Question

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

Answer 1

Given: 

$$f(x)=\left|x\right|+|x-1|$$
${\displaystyle \Rightarrow f\left(x\right)=\{\begin{array}{ll}-x-(x-1)& x<0\\ x-(x-1)& 0\le x<1\\ x+(x-1)& x\ge 1\end{array}}$
${\displaystyle \Rightarrow f\left(x\right)=\{\begin{array}{ll}-2x+1& x<0\\ 1& 0\le x<1\\ 2x-1& x\ge 1\end{array}}$

When 

$$x<0$$ , we have:

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

When

$$0\le x<1$$ , we have: 

$$f(x)=1$$  which, being a constant function is continuous and differentiable on (0,1).

When  

$$x\ge 1$$, we have:

$$f(x)=2x-1$$ 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,
(LHD at x = 0)

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

$$f(x)=-2x+1,x<0$$

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

(RHD at x = 0)

 

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

$$=\underset{x\to 0}{lim}\frac{1-1}{x-1}$$ 
= 0 

$$f(x)=1,0\le x<1$$

Thus, (LHD at x=0) ≠ (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) 

$$\underset{x\to 1^{-}}{lim}\frac{f(x)-f(1)}{x-1}$$

$$=\underset{x\to 1}{lim}\frac{1-1}{x-1}$$
$$=0$$

(RHD at x = 1) 

$$\underset{x\to 1^{+}}{lim}\frac{f(x)-f(1)}{x-1}$$

$$=\underset{x\to 1}{lim}\frac{2x-1-1}{x-1}$$
$$=\underset{x\to 1}{lim}\frac{2(x-1)}{x-1}$$
$$=2$$

Thus, (LHD atx =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.