<s>[INST] Tangent line Exercise. $f (x) = 5e^{−(x−2)^2}$ . Find the coordinates of points where the hill is the most steep. I have this exercise, look easy but I don't know where to start, I think that I need a extra function or value to continue with the calculus.

Imagine that you are riding over a hill having its profile given by
  the function $f(x) = 5e^{−(x−2)^2}$ Find the coordinates of points where
  the hill is the most steep.

Can you help me?. The answer is $x = 2 \pm\frac{1}{\sqrt{2}}$ 
 [/INST] Notice, the slope of the tangent to the curve: $f(x)=5e^{-(x-2)^2}$ is given as 
$$f'(x)=\frac{d}{dx}(f(x))=\frac{d}{dx}(5e^{-(x-2)^2})$$ 
$$=5e^{-(x-2)^2}(-2(x-2))$$$$=-10(x-2)e^{-(x-2)^2}$$
$$f''(x)=-10(x-2)\frac{d}{dx}(e^{-(x-2)^2})-10e^{-(x-2)^2}\frac{d}{dx}(x-2)$$
$$=20(x-2)^2e^{-(x-2)^2}-10e^{-(x-2)^2}$$
$$f'''(x)=-40(x-2)^3e^{-(x-2)^2}+60(x-2)e^{-(x-2)^2}$$
Now, the hill will be most steep at the points where the slope $f'(x)$ is maximum i.e. $f''(x)=0$ hence, $$20(x-2)^2e^{-(x-2)^2}-10e^{-(x-2)^2}=0$$
$$10e^{-(x-2)^2}(2(x-2)^2-1)=0$$
but $e^{-(x-2)^2}\ne 0$, hence
$$2(x-2)^2-1=0$$
$$(x-2)^2=\frac 12$$
$$x-2=\pm \frac{1}{\sqrt 2}$$
$$x=2\pm \frac{1}{\sqrt 2}$$
Now, substituting these values of $x$ in $f'''(x)$, one should get 
$$f'''\left(2+\frac{1}{\sqrt 2}\right)>0, \ \ f'''\left(2-\frac{1}{\sqrt 2}\right)<0$$
hence, the slope $f'(x)$ is maximum at the point where $x=2-\frac{1}{\sqrt 2}$,hence corresponding y-coordinate of the point $$f\left(2-\frac{1}{\sqrt 2}\right)=5e^{-\left(2-2+\frac{1}{\sqrt 2}\right)^2}=\frac{5}{\sqrt e}$$
hence, the coordinates of the point where the hill is most steep i.e. slope is maximum is $\color{red}{\left(2-\frac{1}{\sqrt 2}, \frac{5}{\sqrt e}\right)}$
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