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.