I found this one fun.f(x): R->R is continuous and diferenciable in one point - and one point only.f(x) = ?Alex.

If you think about this - you'll realise that theres no way that the function can be unique!

QuoteOriginally posted by: wannabequantieIf you think about this - you'll realise that theres no way that the function can be unique!Does there exist one? -- I don't know how to define a real-valued function that is continuous at a finite number of points. Nearly all I know of are either continuous at infinite number of points or continous at no point.

bhutes,it does exist.wannabequantie,it is not unique.A

Define f(x)=0 if x is rational and f(x)=x^2 if x is irrational.Then it is differentiable at x=0 only.

Right!

Take any function that is nowhere continuous, and multiply it by x^2, or indeed by x^a with any a>1.

QuoteOriginally posted by: alexandreCI found this one fun.f(x): R->R is continuous and diferenciable in one point - and one point only.f(x) = ?Alex.That one is easy (see many answers below). I actually misread at the beginning thatthe function is "continuos everywhere", but differentiable only in one point.That sounds a bit more difficult. Or may be not?

QuoteOriginally posted by: ZmeiGorynychTake any function that is nowhere continuous, and multiply it by x^2, or indeed by x^a with any a>1.That is not quite right. You need to bound it in some way. Consider, for instance:f(x) = exp(1/x) for x irrationalf(x) = 0 for x rational.Then both f(x) and x^2*f(x) are nowhere continuous, and hence nowhere differentiable.

QuoteOriginally posted by: vit2007QuoteOriginally posted by: alexandreCI found this one fun.f(x): R->R is continuous and diferenciable in one point - and one point only.f(x) = ?Alex.That one is easy (see many answers below). I actually misread at the beginning thatthe function is "continuos everywhere", but differentiable only in one point.That sounds a bit more difficult. Or may be not? For this can you take a function f which is bounded, continuous on R, yet differentiable nowhere, then multiply it by x^2 ?Certainly such functions f exist. (for f, take a brownian sample path B_x on the full real line and multiply it by 1/(1+x^2) for instance.)

SPMars, correct.vit2007, f(x)= W(x) * x^2 does the trick. (W being a motion of Brown.)A

Don't forget that a Brownian path CAN be smooth (with p=0)! So you have to choose a "generic" path, not just an arbitrary one!