Answer
$f$ differentiable in $x=1$
Work Step by Step
We are given the function:
$f(x)=\begin{cases}
(x-1)^3,x\leq 1\\
(x-1)^2,x>1
\end{cases}$
Determine the derivative from the left:
$\lim\limits_{x \to 1^-}\dfrac{f(x)-f(1)}{x-1}
$=\lim\limits_{x \to 1^-}\dfrac{f(x-1)^3-(1-1)^3}{x-1}$
$=\lim\limits_{x \to 1^-}\dfrac{(x-1)^3}{x-1}$
$=\lim\limits_{x \to 1^-} (x-1)^2$
$=(1-1)^2$
$=0$
Determine the derivative from the right:
$\lim\limits_{x \to 1^+}\dfrac{f(x)-f(1)}{x-1}
$=\lim\limits_{x \to 1^+}\dfrac{f(x-1)^2-(1-1)^2}{x-1}$
$=\lim\limits_{x \to 1^+}\dfrac{(x-1)^2}{x-1}$
$=\lim\limits_{x \to 1^+} (x-1)$
$=1-1$
$=0$
We got:
$\lim\limits_{x \to 1^-}\dfrac{f(x)-f(1)}{x-1}=\lim\limits_{x \to 1^+}\dfrac{f(x)-f(1)}{x-1}=0$
So the one-sided limits are equal. This means that the function $f$ is differentiable in $x=1$ and $f'(1)=0$.
