is it possible part 2 (1 Viewer)

[no_arg]

Is it possible for a function to be defined over the entire real line and to have only one point on the graph where a tangent exists


(corrected)

 

[darkliight]

By 'one tangent' I presume you mean differentiable at only one point (otherwise the function f(x) = x has only one tangent line).

Well, I can't write up a proof without using an epsilon/delta type argument (my idea was to consider what it means to not be differentiable, then show that differentiable in a point but nowhere else leads to a contradiction), but I think it's intuitively clear that you can't have this. If a function is differentiable at a point, then lim h->0- blah = lim h->0+ blah. Now the intuition, you can talk about other points arbitrarily close to the differentiable point - and if none of them are differentiable points, then this contradicts your lim h->0 'approximations'.

It might be best to draw a curve and see that. The idea is that, if a function is differentiable in a point, then it is differentiable in some neighborhood of that point.

At least, that's my gut feeling - maybe there is some slick proof without resorting to sequences /shrug.

 

[no_arg]

Yes

What I meant was that there was only one point on the graph where a tangent exists...original corrected

 

[darkliight]

Ok, so does my argument sound reasonable?

 

[Iruka]

How about the function y = x^2 if x is rational, 0 if x is not rational.

I think that would be differentiable at x=0 and nowhere else.

Analysis is full of pathological counter examples.

 

[darkliight]

And that works. Nice.

 

[Affinity]

look up devil's staircase for something just as interesting.

it is also probably possible to construct a continuous function which is differentiable almost nowhere.