Write the number of points where f (x) = |x| + |x − 1| is continuous but not differentiable.
Given:
$f(x)=|x|+|x-1|$
$\Rightarrow f(x)=\left\{\begin{array}{lc}-x-(x-1), & x<0 \\ x-(x-1), & 0 \leq x<1 \\ x+(x-1), & x \geq 1\end{array}\right.$
$\Rightarrow f(x)=\left\{\begin{array}{lc}-2 x+1, & x<0 \\ 1, & 0 \leq x<1 \\ 2 x-1, & x \geq 1\end{array}\right.$
When $x<0$, we have:
$f(x)=-2 x+1$ which, being a polynomial function is continuous and differentiable.
When $0 \leq x<1$, we have:
$f(x)=1$ which, being a constant function is continuous and differentiable on $(0,1)$.
When $x \geq 1$, we have:
$f(x)=2 x-1$ which, being a polynomial function is continuous and differentiable on $x>2$.
Thus, the possible points of non- differentiability of $f(x)$ are 0 and 1 .
Now,
$(\mathrm{LHD}$ at $x=0)$
$\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0} \frac{-2 x+1-1}{x-0}$ $[\because f(x)=-2 x+1, x<0]$
$=\lim _{x \rightarrow 0} \frac{-2 x}{x}$
$=-2$
(RHD at x = 0)
$=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0} \frac{1-1}{x-1}$ $[\because f(x)=1, \quad 0 \leq x<1]$
$=0$
Thus, $(\mathrm{LHD}$ at $x=0) \neq(\mathrm{RHD}$ at $x=0)$
Hence $f(x)$ is not differentiable at $x=0$
Now, $f(x)$ is not differentiable at $x=1$.
(LHD at x = 1)
$\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}$
$=\lim _{x \rightarrow 1} \frac{1-1}{x-1}$
$=0$
(RHD at x = 1)
$=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}$
$=\lim _{x \rightarrow 1} \frac{2 x-1-1}{x-1}$
$=\lim _{x \rightarrow 1} \frac{2(x-1)}{x-1}$
$=2$
Thus, (LHD at x =1) ≠ (RHD at x=1)
Hence $f(x)$ is not differentiable at $x=1$.
Therefore, 0,1 are the points where f(x) is continuous but not differentiable.