Determine all local extremes of $f(x)=\frac{a}{2}\cdot \left(e^{\frac x a}+e^{\frac {-x} a}\right)$

Question

Given is the function $f:[-a,a]\to\mathbb{R},\;a\in\mathbb{R}_{>0}$ with $f(x)=\frac{a}{2}\cdot \left(e^{\frac x a}+e^{\frac {-x} a}\right)$. Determine all local extremes of $f(x)$.

First, we need to find $f'(x)=0$ and decide, if $f''(x_0)<0$, $f''(x_0)>0$ or $f''(x_0)=0$ in order to determine, whether we found minima, maxima or sattle points.

Unfortunately it's hard to find $f'(x)=0$:

$$f'(x)= \frac{\mathrm{e}^{-\frac{x}{a}}\left(\mathrm{e}^\frac{2x}{a}-1\right)}{2},$$ but $$\frac{\mathrm{e}^{-\frac{x}{a}}\left(\mathrm{e}^\frac{2x}{a}-1\right)}{2}=0$$ wouldn't have a solution in $\mathbb{R}$, right? Im stuck here, because I don't seem to find the roots of $f'(x)$. Should we chose a fixed $a$ to get real roots?

Answer 1

Note,

$$f'(x)= \frac{\mathrm{e}^{-\frac{x}{a}}\left(\mathrm{e}^\frac{2x}{a}-1\right)}{2} = \frac{\mathrm{e}^\frac{x}{a}-\mathrm{e}^\frac{x}{a}}{2}=\sinh\frac xa=0$$

which yields $x=0$.

Answer 2

Note that $$e^{x/a}+e^{-x/a}\geq 2\sqrt{e^{x/a}\times e^{-x/a}}$$ by AM-GM

Answer 3

I has solution for $x=0$. $f$ is convex so it is the unique local minimum, $a$ and $-a$ being points of local maximums.