Write the number of points where f (x) = |x| + |x − 1| is continuous but not differentiable.

Solution:

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.