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

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

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

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