Answer
(a) We can see a sketch of $g(x)$ below.
(b) $g'(x)$ is differentiable for all $x$ except for $x=0$
(c) $g'(x) = 0~~~~$ if $x \lt 0$
$g'(x) = 2~~~~$ if $x \gt 0$
Work Step by Step
(a) $g(x) = x + \vert x \vert$
Then:
$g(x) = x+(-x) = 0~~~~$ if $x \lt 0$
$g(x) = x+x = 2x~~~~$ if $x \geq 0$
We can see a sketch of $g(x)$ below.
(b) Suppose $a \lt 0$:
$g'(a) = \lim\limits_{x \to a}\frac{x+\vert x \vert-(a+\vert a \vert)}{x-a}$
$= \lim\limits_{x \to a}\frac{x+(-x)-(a+(-a))}{x-a}$
$= \lim\limits_{x \to a}\frac{0}{x-a}$
$= 0$
Suppose $a \gt 0$:
$g'(a) = \lim\limits_{x \to a}\frac{x+\vert x \vert-(a+\vert a \vert)}{x-a}$
$= \lim\limits_{x \to a}\frac{x+(x)-(a+(a))}{x-a}$
$= \lim\limits_{x \to a}\frac{2x-2a}{x-a}$
$= \lim\limits_{x \to a}\frac{2(x-a)}{x-a}$
$= 2$
Suppose $a = 0$:
$g'(a) = \lim\limits_{x \to a^+}\frac{x+\vert x \vert-(a+\vert a \vert)}{x-a}$
$= \lim\limits_{x \to 0^+}\frac{x+\vert x \vert-(0+\vert 0 \vert)}{x-0}$
$= \lim\limits_{x \to 0^+}\frac{x+x}{x}$
$= 2$
$g'(a) = \lim\limits_{x \to a^-}\frac{x+\vert x \vert-(a+\vert a \vert)}{x-a}$
$= \lim\limits_{x \to 0^-}\frac{x+\vert x \vert-(0+\vert 0 \vert)}{x-0}$
$= \lim\limits_{x \to 0^-}\frac{x-x}{x}$
$= 0$
Since the left limit is not equal to the right limit as $x \to 0$, this limit does not exist.
$g'(x)$ is differentiable for all $x$ except for $x=0$
(c) As shown above:
$g'(x) = 0~~~~$ if $x \lt 0$
$g'(x) = 2~~~~$ if $x \gt 0$
