Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 52

According to the definition, the left hand and right hand derivatives of $f$ at $a$ are defined by



$\displaystyle f'_- (a) = \lim\limits_{h \to 0^-} \frac{f(a+h) - f(a)}{h} \qquad \text{ and } \qquad
f'_+ (a) = \lim\limits_{h \to 0^+} \frac{f(a+h)-f(a)}{h} $


Suppose that these limits exists.
a.) Determine $f'_- (4)$ and $f'_+ (4)$ for the function
$
\displaystyle
f(x) = \left\{
\begin{array}{c}
0 & \text{if} & x \leq 0\\
5-x & \text{if} & 0 < x < 4\\
\frac{1}{5-x} & \text{if} & x \geq 4
\end{array}\right.
$

b.) Sketch the graph of $f$
c.) Where is $f$ discontinuous?
d.) Where is $f$ not differentiable?

a.)
For Left Hand,

$
\begin{equation}
\begin{aligned}
f'_- (x) &= \lim\limits_{h \to 0^-} \left[ \frac{5-(x+h)-(5-x)}{h}\right]\\
f'_- (x) &= \lim\limits_{h \to 0^-} \left[ \frac{\cancel{5}-\cancel{x}-h-\cancel{5}+\cancel{x}}{h}\right]\\
f'_- (x) &= \lim\limits_{h \to 0^-} -\frac{\cancel{h}}{\cancel{h}}\\
f'_- (x) &= \lim\limits_{h \to 0^-} -1\\
f'_- (4) &= -1
\end{aligned}
\end{equation}
$


For Right Hand,

$
\begin{equation}
\begin{aligned}
f'_+ (a) &= \lim\limits_{h \to 0^+} \frac{\left[\frac{1}{5-(x+h)}-\left(\frac{1}{5-x}\right)\right]}{h}\\
f'_+ (a) &= \lim\limits_{h \to 0^+} \frac{\frac{5-x-[5-(x+h)]}{(5-(x+h))(5-x)}}{h}\\
f'_+ (a) &= \lim\limits_{h \to 0^+} \frac{\frac{\cancel{5}-\cancel{x}-\cancel{5}+\cancel{x}+h}{(5-(x+h))(5-x)}}{h}\\
f'_+ (a) &= \lim\limits_{h \to 0^+} \frac{\cancel{h}}{(5-x-h)(5-x)\cancel{h}}\\
f'_+ (a) &= \lim\limits_{h \to 0^+} \frac{1}{(5-x-h)(5-x)}\\
f'_+ (a) &= \lim\limits_{h \to 0^+} \frac{1}{(5-x-0)(5-x)}\\
f'_+ (a) &= \lim\limits_{h \to 0^+} \frac{1}{(5-x)^2}\\
f'_+ (4) &= \frac{1}{(5-4)^2} = \frac{1}{(1)1^2} = 1\\
f'_+ (4) & = 1
\end{aligned}
\end{equation}
$

Therefore, $f'(4)$ doesn't exist since $f'_- (4) \neq f'_+ (4)$

b.)



c.) Based from the graph
The function $f$ is discontinuous at $x=0$ because of jump discontinuity.
Also, $f$ is discontinuous at $x=5$ because of infinite discontinuity

d.) The function $f$ is not differentiable at $x=0$ and $x=5$ because the function is discontinuous
at that points. Also, the function is not differentiable at $x=4$ since $f'_- (4) \neq f'_+ (4)$ based
from definition.