Find the minimum and maximum of $h(x) = \dfrac{1}{1+|x|}+\dfrac{1}{1+|x-a_1|}$ 
Let $a_1 \in \mathbb{R}$. Find the minimum and maximum of $h(x) = \dfrac{1}{1+|x|}+\dfrac{1}{1+|x-a_1|}$.

This question seems hard to solve since we don't know what $a_1$ is. We want both $|x|,|x-a_1|$ to be as small as possible or as large as possible to find the maximum or minimum, so how do I find where that is possible? Maybe it might occur at the median or some other point.
 A: Set $b=a_1/2$ and the translation $x\mapsto x-b$, $y\mapsto y$. This leads to considering the function
$$
f(x)=\frac{1}{1+|x-b|}+\frac{1}{1+|x+b|}
$$
and it's not restrictive to assume $b\ge0$. Leave out the case $b=0$, for the moment. The function is even, so we just need to study it for $x\ge0$ and we can write it as
$$
f(x)=\begin{cases}
\dfrac{1}{1+b-x}+\dfrac{1}{1+x+b} & \text{if $0\le x<b$}\\[6px]
\dfrac{1}{1+x-b}+\dfrac{1}{1+x+b} & \text{if $x\ge b$}
\end{cases}
$$
You can see that $f(0)=2/(1+b)$ and that
$$
\lim_{x\to\infty}f(x)=0
$$
Thus there is no absolute minimum, because $f(x)>0$ for all $x$. We also have
$$
f(b)=1+\frac{1}{1+2b}
$$
The derivative is
$$
f'(x)=\begin{cases}
\dfrac{1}{(1+b-x)^2}-\dfrac{1}{(1+b+x)^2} & \text{if $0\le x<b$}\\[6px]
-\dfrac{1}{(1-b+x)^2}-\dfrac{1}{(1+b+x)^2} & \text{if $x>b$}
\end{cases}
$$
Note that $f$ is decreasing in the interval $(b,\infty)$ (and not differentiable at $b$). In the interval $[0,b)$ we can write
$$
f'(x)=\frac{4x(1+b)}{(1+b-x)^2(1+b+x)^2}
$$
so the function has zero derivative at $0$ and is increasing in the interval $(0,b)$.
This should be enough to finish up.
The case $b=0$ is quite easy:
$$
\frac{2}{1+|x|}\le 2
$$
for all $x$, equality only for $x=0$.
A: Obviously $h(x) > 0$, with $h(x) \to 0$ as $x \to \pm \infty$.  Thus the infimum is $0$, but there is no minimum.
Candidates for the maximum are the two points ($0$ and $a_1$) where the function is non-differentiable, plus any points where the derivative is $0$.  However, the function is convex on intervals not containing $0$ and $a_1$, so points where the derivative is $0$ would be local minima, not maxima.  
A: WLOG, $a_1=2$.
The domain can be decomposed in three pieces:
$$x\le0\implies f(x)=\frac1{1-x}+\frac1{3-x},\\
0\le x\le2\implies f(x)=\frac1{1+x}+\frac1{3-x}=\frac4{(1+x)(3-x)},\\
2\le x\implies f(x)=\frac1{1+x}+\frac1{x-1}.$$
The first (magenta) and third (black) pieces have a derivative which is the sum of two squares and doesn't cancel. The second piece (light green) is the inverse of a downward parabola and has a local minimum at $(1,1)$.
The junction points form two maxima (angular points), at $(0,\frac43)$ and $(2,\frac43)$.

For other values of $a_1$, stretch the plot horizontally.
