## Calculus: Early Transcendentals 8th Edition

$f$ is not differentiable at $x = 6$ $f'(x) = -1~~~~$ if $x \lt 6$ $f'(x) = 1~~~~~$ if $x \gt 6$ We can see a sketch of $f'(x)$ below.
$f(x) = \vert x-6 \vert$ $f'(a) = \lim\limits_{x \to a}\frac{\vert x-6 \vert-\vert a-6 \vert}{x-a}$ $\lim\limits_{x \to 6^-}\frac{\vert x-6 \vert-\vert 6-6 \vert}{x-6}$ $=\lim\limits_{x \to 6^-}\frac{-(x-6)-0}{x-6}$ $= -1$ $\lim\limits_{x \to 6^+}\frac{\vert x-6 \vert-\vert 6-6 \vert}{x-6}$ $=\lim\limits_{x \to 6^+}\frac{(x-6)-0}{x-6}$ $= 1$ Since the left limit does not equal the right limit at $x = 6$, $f'(6)$ does not exist. Then $f$ is not differentiable at $x = 6$ We can find an expression for $f'(a)$ when $a \lt 6$: $\lim\limits_{x \to a}\frac{\vert x-6 \vert-\vert a-6 \vert}{x-a}$ $=\lim\limits_{x \to a}\frac{-(x-6)-(6-a)}{x-a}$ $=\lim\limits_{x \to a}\frac{a-x}{x-a}$ $= -1$ We can find an expression for $f'(a)$ when $a \gt 6$: $\lim\limits_{x \to a}\frac{\vert x-6 \vert-\vert a-6 \vert}{x-a}$ $=\lim\limits_{x \to a}\frac{(x-6)-(a-6)}{x-a}$ $=\lim\limits_{x \to a}\frac{x-a}{x-a}$ $= 1$ Therefore: $f'(x) = -1~~~~$ if $x \lt 6$ $f'(x) = 1~~~~~$ if $x \gt 6$ We can see a sketch of $f'(x)$ below.