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Thread: Differentiable function problem

  1. #1
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    Differentiable function problem

    $\displaystyle f $ is a differentiable function at $\displaystyle [a,\infty)$.
    i need to prove that:
    if there is a constant $\displaystyle m>0$ which maintains that $\displaystyle f'(x)\geq m $ to every $\displaystyle x\in[a,\infty),$
    so$\displaystyle lim_{x\rightarrow\infty}f(x)=\infty $... thx!!
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  2. #2
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    Re: Differentiable function problem

    Quote Originally Posted by orir View Post
    $\displaystyle f $ is a differentiable function on $\displaystyle [a,\infty)$.
    i need to prove that:
    if there is a constant $\displaystyle m>0$ which maintains that $\displaystyle f'(x)\geq m $ to every $\displaystyle x\in[a,\infty),$
    so$\displaystyle lim_{x\rightarrow\infty}f(x)=\infty $... thx!!

    If $\displaystyle x>a$ then $\displaystyle \exists x'\in(a,x)$ such that $\displaystyle \frac{f(x)-f(a)}{x-a}=f'(x')\ge m$. WHY?

    That means that $\displaystyle f(x)\ge m(x-a)+f(a)$. What can you do with that?
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  3. #3
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    Re: Differentiable function problem

    Quote Originally Posted by Plato View Post
    If $\displaystyle x>a$ then $\displaystyle \exists x'\in(a,x)$ such that $\displaystyle \frac{f(x)-f(a)}{x-a}=f'(x')\ge m$. WHY?

    That means that $\displaystyle f(x)\ge m(x-a)+f(a)$. What can you do with that?
    can i say that $\displaystyle \varepsilon=m(x-a)+f(a)$? so then,$\displaystyle f(x)\ge\varepsilon$??
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    Re: Differentiable function problem

    Quote Originally Posted by orir View Post
    can i say that $\displaystyle \varepsilon=m(x-a)+f(a)$? so then,$\displaystyle f(x)\ge\varepsilon$??
    Well, you did not answer the question WHY?

    If you want further help, then you must explain where the inequality comes from.

    Moreover, what does it mean to say that $\displaystyle {\lim _{x \to \infty }}f(x) = \infty~? $ Give the precise definition.
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  5. #5
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    Re: Differentiable function problem

    oh, i know why. it's because the mean value theorem.
    and.. the precise definition is that there is an m>0, and n>0, so that for every x larger than m, f(x) larger than n.
    Last edited by orir; May 8th 2013 at 09:10 AM.
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  6. #6
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    Re: Differentiable function problem

    Quote Originally Posted by orir View Post
    oh, i know why. it's because the mean value theorem.
    and.. the precise definition is that there is an m>0, and n>0, so that for every x larger than m, f(x) larger than n.

    Can you show that $\displaystyle {\lim _{x \to \infty }}\left( {m(x - a) + f(a)} \right) = \infty ~?$

    If so, what does that imply?
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  7. #7
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    Re: Differentiable function problem

    i guess i can show that. but i don't know what does that imply...
    and what about my way? to say that $\displaystyle n=m(x - a) + f(a)$ and that way show the right of the definition above.
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  8. #8
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    Re: Differentiable function problem

    Quote Originally Posted by orir View Post
    i guess i can show that. but i don't know what does that imply...
    and what about my way? to say that $\displaystyle n=m(x - a) + f(a)$ and that way show the right of the definition above.

    Well $\displaystyle \forall x>a$ we have shown that $\displaystyle f(x)\ge m(x-a)+f(a)\to\infty~.$
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  9. #9
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    Re: Differentiable function problem

    oh, i got you..
    but still - does my way work too?
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