Question

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

Answer 1

Given:
$f\left(x\right)=\left|x\right|+\left|x-1\right|$
$$\begin{gathered} \Rightarrow f\left(x\right)=\left\{\begin{array}{l}-x-(x-1),x<0\\ \begin{array}{l}x-(x-1),0\le x<1\\ x+(x-1),x\ge 1\end{array}\end{array}\right. \\ \Rightarrow f\left(x\right)=\left\{\begin{array}{l}-2x+1,x<0\\ \begin{array}{l}1,0\le x<1\\ 2x-1,x\ge 1\end{array}\end{array}\right. \end{gathered}$$

When $x<0$, we have:

$f\left(x\right)=-2x+1$ which, being a polynomial function is continuous and differentiable.

When $0\le x<1$, we have:
$f\left(x\right)=1$ which, being a constant function is continuous and differentiable on (0,1).

When $x\ge 1$, we have:
$f\left(x\right)=2x-1$ which, being a polynomial function is continuous and differentiable on $x>2$.

Thus, the possible points of non- differentiability of $f\left(x\right)$ are 0 and 1.
Now,
(LHD at x = 0)

$\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}$

$=\underset{x\to 0}{\lim }\frac{-2x+1-1}{x-0}$ [∵ $f\left(x\right)=-2x+1,x<0$]

$$\begin{gathered} =\underset{x\to 0}{\lim }\frac{-2x}{x} \\ =-2 \end{gathered}$$

(RHD at x = 0)

= $\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}$

$$\begin{gathered} =\underset{x\to 0}{\lim }\frac{1-1}{x-1} \\ =0 \end{gathered}$$ [∵ $f\left(x\right)=1,0\le x<1$]

Thus, (LHD at x=0) ≠ (RHD at x=0)

Hence $f\left(x\right)$ is not differentiable at $x=0$

Now, $f\left(x\right)$ is not differentiable at $x=1$.

(LHD at x = 1)

$\underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$$\begin{gathered} =\underset{x\to 1}{\lim }\frac{1-1}{x-1} \\ =0 \end{gathered}$$

(RHD at x = 1)
= $\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$$\begin{gathered} =\underset{x\to 1}{\lim }\frac{2x-1-1}{x-1} \\ =\underset{x\to 1}{\lim }\frac{2(x-1)}{x-1} \\ =2 \end{gathered}$$

Thus, (LHD at x =1) ≠ (RHD at x=1)
.
Hence $f\left(x\right)$ is not differentiable at $x=1$.

Therefore, 0,1 are the points where f(x) is continuous but not differentiable.