Let $f:\mathbb{R}\to \mathbb{R}$ be a function $$f(x)= \begin{cases} (x-1)\min(x,x^2)&\text{if $x\geq 0$,}\\ x\min\left(x,\dfrac{1}{x}\right)&\text{if $x< 0$.} \end{cases}$$ Choose the correct option:
a) $f$ is differentiable everywhere;
b) $f$ is not differentiable at one point;
c) $f$ is not differentiable at 3 points;
d) $f$ is not differentiable at two points.
I am getting one point $0$, by considering left and right derivative from $0$ to $1$ and $-1$ to $0$.
Please help.
On
$$f(x)= \begin{cases} (x-1)\min(x,x^2)&\text{if $x\geq 0$,}\\ x\min(x,1/x)&\text{if $x< 0$.} \end{cases}\\ =\begin{cases} (x-1)x&\text{if $x\geq 1$,}\\ (x-1)x^2 &\text{if $0\le x\leq 1$,}\\ x^2&\text{if $x< -1$.}\\ 1&\text{if $-1\le x< 0$.}\end{cases}$$
and f is not continuous at $x=0$ then not differentiable
Your answer is partially correct. Notice that $$f(x)= \begin{cases} (x-1)x&\text{if $1\leq x$,}\\ (x-1)x^2&\text{if $0\leq x<1$,}\\ 1&\text{if $ -1\leq x<0$,}\\ x^2&\text{if $x< -1$.} \end{cases}$$ Since the components are polynomials, you should consider the points where the piecewise-function $f$ switches from one piece to another, that is $-1$, $0$ and $1$.
Is $f$ continuous at $0$? What about the differentiability of $f$ at $x=-1$ and at $x=1$?