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

Explanation

Yes
`f(x)=|x-1|+|x-2|`
This function is continuous for all real values of x.
but it is not differentiable at `x=1` and `x=2`.
at `x=1`
`f(1)=|1-1|+|1-2|=0+1=1`
`L.H.L. =underset(xto 1^(-))(lim)f(x) " Let "x=1-h`
`=underset(h to 0)(lim)f(1-h) " "implies 1-h to 1`
`=underset(h to 0)(lim)|1-h-1|+|1-h-2| " "impliesh to 0`
`=0+|-1|=1`
`R.H.L.=underset(xto 1^('))(lim)f(x) " Let "x=1+h`
`=underset(h to 0)(lim)f(1+h) " "implies 1+h to 1`
`=underset(h to 0)(lim)|1+h-1|+|1+h-2| " "impliesh to 0`
`=0+|-1|=1`
` :' L.H.L. = f(1)=R.H.L.`
` :. f(x) ` is continuous `x=1`.
at `x=2`
`f(2)=|2-1|+|2-2|=1+0=1`
`L.H.L. =underset(xto 2^(-))(lim)f(x) " Let "2-h+x`
`=underset(h to 0)(lim)f(2-h) " "implies 2-h to 2`
`=underset(h to 0)(lim)|2-h-1|+|2-h-2| " "impliesh to 0`
`=1+0=1`
`R.H.L. =underset(xto 2^(+))(lim)f(x) " Let "2+h=x`
`=underset(h to 0)(lim)f(2+h) " "implies 2+h to 2`
`=underset(h to 0)(lim)|2+h-1|+|2+h-2| " "impliesh to 0`
`=1+0=1`
` :' L.H.L=f(2)=R.H.L.`
` :. f(x)` is continuous at `x=2`.
Now `f(x)=|x-1|+|x-2|`
`={(-(x-1)-(x-2)= -2x+3 " if " x lt 1),((x-1)-(x-2)=1 " " if 1 le x le 2.),((x-1)+(x-2)=2x-3 " if " x gt 2 ):}`
at `x=1`
`Lf'(1)=underset(h to 0)(lim)(f(1-h)-f(1))/(-h)`
`=underset(h to 0)(lim)(-2(1-h)+3-1)/(-h)`
`=underset(h to 0)(lim)(2h)/(-h)=underset(h to 0)(lim)(-2)=-2`
`Rf'(1)=underset(h to 0)(lim)(f(1+h)-f(1))/(h)`
`=underset(h to 0)(lim)(1-1)/(h)=underset(h to 0)(lim)(0)=0`
` :' Lf'(1) ne Rf'(1)`
` :. f(x)` is not differentiable at `x=1`.
at `x=2`
`Lf'(2)=underset(h to 0)(lim)(f(2-h)-f(2))/(-h)`
`=underset(h to 0)(lim)(1-(2xx2-3))/(-h)`
`=underset(h to 0)(lim)(0)=0`
`Rf'(2)=underset(h to 0)(lim)(f(2+h)-f(2))/(h)`
`=underset(h to 0)(lim)(2(2+h)-3-1)/(h)`
`=underset(h to 0)(lim)(2h)/(h)=underset(h to 0)(lim)(2)=2`
` :' Lf'(2) ne Rf'(2)`
` :. f(x)` is not differentiable at `x=2`.