Continuous Problem: Show f(x) > 0 on [r,s] of [a,b]

[varignon]

Homework Statement

Let f be a function which is continuous on a closed interval $$[a,b]$$ with $$f(c) > 0$$ for some $$c\in[a,b]$$. Show that there is a closed interval $$[r,s]$$ with $$c\in[r,s]\subseteq[a,b]$$ such that f(x) > 0 for all $$x\in[r,s]$$.

Homework Equations

Hint let epsilon = f(c)/2 and find $$\delta > 0$$ such that |f(x) - f(c)| < $$\epsilon$$ when |x - c| < $$\delta$$

The Attempt at a Solution

If there is a f(c) with c within [a,b], that is greater than zero, then by continuity there must be an interval for which the function is greater than 0, [r,s], that seems fairly obvious.
By completeness [r,s] must have a least upper bound, and a least lower bound this would be x when f(x) = 0 (not sure if that helps or not).

I am not really sure what the hint is telling me at this stage, or why one would pick f(c)/2. I am guessing one could show that there is a range for which f(x) > f(c)/2 or some other constant. It seems as though one could repeat this process and get closer and closer to the edge of the interval.

I am not sure if i am on the right track, but I'd appreciate a bit of input on putting things together, assuming I am even on the right track.

 

[lurflurf]

The hint is confusing in that f(c)/2 works but is in no other way special.
you want f(x)>0 when |x-c|<delta
to make use of continuity all you need is 0<epsilon<f(c)