Question

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

Answer 1

It can be seen from the above graph that, the given function is continuos everywhere but not differentiable at exactly two points which are 0 and 1.

Answer 2

Let the function be

f (x) = |x - 1| + |x - 2|

We reefine f (x) as:

This is continuous at all x ∈ R but not differentiable at x = 1, 2

`f (x) = {(-(x - 1) - (x - 2);, if x<1),((x - 1) - (x - 2);, if 1<= x <=2), ((x - 1) + (x - 2);, if x>2):}`

i.e., `f (x) = {(-2x + 3;, if x<1),(1;, if 1<= x <=2), ((2x - 3);, if x>2):}`

f (x) is clearly continuous at all x except possibly at 1, 2.

At x = 1

`lim_(x->1^-) f (x) = lim_(h->0) (-2(1 - h) + 3)`

= -2 + 3

= 1

`lim_(x->1^+)f (x) = lim_(x->^+) (1) = 1`

Also, f (1) = 1

Thus, `lim_(x->1^-) f (x) = lim_(x->1^+) f (x) = f (1)`

Hence, f (x) is continuous at x = 1

At x = 2

`lim_(x->2^-) f (x) = lim_(x->2^-) 1 = 1`

`lim_(x->2^+) f (x) = lim_(x->2^+) (2x - 3) = lim_(h->0) (2(2 + h) -3)`

= `2 (2) - 3`

= 1

Also, f (2) = 1.

Thus `lim_(x->2^-) f(x) = lim_(x->2^+) f(x) = f(2)`

Hence f(x) is contnuous at x = 2

Hence, 'f' is continuous at all x ∈ R.

Now, `f' (x) = {(-2;, if x<1),(0;, if  1< x <2), (2;, if x>2):}`

Derivability at x = 1

`Lf' (1) = lim_(h->0) (f (1-h) - f (1))/(-h)`

`= lim_(h->0) (-2 (1 - h) + 3 - 1)/-h = lim_(h->0) (2h)/-h`

`lim_(h->0) (-2) = -2`

`Lf' (2) = lim_(h->0) (f(2 - h) - f (2))/h = lim_(h->0) (1 - 1)/h = 0`

Thus, Lf' (1) ≠ Rf' (1)

= 'f' is not derivable.

Derivability at x = 2

`Lf' (2) = lim_(h->0) (f (2 - h) - f(2))/h = lim_(h->0) (1 - 1)/h = 0`

`Rf' (2) = lim_(h->0) (f (2 + h) - f (2))/h`

`= lim_(h->0) (2 (2 + h) - 3 - 1)/h = lim_(h->0^+) (2h)/h = lim_(h->0^+) 2 = 2`

= Lf' (2) ≠ Rf' (2)

= f is not derivable at x = 2

Hence f (x) = |x - 1| + |x - 2| is continuous every where and differentiable at all x ∈ R except at 1, 2