Solved: Analysis Question - Show Exists Interval f(x)>0

[Math_Geek]

[SOLVED] Analysis Question

Homework Statement

Let f:Reals to Reals be a continuous at x=a, and further suppose f(a)>0. Show there exists an interval I about x=a such that f(x)>0 for all x in I.

Homework Equations

none

The Attempt at a Solution

i know the defn of continuity but I am not sure how to show this interval exists.
I need HELP!

 

[sutupidmath]

well, this is quite easy to show, just use the definition of the limit, and chose epsylon in such a manner that it will allow you to do this. i.e $$0<\epsilon<f(a)-0$$.

$$\forall\epsilon>0, \ \ also \ \ for 0<\epsilon<f(a),\exists\delta(\epsilon)>0$$ such that

$$|f(x)-f(a)|<\epsilon, \ \ whenever \ \ 0<|x-a|<\delta$$ so from here we have:

$$-\epsilon<f(x)-f(a)<\epsilon =>f(a)-\epsilon<f(x)<f(a)+\epsilon$$ but look now how we chose our epsilon, $$\epsilon<f(a)=>f(a)-\epsilon>0$$ so, eventually

$$0<f(a)-\epsilon<f(x), \ \ whenever, \ \ \ xE(a-\delta,a+\delta$$.

hope this helps.

 

[Math_Geek]

wow thanks

 

[sutupidmath]

wow thanks

Yeah, but make sure next time to show some work of yours!.

 

[Math_Geek]

how do I know it is in the interval around a?

 

[sutupidmath]

how do I know it is in the interval around a?

But notice that x is from the interval $$(a-\delta,a+\delta)$$ which actually includes a. MOreover, it works only for that interval around a, for other intervals we are not sure.