The values of x for which the function $f(x)= \begin{cases}1-x \quad ~~~~, & x<1 \\ (1-x)(2-x), & 1 \leq x \leq 2 \\ 3-x \quad ~~~~, & x>2\end{cases} $ fails to be continuous or differentiable are

x = 2

We have,

$\lim\limits_{x \rightarrow 1^{-}} f(x)=\lim\limits_{x \rightarrow 1^{-}}(1-x)=0$

$\lim\limits_{x \rightarrow 1^{+}} f(x)=\lim\limits_{x \rightarrow 1^{+}}(1-x)(2-x)=0$

and, f(1) = 0

∴  $\lim\limits_{x \rightarrow 1^{-}} f(x)=\lim\limits_{x \rightarrow 1^{+}} f(x)=f(1)$

So, f(x) is continuous at x = 1.

Now,

(LHD at x = 1) = $\left(\frac{d}{d x}(1-x)\right)_{x=1}=-1$

(RHD at x = 1) = $\left(\frac{d}{d x}(1-x)(2-x)\right)_{x=1}=-1$

∴  (LHD at x = 1) = (RHD at x = 1)

So, f(x) is differentiable at x = 1.

Now,

$\lim\limits_{x \rightarrow 2^{-}} f(x)=\lim\limits_{x \rightarrow 2^{-}}(1-x)(2-x)=0$

and,

$\lim\limits_{x \rightarrow 2^{+}} f(x)=\lim\limits_{x \rightarrow 2^{+}} 3-x=1$

∴  $\lim\limits_{x \rightarrow 2^{-}} f(x) \neq \lim\limits_{x \rightarrow 2^{+}} f(x)$

Thus, f(x) is discontinuous and non-differentiable at x = 2. Hence, the only point of discontinuity and non-differentiability of f(x) is x = 2.