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Thread: Function

  1. #1
    Junior Member
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    Function

    Hi people,

    f is a coninuous function on [a, b], and differentiable on ]a, b[ such as: $\displaystyle f(1) \neq f(0)$ and $\displaystyle (\forall x \in ]0;1[): f'(x) \neq 0$

    I must show that $\displaystyle (\exists ! \ \alpha \in ]0;1[): 2f(\alpha)-f(0)-f(1)=0$

    can you help me please???
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by lehder View Post
    Hi people,

    f is a coninuous function on [a, b], and differentiable on ]a, b[ such as: $\displaystyle f(1) \neq f(0)$ and $\displaystyle (\forall x \in ]0;1[): f'(x) \neq 0$

    I must show that $\displaystyle (\exists ! \ \alpha \in ]0;1[): 2f(\alpha)-f(0)-f(1)=0$

    can you help me please???
    That there exists at least one $\displaystyle \alpha$ with $\displaystyle 2f(\alpha)-f(0)-f(1)=0$ follows from the intermediate value theorem for the continuous function f, since this is equivalent to $\displaystyle f(\alpha)=(f(1)+f(0))/2$ and, surely, $\displaystyle (f(1)+f(0))/2$ lies between f(1) and f(0).

    The remaining problem is to show that there exists only one such $\displaystyle \alpha$. But that follows from the mean value theorem: since if there existed two $\displaystyle \alpha_1,\alpha_2 \in \;]0;1[, \alpha_1<\alpha_2$ with $\displaystyle f(\alpha_1)=f(\alpha_2)=(f(1)+f(0))/2$, then there would exist an $\displaystyle x\in \;]\alpha_1;\alpha_2[$ such that $\displaystyle f'(x)=0$, contradicting the givens.
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