Need help on this proof

**binkypoo** · October 22nd 2009, 10:31 PM

I have to prove the following:

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

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

**el_manco** · October 22nd 2009, 11:53 PM

Hello

First you can proof this result.

If $f<img src=$ a,b)\longrightarrow R" alt="f

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

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

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

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

Best regards.

**binkypoo** · October 25th 2009, 08:18 PM

[quote=el_manco;389581]taking $\epsilon=f(x)$
do you mean $\epsilon=f(c)$ ??

**el_manco** · October 25th 2009, 11:40 PM

[quote=binkypoo;391521]

Originally Posted by el_manco

taking $\epsilon=f(x)$
do you mean $\epsilon=f(c)$ ??

Yes, of course. I meant $\epsilon=f(c)$ . Sorry for the typo.

Best regards.

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