Proving Existence of a Point Using Rolle's Theorem

[killpoppop]

Homework Statement

Let f be continuous on [a,b] and differentiable on (a,b) Suppose that:

f²(b) f²(a) = b² - a²:

Prove (using Rolle's theorem) that:( exists x belonging to (a, b) ) ( f'(x)f(x) = x )I just don't know where to start I've done basic proofs with the theorem but only when f(a) = f(b). Any help would be greatly appreciated.

Thanks

 

[Dick]

I think you mean f(b)^2-f(a)^2=b^2-a^2. Define g(x)=f(x)^2-x^2. Can you apply Rolle's theorem to g(x) on the interval [a,b]?

 

[killpoppop]

I think you mean f(b)^2-f(a)^2=b^2-a^2. Define g(x)=f(x)^2-x^2. Can you apply Rolle's theorem to g(x) on the interval [a,b]?

No the notation i used is what is displayed in the question. I've never seen it before. Is it just the second derivative?

 

[Dick]

No. '^2' just means squared. E.g. f(b)^2=f(b)*f(b). Which is the same as f^2(b). The problem I'm having is that what you wrote looks like f(b)^2*f(a)^2=b^2-a^2. I really think it should be f(b)^2-f(a)^2. With a minus sign between the two squares, not a product of them.

 

[killpoppop]

No. '^2' just means squared. E.g. f(b)^2=f(b)*f(b). Which is the same as f^2(b). The problem I'm having is that what you wrote looks like f(b)^2*f(a)^2=b^2-a^2. I really think it should be f(b)^2-f(a)^2. With a minus sign between the two squares, not a product of them.

Yep. So sorry your spot on =]