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!! :)

Re: Differentiable function problem

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**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?

Re: Differentiable function problem

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**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,$\displaystyle f(x)\ge\varepsilon$??

Re: Differentiable function problem

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**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.

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.

Re: Differentiable function problem

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**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?

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.

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 $\displaystyle f(x)\ge m(x-a)+f(a)\to\infty~.$

Re: Differentiable function problem

oh, i got you..

but still - does my way work too?