Can a Continuous Function Satisfy f'(f(x)) = x?

[BugMeNot_dude]

Homework Statement

Is there a continuous function f(x), which has the whole real line as its domain,
such that f'(f(x))=x?

Homework Equations

The Attempt at a Solution

I mentally thought of the inputs that x goes into f(x) that goes into f'(x) which inturn equals x.
so i thought that
[f(x) should = 1/f'(x)]

i solved the equation to get f(x) = sqrt(2x)

which does not satisfy the initial condition of

f'(f(x))=x

all help is appreciated!

 

[cepheid]

Welcome to PF!

I think you are thinking about this in the right way. You basically have a composition of functions g(x) and f(x) to make g(f(x)). In this example, g(x) just happens to be f'(x).

I mentally thought of the inputs that x goes into f(x) that goes into f'(x) which inturn equals x.
so i thought that
[f(x) should = 1/f'(x)]

Not quite! You are correct that g(x) = f'(x) essentially "undoes" whatever f(x) does. But that does NOT mean that they are reciprocals as you have written. It means that they are inverses. The function g(x) is the inverse function to the function f(x). I know it's confusing especially since the notation f^-1(x) is used to mean "the inverse function of f(x)." But here, the superscript -1 does not mean "raise to the power of -1 (reciprocal)." It denotes the inverse function.

So the question is, is there a function f(x) whose derivative is equal to its own inverse?

 

[fzero]

Does

$$f'(f(x)) = \frac{ df(f(x))}{ dx} or \frac{ df(f(x))}{ df(x)}?$$

In the first case, we can directly integrate to eliminate the derivative. In the second case, we can integrate by parts.

 

[cepheid]

I think that f'(f(x)) means "evaluate the derivative of the function f(y) at y = f(x)"

In other words:

$$f^\prime (f(x)) = \frac{df}{dy}\bigg |_{y=f(x)}$$

I edited my post a few times, because I realized that the letter 'x' was doing double duty as a generic symbol for the arguments of two different functions.

 

[BugMeNot_dude]

Hey thank you for replying i realize you are correct and that
f-1(x) = f'(x) but I was unable to differentiate it. I know that the theorum
(f-1)'(x) = 1/(f'(f-1(x))), but I don't know how to go about using the two equations...

Can you guys help me and explain the answer?

 

[ZioX]

This is an interesting question.

If $$f^\prime(f(x))=x$$ for all x in R then $$f^\prime(x)=f^{-1}(x)$$ so that

$$(f(f(x))^\prime=f^\prime (f(x))=f^\prime(f(x))f^\prime(x)=xf^\prime(x)=xf^{-1}(x)$$

Can you take it from here?

Edit: made a mistake, will fix it up

 

[hupert]

silly eng sci frosh, tsc tsc, cheating is no good