F(x) differentiable at / near x=0?

[Redbelly98]

This came up in a recent discussion about l'Hopital's Rule.

Suppose f(x) has a derivative at x=0, that is f'(0) exists.
Is it necessarily true that f(x) is differentiable in some open interval containing x=0?
Others--who know calculus better than I--say no, f(x) is not necessarily differentiable for x≠0. So my question is, what is an example function where that is the case? That is,

f'(x) exists at x=0
f'(x) does not exist for x close to zero
​

I'm unable to think of an example, but am quite curious about this.

 

[slider142]

Similar to constructing the example of a function which is continuous at only a single point, have a look at the http://www.math.tamu.edu/~tvogel/gallery/node4.html which is differentiable at only a single point. It just takes the fact that everywhere else, the tangent line has two disparate values, but at 0, all the tangent lines approach the same tangent line.

 

[Redbelly98]

Thanks! By the way, one of the other people I was discussing this with came up with this example:

f(x) = x² sin(1/x²) for x≠0,
f(0) = 0

 

[CompuChip]

I cannot think of one immediately as well, so probably the counterexample is pretty pathological.

Maybe you can take the f(x) from this webpage and define g(x) by
g(x) = -f(x) if x < 0
g(x) = f(x) if x > 0
g(0) = 0.

I don't really feel like proving that that's differentiable at x = 0 though
It might be a good example, because around x = 0 it just looks like x |--> -x for all partial sums.

[edit]Lot of people posting while I was thinking about this
Very nice example slider! And the other one from Dick RedBelly is even nicer because it is continuous everywhere :)[/edit]

 

[slider142]

Thanks! By the way, one of the other people I was discussing this with came up with this example:

f(x) = x² sin(1/x²) for x≠0,
f(0) = 0

Hmm, this derivative is defined everywhere, though, not just at x=0. (Before I was looking at the second derivative at x=0. My bad. :D)

 

[Redbelly98]

Hmm, this derivative is defined everywhere, though, not just at x=0.

Aarrh, you're right. Hmmm.

The original discussion involved l'Hopital's Rule. Thinking about it some more, the actual conditions should be

f'(x) exists at x=0
lim[x→0] f'(x) does not exist​

That is, even though f'(0) exists, one cannot apply l'Hopital's Rule to problems involving f(x) that meets the above conditions.

slider, thanks for providing another example.