Determine if the piecewise defined function is differentiable at the origin.
-$$g ( x ) = \left\{ \begin{array} { l l l }
x ^ { 7 / 8 } & \text { if } x \geq 0 & \
x ^ { 1 / 8 } & & \text { if } x

Question

Determine if the piecewise defined function is differentiable at the origin.
-$$g ( x ) = \left\{ \begin{array} { l l l } 
x ^ { 7 / 8 } & \text { if } x \geq 0 & \
x ^ { 1 / 8 } & & \text { if } x < 0
\end{array} \right.$$
A) Differentiable
B) Not differentiable

Quizplus · Accepted Answer

To show non-differentiability at the origin we need to show that the limit of the difference quotient does not exist or is infinite as $x\rightarrow0$. 
For $x>0$, we have 
$$g'(x)=\frac{7}{8}x^{-1/8}$$
Thus, the right limit of the difference quotient is 
$$\lim_{h\rightarrow0^+}\frac{g(0+h)-g(0)}{h}=\lim_{h\rightarrow0^+}\frac{h^{7/8}}{h}=\lim_{h\rightarrow0^+}h^{-1/8}=0$$
For $x<0$, we have 
$$g'(x)=\frac{1}{8}x^{-7/8}$$
Thus, the left limit of the difference quotient is 
$$\lim_{h\rightarrow0^-}\frac{g(0+h)-g(0)}{h}=\lim_{h\rightarrow0^-}\frac{h^{1/8}}{h}=\lim_{h\rightarrow0^-}h^{-7/8}=\infty$$
Since the left and right limits are not equal (i.e., they don't exist) then $g$ is not differentiable at $x=0$.

Determine If the Piecewise Defined Function Is Differentiable at the Origin