sigma^2)=-\frac{1}{\sigma^2}<0$$
for all $x\in\mathbb{R}$ then we can conclude that this function is concave with a global maximum at $x=\mu$.
A: The mode of any symmetrical, unimodal distribution is always equal to the mean, and also to the median of that distribution. A unimodal distribution is one with only one peak. All normal distributions fall in that category, so their mode and mean are identical. 
