# Thread: Need help on this proof

1. ## 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!

2. 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.

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

4. [quote=binkypoo;391521]
Originally Posted by el_manco
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.