Question

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

Answer 1

Answer:

Yes, there are some function which are continuous everywhere but not differentiable at exactly two points.

Step-by-step explanation:

Yes, there are some function which are continuous everywhere but not differentiable at exactly two points.

Let us take an example.

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

Since we know that modulus functions are continuous at every point,

So there sum is also continuous at every point. But it is not differentiable at every point.

Let x = 1, 2

Now at x = 1

LHD = limx->1- [{f(x) - f(1)}/(x-1)]

= limh->0 [{f(1-h) - f(1)}/-h]

= limh->0 [{|1-h-1| + |1-h-2| - |1-1|-|1-2|}/-h]

= limh->0 [{|1-h-1| + |1-h-2| - |0|-|-1|}/-h]

= limh->0 [{|-h| + |-h-1| - 1}/-h]

= limh->0 [{h - (-h-1) - 1}/-h]

= limh->0 [{h + h + 1 - 1}/-h]

= limh->0 [{2h}/-h]

= -2

RHD = limx->1+ [{f(x) - f(1)}/(x-1)]

RHD = limh->0 [{f(1+h) - f(1)}/h]

= limh->0 [{|1+h-1| + |1+h-2| - |1-1|-|1-2|}/h]

= limh->0 [{|1+h-1| + |1+h-2| - |0|-|-1|}/h]

= limh->0 [{|h| + |h-1| - 1}/h]

= limh->0 [{h - (h-1) - 1}/h]

= limh->0 [{h - h + 1 - 1}/h]

= limh->0 [0/h]

= 0

Since LHD ≠ RHD

So given function is not diffenetiable at x = 1.

Similarly, we can show that the given function is not differentiable at x = 2.