Existence of a Critical Point for Differentiable Functions on Closed Intervals

[Samuelb88]

Homework Statement

If f and g are continuous real functions on [a,b] which are differentiable on (a,b), then there exists a point x \in (a,b) such that [f(b)-f(a)]g'(x) = [g(b)-g(a)]f'(x).

The Attempt at a Solution

Not sure if my reasoning is correct here... I can assume that closed intervals are compact, and if a function a real function f defined on an interval [a,b] obtains a maximum value at a point x such that a<x<b, and if f' exists, then f'(x) = 0.

Proof: Define \gamma : [a,b] \rightarrow \mathbb{R} by the rule \gamma(t) = [f(b)-f(a)]g(t) - [g(b) - g(a)]f(t). Want to show their exists a point x \in (a,b) such that \gamma&#039;(x) = 0. Observe that \gamma(a) = \gamma(b). Hence if t \in (a,b) such that \gamma(a) = \gamma(b) &lt; \gamma(t), then since [a,b] is compact, f obtains a maximum value on [a,b]. Call this point at which \gamma obtains a maximum value x. x\neq a since \gamma(a) &lt; \gamma(t). Similar reasoning shows that x \neq b. Hence since \gamma obtains a maximum value at x, it follows that \gamma&#039;(x) = 0. This completes the proof. If instead t \in (a,b) such that \gamma(t) &lt; \gamma(a) = \gamma(b), then same conclusion holds following similar reasoning.

I'd like to know whether my reasoning that x can equal neither a nor b is correct. (and if this is sufficiently rigorous by your standards.)

 

[rat bass]

I think it looks good.

 

[Rayquesto]

Wait I have an individual question for you. I admit to my noobiness, but I just want to ask you a question. So, suppose you could treat this as a form of a chain rule. Could you use inverse function to prove this exists as well such that finverseb-finversea=g'(x)? I mean sorry I didn't symbolize it correctly, but when I see this, I automatically recognize some inverse relations applied with the chain rule. I could be wrong though. You care to elaborate?