Question

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

Answer 1

Let the function be f(x) = |x − 1| + |x − 2|

We redefine 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 continuous 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 everywhere and differentiable at all x ∈ R except at 1, 2.