Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 51

a.) Illustrate the graph of the function $f(x) = x|x|$.
b.) For what values of $x$ is $f$ differentiable.
c.) Find a formula for $f'$

a.)



b.) Using the definition of absolute value,

$
f(x) = \left\{
\begin{array}{c}
x(x) & \text{if} & x \geq 0\\
x(-x) & \text{if} & x < 0
\end{array}\right.

\qquad
\Longrightarrow
\qquad

f(x) = \left\{
\begin{array}{c}
x^2 & \text{if} & x \geq 0\\
-x^2 & \text{if} & x < 0
\end{array}\right.
$


Let's Check if the function is differentiable at $x=0$
By definition,
$\quad \displaystyle f'_- (x) = \lim\limits_{h \to 0^-} \frac{f(x+h)-f(x)}{h}
\qquad \text{ and } \qquad
\displaystyle f'_+ (x) = \lim\limits_{h \to 0^+} \frac{f(x+h)-f(x)}{h}$

For Right Hand,

$
\begin{equation}
\begin{aligned}
f'_+ (x) &= \lim\limits_{h \to 0^+} \frac{(x+h)^2-x^2}{h}\\
f'_+ (x) &= \lim\limits_{h \to 0^+} \frac{\cancel{x^2}+2xh+h^2-\cancel{x^2}}{h}\\
f'_+ (x) &= \lim\limits_{h \to 0^+} \frac{\cancel{h}(2x+h)}{\cancel{h}}\\
f'_+ (x) &= \lim\limits_{h \to 0^+} 2x+h\\
f'_+ (x) &= 2x+0\\
f'_+ (x) &= 2x\\
f'_+ (0) &= 2(0) = 0\\
f'_+ (0) &= 0
\end{aligned}
\end{equation}
$


For Left Hand,


$
\begin{equation}
\begin{aligned}
f'_- (x) &= \lim\limits_{h \to 0^-} \frac{-(x+h)^2-(-x^2)}{h}\\
f'_- (x) &= \lim\limits_{h \to 0^-} \frac{-\cancel{x^2}-2xh-h^2+\cancel{x^2}}{h}\\
f'_- (x) &= \lim\limits_{h \to 0^-} \frac{\cancel{h}(-2x-h)}{\cancel{h}}\\
f'_- (x) &= \lim\limits_{h \to 0^-} (-2x-h)\\
f'_- (x) &= -2x-0\\
f'_- (x) &= -2x\\
f'_- (0) &= -2(0) = 0\\
\end{aligned}
\end{equation}
$


The derivative from the right hand is equal to the derivative of left hand. $f'_+(0) = f'_-(0)$

Therefore,
The function $f(x) = x|x|$ is differentiable everywhere.