Question

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

Answer 1

Yes it can exist. If you have a piecewise function with three parts, the endpoints can be continuous but not differentiable. For instance, say that you have a linear function that goes from negative infinity to negative 2 and stops. Then for the second part of the piecewise function, from negative 2 to positive 2 you would have a quadratic function. And from positive 2 to infinity, you could have another linear function. At x-values of negative and positive 2, the graphs intersect; therefore, they are continuous. However, at these points, cusps could exist, thereby making the function/curve not differentiable at those two points.

Answer 2

So, if there is a function like \[f(x)=\sqrt{x^{2}-4}\]then \[x \epsilon [-2,2]\]and at -2 and 2 the function is not differentiable.

Answer 3

@merchandize ?

Answer 4

Yes but it is not continuous everywhere.

Answer 5

Think of setting up a piecewise function as merchandize suggested using the absolute value function.

Answer 6

yes.

Answer 7

what is a piecewise function actually ?

Answer 8

Example:
\(f(x) = \left\{\begin{array}{ccc}0 & ; & x \leq 0 \\ 1 & ; & x > 0\end{array}\right.\)

Answer 9

You can create functions that behave differently depending on the input (x) value

Answer 10

it is same in the form as Alchemista mentioned in the example.

Answer 11

then the graph would be this way right ? @Alchemista

Answer 12

Correct can you think of a way to modify my example using the absolute value function?

Answer 13

We want to use the absolute value function to create two points where the function is not differentiable.

Answer 14

I will give you half of what you need.
$$f(x) = \left\{\begin{array}{ccc}|x + 1| & ; & x \leq 0 \\ 1 & ; & x > 0\end{array}\right.$$

Answer 15

Actually that's all you need.

Answer 16

Graph this function carefully and try to understand why its not differentiable at -1 and 0
$$f(x) = \left\{\begin{array}{ccc}|x + 1| & ; & x \leq 0 \\ 1 & ; & x > 0\end{array}\right.$$

Answer 17

these are the points where the graph turns.

Answer 18

And they are cusps. In other words they are not smooth so the limit of the difference function will not exist.

Answer 19

oh yes the left and the right limits won't coincide.

Answer 20

In less rigorous terms you will end up with a limit where the slope is 0 in one direction and nonzero in the other direction.

Answer 21

At zero that is.

Answer 22

now i get it :D

Answer 23

thank you so much @Alchemista and @merchandize :)