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Math Help - 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: f(1) \neq f(0) and (\forall x \in ]0;1[): f'(x) \neq 0

    I must show that (\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: f(1) \neq f(0) and (\forall x \in ]0;1[): f'(x) \neq 0

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

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

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