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

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SUMMARY

The discussion centers on proving that for a continuous function f on the interval [a,b], if f(c) > 0 for some c in [a,b], then there exists a closed interval [r,s] containing c where f(x) > 0 for all x in [r,s]. The key approach involves selecting epsilon as f(c)/2 and finding a delta such that |f(x) - f(c)| < epsilon when |x - c| < delta. This guarantees that f(x) remains positive within the interval [r,s], leveraging the properties of continuity.

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Homework Statement



Let f be a function which is continuous on a closed interval [a,b] with f(c) &gt; 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 &gt; 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.
 
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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)
 

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