Maximize this parametric function

Question

I have a question:

For what values of $a>0$ is $$\frac1{1+\vert x\vert}+\frac1{1+\vert x-a\vert}$$ is maximized?

I have no idea on how to work with derivatives of absolute values. Any help is appreciated. Thank you all.

Answer 1

Set $$f(x)=\frac1{1+\vert x\vert}+\frac1{1+\vert x-a\vert}$$ And split it in the following way $$f(x)=\begin{cases} \frac1{1-x}+\frac1{1-x+a} &\text{if $x<0$}\\ \frac1{1+x}+\frac1{1-x+a} &\text{if $0\leq x<a$}\\ \frac1{1+x}+\frac1{1+x-a} &\text{if $x\geq a$}. \end{cases}$$ and calculating the first derivative you obtain $$f'(x)=\begin{cases} \frac{1}{(a-x+1)^2}+\frac{1}{(1-x)^2} &\text{if $x<0$}\\ \frac{1}{(a-x+1)^2}-\frac{1}{(x+1)^2} &\text{if $0< x<a$}\\ -\frac{1}{(-a+x+1)^2}-\frac{1}{(x+1)^2} &\text{if $x> a$}. \end{cases}$$ You can immediately notice that $f(x)$ is increasing for $x<0$ since $$\left\{\frac{1}{(a-x+1)^2}+\frac{1}{(1-x)^2}>0\,\Big{|}\,\forall x \in \mathbb{R}\right\}.$$ Also, $$\frac{1}{(a-x+1)^2}-\frac{1}{(x+1)^2}=\frac{-a^2+2 a x-2 a+4 x}{(x+1)^2 (a-x+1)^2}$$ which is negative for $x<a/2$, therefore the function $f(x)$ is decreasing in $\left(0,\frac{a}{2}\right)$. This implies that $x=0$ is a maximum. Also, being that $f(x)$ is symmetric with respect to $x=a/2$, there is another maximum point, with the same value of the one found previously, at $x=a$.The function at these critical points has value: $$f(0)=f(a)=\frac{2+a}{1+a}.$$