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Math Help - Using Exponentials

  1. #1
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    Using Exponentials

    I could use some help getting started on this proof. Any help would be appreciated.

    Prove that if f:[0,1]\rightarrow \mathbb{R} is differentiable and f(0)=f(1)=0, then for any a\in \mathbb{R},\exists c\in (0,1) such that f'(c)=af(c).

    I was told a hint that I have to apply Rolle's Theorem to f(x)e^{ax} but I'm not sure how to work that in yet...
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    If a\in \mathbb{R}, let b=-a. By Rolle's Theorem, there is a c in (0,1) with f'(c)e^{bc}+bf(c)e^{bc}=0.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by zebra2147 View Post
    I could use some help getting started on this proof. Any help would be appreciated.

    Prove that if f:[0,1]\rightarrow \mathbb{R} is differentiable and f(0)=f(1)=0, then for any a\in \mathbb{R},\exists c\in (0,1) such that f'(c)=af(c).

    I was told a hint that I have to apply Rolle's Theorem to f(x)e^{ax} but I'm not sure how to work that in yet...
    First note that if g(x)=f(x)e^{ax}, then g(0)=f(0)=0 and g(1)=f(1)e^{a}=0 since f(0)=f(1)=0. Now, by Rolle's theorem, \exists c\in(0,1):g^{\prime}(c)=0.

    Note that g^{\prime}(x)=f^{\prime}(x)e^{ax}+af(x)e^{ax}. So g^{\prime}(c)=0\implies\ldots

    Can you finish this problem?
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  4. #4
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    This implies that f^{\prime}(x)e^{ax}+af(x)e^{ax}=0. Therefore,
    f^{\prime}(x)e^{ax}=-af(x)e^{ax} Which implies f^{\prime}(x)=-af(x)
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  5. #5
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    Quote Originally Posted by zebra2147 View Post
    Prove that if f:[0,1]\rightarrow \mathbb{R} is differentiable and f(0)=f(1)=0, then for any a\in \mathbb{R},\exists c\in (0,1) such that f'(c)=af(c).
    I was told a hint that I have to apply Rolle's Theorem
    Use \displaystyle g(x)=f(x)\cdot e^{-ax}
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