Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 49

Prove that the function $f(x) = |x-6|$ is not differentiable at 6. Find formula for $f'$ and sketch its graph.

From the definition of absolute value,

$
f(x) = |x-6|= \left\{
\begin{array}{c}
x-6 & \text{if} & x > 0\\
-(x-6) & \text{if} & x < 0
\end{array}\right.
\qquad
\Longrightarrow
\qquad
f(x) = \left\{
\begin{array}{c}
x - 6 & \text{if} & x > 0\\
-x+6 & \text{if} & x < 0
\end{array}\right.
$


From the definition of derivative,
$\displaystyle f'_+ (x) = \lim\limits_{h \to 0^+} \frac{f(x+h) - f(x)}{h} \qquad \text{ and }
\qquad f'_- (x) = \lim\limits_{h \to 0^-} \frac{f(x+h)-f(x)}{h}$

For Right Hand,

$
\begin{equation}
\begin{aligned}
f'_+ (x) & = \lim\limits_{h \to 0^+} \frac{x+h-6-(x-6)}{h}\\
f'_+ (x) & = \lim\limits_{h \to 0^+} \frac{\cancel{-6}-\cancel{x}+\cancel{6}}{h}\\
f'_+ (x) & = \lim\limits_{h \to 0^+} \frac{\cancel{h}}{\cancel{h}}\\
f'_+ (x) & = \lim\limits_{h \to 0^+} 1\\
f'_+ (6) & = 1
\end{aligned}
\end{equation}
$


For Left Hand,

$
\begin{equation}
\begin{aligned}
f'_- (x) & = \lim\limits_{h \to 0^-} \frac{-(x+h)+6-(-x+6)}{h}\\
f'_- (x) & = \lim\limits_{h \to 0^-} \frac{\cancel{-x}-h+\cancel{6}+\cancel{x}-\cancel{6}}{h}\\
f'_- (x) & = \lim\limits_{h \to 0^-} \frac{\cancel{h}}{\cancel{h}}\\
f'_- (x) & = \lim\limits_{h \to 0^-} -1\\
f'_- (6) & = -1
\end{aligned}
\end{equation}
$


$f'_+ (6) \neq f'_- (6)$; Therefore, $f$ is not differentiable at 6.
We can find the formula for $f'$ using $f'_- (x) = -1$ and $f'(x) = 1$
Therefore, we can rewrite $f'(x)$ as

$
f'(x) = \left\{
\begin{array}{c}
1 & \text{if} & x \geq 6\\
-1 & \text{if} & x < 6
\end{array}\right.
$