Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$ Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$.
My try :
$$\begin{align}f(-x)&=|-x+a|-|-x+2|+b|-x+5| \\ &=-f(x) \\&= -\big(|x+a|-|x+2|+b|x+5| \big)  \end{align}$$
$$|-x+a|-|-x+2|+b|-x+5|+ \big(|x+a|-|x+2|+b|x+5|) =0$$
Now what do I do?

The function $\phi(x):=|x+a|-|x+2|$ is $\ \equiv a-2$ when $x\gg1$ and is $\ \equiv2-a$ when $x\ll-1$. On the other hand $|x|\gg1$ implies ${\rm sgn}\bigl(b|x+5|\bigr)={\rm sgn}(b)$ when $b\ne0$, which is forbidden for an odd function. It follows that necessarily $b=0$.
The graph of $\phi$ has corners at $x=-a$ and at $x=-2$. In order to make $\phi$ odd we can choose $a=2$, which results in $f(x)=\phi(x)\equiv0$, or we can choose $a=-2$ in the hope of obtaining a nontrivial solution of the problem. This second choice results in
$$f(x)=\phi(x)=|x-2|-|x+2|=|(-x)+2|-|x+2|\ ,$$
which is obviously odd.
It follows that the possible values of $a+b$ are $\pm2$.
