(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

3. 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 [itex]\gamma : [a,b] \rightarrow \mathbb{R}[/itex] by the rule [itex]\gamma(t) = [f(b)-f(a)]g(t) - [g(b) - g(a)]f(t)[/itex]. Want to show their exists a point [itex]x \in (a,b)[/itex] such that [itex]\gamma'(x) = 0[/itex]. Observe that [itex]\gamma(a) = \gamma(b)[/itex]. Hence if [itex] t \in (a,b) [/itex] such that [itex]\gamma(a) = \gamma(b) < \gamma(t)[/itex], then since [a,b] is compact, f obtains a maximum value on [a,b]. Call this point at which [itex]\gamma[/itex] obtains a maximum value x. [itex]x\neq a[/itex] since [itex]\gamma(a) < \gamma(t)[/itex]. Similar reasoning shows that [itex]x \neq b[/itex]. Hence since [itex]\gamma[/itex] obtains a maximum value at x, it follows that [itex]\gamma'(x) = 0[/itex]. This completes the proof. If instead [itex]t \in (a,b)[/itex] such that [itex]\gamma(t) < \gamma(a) = \gamma(b)[/itex], 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.)

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# Homework Help: Need someone to check my proof

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