Let $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x,\mathrm{x}<1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2-x,\hspace{0.17em}1\le \mathrm{x}\le 2\\ -2+3x-x^2,\mathrm{x}>2\end{array}\right.$then f(x) is 

$\begin{array}{l}\\ f\text{'}\left(x\right)=\left\{\begin{array}{l}1,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}<1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\\ -1,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1\le \mathrm{x}\le 2\\ 3-2x,\mathrm{x}>2\hspace{0.17em}\end{array}\right. \\ \mathrm{Where} {\mathrm{f}}^{\text{'}}\left(1^{-}\right)\ne {\mathrm{f}}^{\text{'}}\left(1^{+}\right)\text{\hspace{0.17em} so f is n}\mathrm{ot}\mathrm{differentiable}\mathrm{at}\mathrm{x}=1\\ \text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{f}}^{\text{'}}\left(2^{-}\right)={\mathrm{f}}^{\text{'}}\left(2^{+}\right)=-1 so f is \mathrm{diffrentiable}\mathrm{at}\mathrm{x}=2\\ \end{array}$