Problem 19 - Thomas Calculus

Question

Each of Exercises \(15-30\) gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon>0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. $$ f(x)=\sqrt{19-x}, \quad L=3, \quad c=10, \quad \epsilon=1 $$

Step-by-Step Solution

Answer

The interval is (3, 15), and \(\delta = 5\).

Step 1: Identify the Inequality

The inequality to solve is \(|f(x) - L| < \epsilon\). Here, \(f(x) = \sqrt{19-x}\), \(L = 3\), and \(\epsilon = 1\). Thus, the inequality becomes \(|\sqrt{19-x} - 3| < 1\).

Step 2: Solve the Inequality for x

To solve \(|\sqrt{19-x} - 3| < 1\), we split it into two inequalities: \(\sqrt{19-x} - 3 < 1\) and \(\sqrt{19-x} - 3 > -1\). Solve each one separately.

Step 3: Solve \(\sqrt{19-x} - 3 < 1\)

Add 3 to both sides: \(\sqrt{19-x} < 4\). Square both sides to eliminate the square root: \(19-x < 16\). Simplify further to get \(x > 3\).

Step 4: Solve \(\sqrt{19-x} - 3 > -1\)

Add 3 to both sides: \(\sqrt{19-x} > 2\). Square both sides: \(19-x > 4\). Simplify to find \(x < 15\).

Step 5: Find Interval for x

Combining Step 3 and Step 4, we get the interval: \(3 < x < 15\). This is the open interval about \(c = 10\) on which the inequality holds.

Step 6: Determine \(\delta\)

Since \(c = 10\), we need \(0 < |x - 10| < \delta\). The interval \(3 < x < 15\) suggests that \(\delta\) can be the smaller distance to the endpoints. The distance from 10 to 3 is 7 and from 10 to 15 is 5, thus \(\delta = 5\).

Key Concepts

Inequality Solutions

Understanding how to solve inequalities is fundamental in calculus and analytical work. When given an inequality like \(|f(x) - L| < \epsilon\), it involves ensuring that the function value \(f(x)\) stays within a particular bound, \(\epsilon\), from a target value \(L\). Solving these inequalities involves logical steps:

• Break down the absolute value inequality \(|A| < B\) into two separate parts: \(A < B\) and \(A > -B\).
• Solve each part individually to find the solution set, usually in terms of \(x\).
• Combine these solutions to determine the range of values for \(x\) that satisfy both conditions.

In our exercise, we dealt with the expression \(|\sqrt{19-x} - 3| < 1\) which was solved by navigating through two inequalities: \(\sqrt{19-x} - 3 < 1\) and \(\sqrt{19-x} - 3 > -1\). Solving these gave us an interval for \(x\), indicating where the function adheres to our restriction.

Open Intervals

Open intervals are intervals that do not include their endpoint values. In mathematical terms, an open interval \((a, b)\) consists of all numbers between \(a\) and \(b\), but not including \(a\) and \(b\) themselves. These are represented as \((a, b)\) rather than \([a, b]\) which would indicate a closed interval.In this particular problem, our solution led to the interval \((3, 15)\). This shows us which part of the domain of \(x\) allows the function \(f(x) = \sqrt{19-x}\) to stay within our desired range from the value \(L = 3\). Open intervals are very useful in calculus because they help identify where certain conditions are strictly met without including the boundaries, which is particularly important in limit and continuity discussions.

Delta-Epsilon Definition

The delta-epsilon definition is a formal method to rigorously establish the concept of a limit in calculus. It states that for every number \(\epsilon > 0\), there exists a number \(\delta > 0\) such that if \(0 < |x - c| < \delta\), then \(|f(x) - L| < \epsilon\). This definition is what secures the logical underpinning for limits to be precise and accurate.Applying this to our exercise:

• We concluded with an open interval \((3, 15)\) which determines where the inequality holds.
• To ensure this condition strictly around a particular center \(c = 10\), the \(\delta\) value is chosen as the smallest distance from \(c\) to any of the interval's endpoints.
• Here, we computed \(\delta = 5\) because 5 was the minimum distance from 10 to either reversed boundary of the open interval.

This method allows us to precisely define the bounds within which our function behaves as expected, which is crucial for studying limits and continuity.