# Differentiable function problem

• May 8th 2013, 07:53 AM
orir
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!! :)
• May 8th 2013, 08:16 AM
Plato
Re: Differentiable function problem
Quote:

Originally Posted by orir
$\displaystyle f$ 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?
• May 8th 2013, 08:33 AM
orir
Re: Differentiable function problem
Quote:

Originally Posted by Plato
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,$\displaystyle f(x)\ge\varepsilon$??
• May 8th 2013, 08:39 AM
Plato
Re: Differentiable function problem
Quote:

Originally Posted by orir
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.
• May 8th 2013, 09:04 AM
orir
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.
• May 8th 2013, 09:13 AM
Plato
Re: Differentiable function problem
Quote:

Originally Posted by orir
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?
• May 8th 2013, 09:19 AM
orir
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.
• May 8th 2013, 09:23 AM
Plato
Re: Differentiable function problem
Quote:

Originally Posted by orir
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 $\displaystyle f(x)\ge m(x-a)+f(a)\to\infty~.$
• May 8th 2013, 10:11 AM
orir
Re: Differentiable function problem
oh, i got you..
but still - does my way work too?