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Homework Help: Need someone to check my proof

  1. Jan 5, 2012 #1
    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.:smile: (and if this is sufficiently rigorous by your standards.)
    Last edited: Jan 5, 2012
  2. jcsd
  3. Jan 5, 2012 #2
    I think it looks good.
  4. Jan 5, 2012 #3
    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?
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