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Thread: Need help on this proof

  1. #1
    Junior Member
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    Need help on this proof

    I have to prove the following:

    Suppose that functions f and g are continuous at $\displaystyle x=c\in (a,b)$ and f(c) > g(c). Prove there exists $\displaystyle \delta$ > 0 such that for all $\displaystyle x\in (a,b)$ with $\displaystyle |x-c|<\delta$, we have f(x) > g(x).

    I really have no idea. any help on this would be great thanks!
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  2. #2
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    Hello

    First you can proof this result.

    If $\displaystyle fa,b)\longrightarrow R$ is a continuous function in $\displaystyle c\in (a,b)$ then and $\displaystyle f(c)>0$, then there exists $\displaystyle \delta>0$ such that $\displaystyle |x-c|<\delta, x\in (a,b)$ implies $\displaystyle f(x)>0$.

    This follows inmediately from continuous definition; taking $\displaystyle \epsilon=f(x)$ there exists $\displaystyle \delta>0$ such that $\displaystyle |x-c|<\delta, x\in (a,b)$ implies:

    $\displaystyle |f(x)-f(c)|<f(c)\quad \Rightarrow\quad f(x)-f(c)>-f(c)\quad \Rightarrow\quad f(x)>0$

    Then, for your exercise, apply this result to the funcion $\displaystyle f-g$. Note that the difference of continuous function is continuous.

    Best regards.
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  3. #3
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    [quote=el_manco;389581]taking $\displaystyle \epsilon=f(x)$
    do you mean $\displaystyle \epsilon=f(c)$??
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  4. #4
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    [quote=binkypoo;391521]
    Quote Originally Posted by el_manco View Post
    taking $\displaystyle \epsilon=f(x)$
    do you mean $\displaystyle \epsilon=f(c)$??
    Yes, of course. I meant $\displaystyle \epsilon=f(c)$. Sorry for the typo.

    Best regards.
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