Interesting Range

We know that

$|x| = \begin{cases} x &\quad x\geq 0\\ -x&\quad x<0\\ \end{cases}$

Therefore, for $f(x) = x+|x|$ , we have

$f(x) = \begin{cases} 2x & \quad x\geq 0\\ 0 &\quad x < 0\\ \end{cases}$

For the domain $-5\leq x < 0$ , we have only one exact value for the range: $\{0\}$

For the domain $0 \leq x \leq 5$ , we have a straight line, which gives us the range $[0,10]$

Combining these two, the range of $f(x)$ for $-5 \leq x \leq 5$ is $[0,10]$

$a=0,\;b=10,\;a+b=0+10=\boxed{10}$

min value is 0 because f(-5) = -5 + |-5| = 0 max value is 10 because f(5) = 5 + |5| = 10 then [0,10] . a+b = 10