Differentiation

$f\left(x\right) = x^5 \text{sgn} \; x$, where $\text{sgn} \; x = \begin{cases} \dfrac{\left|x\right|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ $\;\;$ ($\text{sgn} \; x$ is the signum function) Examine the continuity and differentiability of $f\left(x\right)$ at $x=0$.

$\begin{aligned} \left|x\right| = \begin{cases} x, & x > 0 \\ -x, & x < 0 \end{cases} \end{aligned}$

$\begin{aligned} \therefore \; \text{sgn} \; x = \begin{cases} \dfrac{x}{x}=1, & x > 0 \\ & \\ -\dfrac{x}{x}=-1, & x < 0 \\ & \\ 0, & x=0 \end{cases} \end{aligned}$

$\begin{aligned} \therefore \; f\left(x\right) = x^5 \text{sgn} \; x = \begin{cases} x^5, & x > 0 \\ -x^5, & x < 0 \\ 0, & x=0 \end{cases} \end{aligned}$

$\begin{aligned} \text{Left Hand Derivative } \left(\text{at }x=0\right) = Lf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0-h\right)-f\left(0\right)}{-h} \\ & = \lim\limits_{h \to 0} \; \dfrac{-\left(-h\right)^5 - 0}{-h} \\ & = \lim\limits_{h \to 0} \; \left(-h^4\right) = 0 \end{aligned}$

$\begin{aligned} \text{Right Hand Derivative } \left(\text{at }x=0\right) = Rf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0+h\right)-f\left(0\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{h^5 - 0}{h} \\ & = \lim\limits_{h \to 0} \; \left(h^4\right) = 0 \end{aligned}$

$\therefore$ $\;$ $Lf'\left(0\right) = Rf'\left(0\right) = 0$

$\implies$ $f\left(x\right)$ is differentiable at $x=0$

$\because$ $\;$ $f\left(x\right)$ is differentiable at $x=0$, $f\left(x\right)$ is also continuous at $x=0$.