Problem 73 - Calculus: Early Transcendentals

Question

A continuity proof Suppose \(f\) is continuous at \(a\) and defined for all \(x\) near \(a\). If \(f(a)>0,\) show that there is a positive number \(\delta>0\) for which \(f(x)>0\) for all \(x\) in \((a-\delta, a+\delta) .\) (In other words, \(f\) is positive for all \(x\) in some interval containing \(a .\) )

Step-by-Step Solution

Answer

Question: Prove that if a function \(f\) is continuous at a point \(a\), and \(f(a) > 0\), then there exists a positive number \(\delta > 0\) such that \(f(x) > 0\) for all \(x\) in the interval \((a-\delta, a+\delta)\). Answer: To show that \(f\) is continuous at \(a\), we chose an appropriate \(\epsilon = \frac{f(a)}{2}\), which is positive since \(f(a)>0\). By definition of continuity, there exists a \(\delta > 0\) such that \(|f(x)-f(a)| < \epsilon\) for all \(x\) satisfying \(0 < |x-a| < \delta\). From this, we showed that \(f(x) > 0\) for all such \(x\) in the interval \((a-\delta, a+\delta)\).

Step 1: Review the definition of continuity

A function \(f\) is continuous at a point \(a\) if for each \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all points \(x\) satisfying \(0 < |x-a| < \delta\), we have \(|f(x) - f(a)| < \epsilon\). In our case, the condition is already given, meaning that \(f\) is continuous at \(a\).

Step 2: Select an appropriate \(\epsilon\)

Since it is given in the problem that \(f(a) > 0\), we can choose an appropriate \(\epsilon\) value. Let's choose: $$\epsilon = \frac{f(a)}{2}$$ Since \(f(a)>0\), \(\epsilon\) will be also positive.

Step 3: Find the corresponding \(\delta\)

As we already know that \(f\) is continuous at \(a\), there must exist a \(\delta > 0\) such that, for every \(x\) satisfying \(0 < |x-a| < \delta\), we also have $$|f(x) - f(a)| < \epsilon $$

Step 4: Check if \(f(x) > 0\) for \(x\) in \((a-\delta, a+\delta)\)

Now, we need to show that \(f(x) > 0\) for all \(x\) in the interval \((a-\delta, a+\delta)\). Let's check if \(f(x) - f(a) > -\epsilon\). If \(|f(x) - f(a)| < \epsilon\) is equivalent to $$-\epsilon < f(x) - f(a) < \epsilon$$ Now, substituting \(\epsilon = \frac{f(a)}{2}\) gives us the inequality: $$-\frac{f(a)}{2} < f(x) - f(a) < \frac{f(a)}{2}$$ Adding \(f(a)\) to the entire inequality, we have: $$\frac{f(a)}{2} < f(x) < \frac{3}{2}f(a)$$ Hence, we can conclude that for all \(x\) in \((a-\delta, a+\delta)\), \(f(x)>0\). In conclusion, there exists a positive number \(\delta > 0\) such that \(f(x) > 0\) for all \(x\) in the interval \((a-\delta, a+\delta)\).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a formal way to prove the continuity of a function at a point. This concept revolves around two small positive quantities: \(\epsilon\) and \(\delta\). Here is how it works:

• Given any small positive number \(\epsilon > 0\), we want to ensure that the function \(f\) stays close to its value at a point \(a\).
• Specifically, \(f(x)\) should stay within \(\epsilon\) units from \(f(a)\), i.e., \(|f(x) - f(a)| < \epsilon\).

To make this happen, we find a distance \(\delta > 0\) such that for all \(x\) in the interval close to \(a\), defined by \(0 < |x - a| < \delta\), this condition holds.In essence, the epsilon-delta definition confirms that as \(x\) approaches \(a\), \(f(x)\) approaches \(f(a)\) as well. This foundation allows us to prove more complex properties of functions, like positivity within an interval.

Interval

An interval is a set of numbers lying between two endpoints. In the context of function continuity, we are often concerned with a special kind of interval around a point \(a\). A typical interval can be written as \((a - \delta, a + \delta)\), capturing all numbers that are closer to \(a\) than \(\delta\).Why does this interval matter?

• It's where we test if the function remains close to \(f(a)\).
• The interval ensures we check all relevant \(x\) values where the function's behavior needs examination for continuity and other properties.

Understanding intervals helps identify when a function is reliably behaving in a specific way over a range, which aids in proving positivity or other features locally around a particular point.

Positive Function

A positive function simply means that the function's value is greater than zero for the inputs considered. Demonstrating that a function remains positive over an interval is an essential step in many proofs, especially when establishing certain properties of a function within that interval.In our context:

• It is provided that \(f(a) > 0\), meaning the function value at \(a\) is positive.
• Our goal was to show that this positivity holds for all \(x\) in an interval \((a-\delta, a+\delta)\) around \(a\).

By leveraging the concept of continuity (from the epsilon-delta definition), you can secure a small \(\delta\) making sure \(f(x)\) stays positive around \(a\).Therefore, proving a function is positive over a specific interval implies a certain stability of its output, confirming it is above zero over that range.

Limit

The concept of a limit is fundamental in understanding how a function behaves as \(x\) approaches a particular value. A limit describes the value that \(f(x)\) gets closer and closer to as \(x\) gets closer and closer to \(a\).In continuity, limits help establish that:

• As \(x\) nears \(a\), \(f(x)\) gets closer to \(f(a)\).
• If the limit exists and equals \(f(a)\), the function doesn't "jump" but smoothly transitions through \(a\).

For our proof, knowing \(f\) is continuous at \(a\) implies \(f(x)\) stays close to \(f(a)\) as \(x\) enters the interval defined by \(\delta\). This commitment to staying near \(f(a)\) ensures that positive values at \(a\) can be safely expected in this region, supporting proofs of positivity or other behaviors within small ranges.