Question

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

Answer 1

Given: 

\[f(x) = \left| x \right| + \left| x - 1 \right|\]
`⇒ f(x) = {(-x-(x-1), x<0),(x-(x-1),0le x<1),(x +(x-1),xge1):}`
`⇒ f(x) = {(-2x+1, x < 0 ),(1, 0le x <1),(2x -1, xge1):} `

When 

\[x < 0\] , we have:

\[f(x) = - 2x + 1\] 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:

\[f(x) = 2x - 1\] 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)

\[\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0}\]

\[f(x) = - 2x + 1, x < 0\]

\[= \lim_{x \to 0} \frac{- 2x}{x} \]

(RHD at x = 0)

 

\[\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}\]

\[= \lim_{x \to 0} \frac{1 - 1}{x - 1}\] 
= 0 

\[f(x) = 1, 0 \leq x < 1\]

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) 

\[\lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1}\]

\[= \lim_{x \to 1} \frac{1 - 1}{x - 1} \]
\[ = 0\]

(RHD at x = 1) 

\[\lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1}\]

\[= \lim_{x \to 1} \frac{2x - 1 - 1}{x - 1}\]
\[ = \lim_{x \to 1} \frac{2(x - 1)}{x - 1} \]
\[ = 2\]

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.