$f(x)=\begin{cases} -x-x+1, & x>0 \\ x-(x-1), & 0 \le x < 1 \\ x+x-1, & x>1 \end{cases}$$f(x)=\begin{cases} -2x+1, & x<0 \\ 1, & 0 \le x < 1 \\ 2x-1, & x \ge 1 \end{cases}$Now: If f(x) is continuous at x = 0$\lim\limits_{x \to 0^{+}} f(x)=\lim\limits_{x \to 0^{-}} f(x)=f(0)$$\Rightarrow 1=\lim\limits_{h \to 0} f(0-h)=1$$\Rightarrow 1=\lim\limits_{h \to 0} (-2(-h)+1)=1$$\Rightarrow 1=1=1$Which is true$\Rightarrow f(x)$ is continuous at x = 0Similarly for continuity at x = 1$\lim\limits_{h \to 0} f(x)=\lim\limits_{x \to 1^{+}} f(x)=f(1)$$\lim\limits_{h \to 0} f(1-h)=\lim\limits_{h \to 0} f(1+h)=f(1)$$1=\lim\limits_{h \to 0} (2(1+h)-1)=1$$\Rightarrow 1=1=1$$\Rightarrow f(x)$ is continuous at x = 1When x >0$\lim\limits_{x \to 0^{+}} f'(x)=\lim\limits_{x \to 0} \frac{f(0+h)-f(0)}{h}$$=\lim\limits_{h \to 0} \frac{1-1}{h}=0$$\lim\limits_{x \to 0^{-}} f'(x)=\lim\limits_{h \to 0} \frac{f(0-h)-f(0)}{-h}$$=\lim\limits_{h \to 0} \frac{-2(-h)+1-1}{-h}=-2$$\therefore f'(0^{+}) \neq f'(0^{-})$∴ f(x) is not differentiable at x = 0Now $f'(1^{+})=\lim\limits_{h \to 0} \frac{f(1+h)-f(1)}{h}$$=\lim\limits_{h \to 0} \frac{2(1+h)-1-1}{h}$$=\lim\limits_{h \to 0} \frac{2h}{h}=2$$\therefore f'(1^{+}) \neq f'(1^{-})$$\Rightarrow f(x)$ is not differentiable at x = 1∴ f(x) is continuous at x = 0,1 and non-differentiable at x = 0,1