Does there exists a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Consider the function

$f\left(x\right)=|x|+|x-1|$

Since, the $f\left(x\right)=|x|+|x-1|$ is continuous everywhere, but it is not differentiable at
$x=0$ and $x=1$

$f\left(x\right)=\{\begin{array}{cc}-x-(x-1)& x\le 0\\ x-(x-1)& 0<x<1\\ x+(x-1)& x\ge 1\end{array}$

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

Check continuity when $x=0$

If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)$

$L.H.L.=\underset{x\to 0^{-}}{lim}-2x+1$
$L.H.L.=\underset{h\to 0}{lim}-2(0-h)+1$
$L.H.L.=\underset{h\to 0}{lim}2h+1$
Substituting $h=0$, we get

$L.H.L.=2\left(0\right)+1=0+1=1$

$R.H.L.=\underset{x\to 0^{+}}{lim}1=1$

Find $f\left(0\right)$

$f\left(x\right)=-2x+1$ at $x=0$
$f\left(0\right)=-2\left(0\right)+1=0+1=1$

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

Therefore, the function
$f\left(x\right)=\{\begin{array}{cc}-2x+1& x\le 0\\ 1& 0<x<1\\ 2x+1& x\ge 1\end{array}$
is continuous at $x=0$

Now, checking differentiability when $x=0$

$R.H.D.=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$

$R.H.D.=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$

$R.H.D.=\underset{h\to 0}{lim}\frac{(1-(-2\left(0\right)+1)}{h}$

$R.H.D.=\underset{h\to 0}{lim}\frac{(1-1)}{h}=0$

$L.H.D.=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$

$L.H.D.=\underset{h\to 0}{lim}\frac{f(-h)-f\left(0\right)}{-h}$

$L.H.D.=\underset{h\to 0}{lim}\frac{\left(2\right(-h)+1)-(-2(0)+1)}{-h}$

$L.H.D.=\underset{h\to 0}{lim}\frac{1-1-2h}{-h}$

$L.H.D.=\underset{h\to 0}{lim}\frac{2h}{h}=2$

Thus, $L.H.D.\ne R.H.D.$

Hence, $f\left(x\right)=\{\begin{array}{cc}-2x+1& x\le 0\\ 1& 0<x<1\\ 2x-1& x\ge 1\end{array}$
is not differentiable at $x=0$

Check continuity when $x=1$

Given function is
$f\left(x\right)=\{\begin{array}{cc}-2x+1& x\le 0\\ 1& 0<x<1\\ 2x-1& x\ge 1\end{array}$

If $f\left(x\right)$ is continuous at $x=1$, then

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=f\left(1\right)$

$L.H.L.=\underset{x\to 1^{-}}{lim}1=1$

$R.H.L=\underset{x\to 1^{+}}{lim}2x-1$

$R.H.L=\underset{h\to 0}{lim}2(1+h)-1$

$R.H.L=\underset{h\to 0}{lim}2+2h-1$

$R.H.L=\underset{h\to 0}{lim}1+2h$

Substituting $h=0$, we get
$R.H.L.=1+2\left(0\right)=1$

Find $f\left(1\right)$

$f\left(x\right)=2x-1$ at $x=1$
$f\left(x\right)=2\left(1\right)-1=2-1=1$
Hence,
$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=f\left(1\right)$

Therefore, the function

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

is continuous at $x=1$
Now, checking differentiability when $x=1$

$R.H.D.=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}$

$R.H.D.=\underset{h\to 0}{lim}\frac{\left(2\right(1+h)-1)-\left(2\right(1)-1)}{h}$

$R.H.D.=\underset{h\to 0}{lim}\frac{(2+2h-1)-(2-1)}{h}$

$R.H.D.=\underset{h\to 0}{lim}\frac{(1+2h)-1}{h}$

$R.H.D.=\underset{h\to 0}{lim}\frac{2h}{h}=2$

$L.H.D.=\underset{h\to 0}{lim}\frac{f(1-h)-f\left(1\right)}{-h}$

$L.H.D.=\underset{h\to 0}{lim}\frac{1-1}{-h}=0$

Thus, $L.H.D.\ne R.H.D.$

Hence, $f\left(x\right)=\{\begin{array}{cc}-2x+1& x\le 0\\ 1& 0<x<1\\ 2x-1& x\ge 1\end{array}$

is not differentiable at $x=1$

When $x<0$

For $x<0,f\left(x\right)=-2x+1$

Since the function $f\left(x\right)=-2x+1$ is a polynomial function of degree ‘one’, so it is continuous and differentiable. Therefore, $f\left(x\right)$ is continuous and differentiable for $x<0$

When $x>1$

For $x>1,f\left(x\right)=2x-1$

Since the function $f\left(x\right)=2x-1$ is a polynomial of degree ‘one’, so it is continuous and differentiable. Therefore, $f\left(x\right)$ is continuous and differentiable for $x>1$

When $0<x<1$

For $0<x<1,f\left(x\right)=1$

Since the function $f\left(x\right)=1$ is a constant polynomial so it is continuous and differentiable. . Therefore, $f\left(x\right)$ is continuous and differentiable for $0<x<1$
Therefore, the function
$f\left(x\right)=|x|+|x-1|=\{\begin{array}{cc}-2x+1& x\le 0\\ 1& 0<x<1\\ 2x-1& x\ge 1\end{array}$
is continuous everywhere but not differentiable at two points, $x=0\&x=1$