Do such functions exist?

Question

I'm having difficulty with this, would like help, please:

Is it possible to find a function defined on an Interval I of R and values in R such that f^(-1)(x) = f'(x)? ( f^(-1) means the inverse of f, f(f^(-1)(x)) = x)?

Is it possible that I is the whole R?

Thank you

Answer 1

Let's start at part 2, seems easier. The answer is no. Suppose there exists f:R --> R such that f^(-1)(x) = f'(x) on all of R. This implies f is differentiable and, therefore, continuous. And since it has an inverse, it's strictly monotone on R.

If f is monotone increasing, then f'(x) >= 0 (1) for every real x. According to the given conditions, for every real x we also have f'(f(x)) = x (2). But, in virtue of (1), f'(f(x) >= 0 for every real x, so that (2) cannot be satisfied for x<0, a contradiction. So, in this case this function can't be define on (-oo, 0)


By a similar reasoning, we see that, if f is monotone decreasing, then it can't be defined on (0, oo).

But maybe this f exists on a smaller domain. A candidate for this f is f(x) = a x^b, with a and b positive, because it's inverse and it's derivative are of the same kind. Let's check if this is possible.

We have f^(-1)(x) = (x/a)^(1/b) and f'(x) = a b x^(b-1). If they are the same function, then we must have

1/b = b-1 => 1 = b^2 -b => b^2 - b - 1 = 0 and the only positive root is b = (1 + sqrt(5))/2. Hey, just one of the golden ratios! Math wonders!

In addition, we must have 1/a = ab, so that a = 1/sqrt(b), and we have just proved the function f(x) = a x^b , with b = (1 + sqrt(5))/2 and a = 1/b satisfies your condition on (0, oo).

So there exists I and a function f:I--> R such that f'(x) = f^(-1)(x) on I. There can be more I's and more f's like that. Anyway, the answer to the first question is yes. And to the second is no.

EDIT:
To most people who answered before me: It's f^(-1)(x) = f'(x), in the RHS it's the DERIVATIVE of f!

EDIT2
Thank you, Lobosito! I corrected my mistake, it's OK now.

Answer 2

Is it possible to find a function defined on an Interval I of R and values in R such that f^(-1)(x) = f'(x)?

Yes, there is an infinity of such functions.Try to take:
f(x)=(ax+b)/(cx+d), compose its inverse and equate it with the inverse. You will be surprised how many are. Actually, if I am not wrong, you need only a=d Just check
So I will be R except one point.


Is it possible that I is the whole R?
Well, f(x)=x satisfies this

Question: Is it possible that I is the whole R and f is not the identity f(x)=x?

Edit: Ooh I didn't notice the " ' ". About Steiner's proof, an error:

"But, according to the given conditions, for every x, f'(x) = f(f^(-1)(x)) = x, "

No, f' was f^(-1), not f composed with f^(-1)

Correct: f'(f(x)) = f^(-1)(f(x)) = x and you get the contradiction

Answer 3

The first thing that pops into my mind is f(x)=x and it is defined in (- infinity, + infinity), so yes the interval is the whole of R

edit: to thomasoa: if f(x) = 1 then x = 1 aka f(1) = 1, i don't see your point?
edit 2: OOOOH I didn't see the ' sign :D hehe, that changes the things a bit... Steiner answered it already...

Answer 4

Why is everybody saying the identity works? If f(x)=x, then f'(x)=1, which is not the inverse of f(x).

Answer 5

it is derivative on right hand side
it is not possible for whole R
but i think it is possible for some intervals in R.