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

• varignon
In summary, to show that there is a closed interval [r,s] within [a,b] where f(x) is greater than 0 for all x, we can use the fact that f(c) is greater than 0 for some c within [a,b] due to continuity. By choosing epsilon as f(c)/2 and finding a suitable delta, we can show that f(x) is greater than 0 for all x within the interval [r,s]. This process can be repeated to get closer and closer to the edge of the interval. The hint suggests using f(c)/2 as a constant to demonstrate this.
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.

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)

## 1. What does it mean to show f(x) > 0 on [r,s] of [a,b]?

Showing f(x) > 0 on [r,s] of [a,b] means to prove that for all values of x within the interval [r,s], the function f(x) outputs a positive value. This also applies to the larger interval [a,b], meaning that the function must output a positive value for all values of x within this interval as well.

## 2. Why is it important to prove that f(x) > 0 on [r,s] of [a,b]?

Proving that f(x) > 0 on [r,s] of [a,b] is important because it provides evidence that the function is always positive within the given interval. This information is useful in many applications, such as optimization problems and determining the behavior of a function.

## 3. What methods can be used to show f(x) > 0 on [r,s] of [a,b]?

There are several methods that can be used to show f(x) > 0 on [r,s] of [a,b]. These include using the Intermediate Value Theorem, analyzing the sign of the derivative of the function, and evaluating the function at specific values within the interval.

## 4. Can f(x) ever be equal to 0 on [r,s] of [a,b]?

No, the statement f(x) > 0 on [r,s] of [a,b] implies that the function must output values that are strictly greater than 0 within the given interval. If the function were to output a value of 0, then it would not satisfy the condition of being greater than 0.

## 5. Is showing f(x) > 0 on [r,s] of [a,b] sufficient to prove that the function is always positive?

Not necessarily. While showing f(x) > 0 on [r,s] of [a,b] provides strong evidence that the function is always positive within the given interval, it is not always sufficient. It is important to also consider the behavior of the function outside of the given interval and to check for any points where the function may become negative.

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