Problem 30 - 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. $$ \begin{array}{ll}{f(x)=m x+b,} & {m>0, \quad L=m+b, \quad c=1} \\\ {\epsilon=0.05}\end{array} $$

Step-by-Step Solution

Answer

An open interval is \( (1 - \frac{0.05}{m}, 1 + \frac{0.05}{m}) \) and \( \delta = \frac{0.05}{m} \).

Step 1: Understand the Problem

We are given a linear function \( f(x) = mx + b \), a point \( c = 1 \), limit \( L = m + b \), and a small positive number \( \epsilon = 0.05 \). We need to find an interval around \( c \) such that the inequality \( |f(x) - L| < \epsilon \) holds, and a \( \delta > 0 \) such that for \( 0<|x-c|<\delta \), the inequality also holds.

Step 2: Set Up the Inequality

The inequality \( |f(x) - L| < \epsilon \) translates to \( |mx + b - (m+b)| < 0.05 \). Simplifying, we get \( |mx - m| < 0.05 \). This can be rewritten as \( |m(x - 1)| < 0.05 \).

Step 3: Solve for the Interval

Since \( m > 0 \), we divide both sides of \( |m(x-1)| < 0.05 \) by \( m \) to isolate \( |x-1| \). This gives us \( |x-1| < \frac{0.05}{m} \). Thus, the interval is \( (1 - \frac{0.05}{m}, 1 + \frac{0.05}{m}) \).

Step 4: Determine \( \delta \)

From the interval \( |x-1| < \frac{0.05}{m} \), we can see that \( \delta = \frac{0.05}{m} \) will satisfy \( |f(x) - L| < \epsilon \) for all \( x \) such that \( 0<|x-1|<\delta \).

Key Concepts

Linear Functions

Linear functions are a fundamental concept in mathematics that describe a straight line on a graph. These functions have the general form:

• \( f(x) = mx + b \)
• where \( m \) is the slope of the line and \( b \) is the y-intercept.

The slope \( m \) determines the steepness of the line. A positive slope means the line goes upwards, while a negative slope means it goes downwards. The y-intercept \( b \) is the point where the line crosses the y-axis.
Linear functions are essential in elementary algebra and form the basis for understanding more complex functions and mathematical concepts. They are easy to work with and allow for straightforward computations and predictions of value changes.

Inequality Solutions

Solving inequalities involves finding the range of values that satisfy a particular condition. When working with inequalities, like our exercise's \(|f(x)-L|<\epsilon\), we need to determine the values of \( x \) that make the inequality true. The steps generally involve:

• Isolating the variable from the rest of the terms.
• Consider the direction of the inequality sign. If you multiply or divide by a negative number, remember to flip the inequality sign!
• Representing the solution, preferably using interval notation or by shading a number line.

When inequalities involve absolute values, such as \( |m(x-c)| \), the solution contains two cases because absolute values measure distance and are always non-negative:

• The expression inside the absolute value is equal to the positive form.
• The expression inside the absolute value is equal to the negative form.

Combining these cases provides the range for \( x \) that solves the inequality.

Intervals and Neighborhoods

In mathematics, an interval is a range of numbers between two endpoints along the number line. An interval can be open, closed, or half-open:

• Open interval \((a, b)\): \( x \) values such that \( a < x < b \)
• Closed interval \([a, b]\): \( x \) values such that \( a \leq x \leq b \)
• Half-open \((a, b]\) or \([a, b)\): a mix of the two previous types.

Intervals help define the domain of functions and solutions to inequalities.
In calculus, neighborhoods are intervals that are centered around a particular point, \( c \). A neighborhood of a point \( c \) with radius \( \delta \) would be an interval \((c - \delta, c + \delta)\).
Finding the correct interval or neighborhood where a function meets certain criteria—like in our exercise—ensures the behavior of the function adheres to given limits. Understanding intervals and neighborhoods provides a solid foundation for analyzing how functions behave globally and locally.