Calculus: Early Transcendentals 8th Edition

(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 $x \gt 0$
(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 $x \gt 0$