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Math Help - f(0)=f(0)=0, f(x)>0, x>0 -> f(x)>0, x>0

  1. #1
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    f(0)=f(0)=0, f(x)>0, x>0 -> f(x)>0, x>0

    Given f(0)=f(0)=0, f(x)>0, x>0
    Show: f(x)>0, x>0 using std calculus (no abstract symbolism please).
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  2. #2
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    Re: f(0)=f(0)=0, f(x)>0, x>0 -> f(x)>0, x>0

    Question: Is f(x) continuous on [0,\infty] and differentiable on (0,\infty)? (This is true for f(x) = x - \ln(1+x)).

    If so, then we can apply the Mean Value Theorem. Let a>0. By assumption, f(x) is continuous on [0,a] and differentiable on (0,a), so by the Mean Value Theorem, there exists c\in(0,a) such that:

    f'(c) = \dfrac{f(a)-f(0)}{a-0} = \dfrac{f(a)}{a}

    Solving for f(a), we have f(a) = af'(c). Since c>0, we are given that f'(c)>0. The product of two positive real numbers is positive, so f(a)>0.
    Thanks from Hartlw
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