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

    I was told a hint that I have to apply Rolle's Theorem to $\displaystyle f(x)e^{ax}$ but I'm not sure how to work that in yet...
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    If $\displaystyle a\in \mathbb{R}$, let $\displaystyle b=-a$. By Rolle's Theorem, there is a c in (0,1) with $\displaystyle f'(c)e^{bc}+bf(c)e^{bc}=0$.
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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 $\displaystyle f:[0,1]\rightarrow \mathbb{R}$ is differentiable and $\displaystyle f(0)=f(1)=0$, then for any $\displaystyle a\in \mathbb{R},\exists c\in (0,1)$ such that $\displaystyle f'(c)=af(c)$.

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

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

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