Problem 16 - 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. $$ f(x)=2 x-2, \quad L=-6, \quad c=-2, \quad \epsilon=0.02 $$

Step-by-Step Solution

Answer

The open interval is \((-2.01, -1.99)\), with \(\delta = 0.01\).

Step 1: Identify the function and expression

The function provided is \( f(x) = 2x - 2 \). We need to ensure the inequality \( |f(x) - L| < \epsilon \) holds true. Substitute the given values, \( L = -6 \) and \( \epsilon = 0.02 \), to rewrite the equation as \( |2x - 2 + 6| < 0.02 \).

Step 2: Simplify the inequality

Simplify the expression inside the absolute value: \( |2x + 4| < 0.02 \). This can be transformed into two separate inequalities without the absolute value: \( -0.02 < 2x + 4 < 0.02 \).

Step 3: Solve the inequality for \(x\)

For \(-0.02 < 2x + 4\), subtract 4 from all parts of the inequality: \(-4.02 < 2x\). Then divide every term by 2 to solve for \(x\): \(-2.01 < x\). Similarly, solve \(2x + 4 < 0.02\) by subtracting 4: \(2x < -3.98\). Dividing by 2 gives \(x < -1.99\). Combine these to get \(-2.01 < x < -1.99\).

Step 4: Determine \(\delta\)

The next task is to find a \( \delta > 0 \) such that the inequality \( 0 < |x - c| < \delta \) satisfies the inequality \( |2x + 4| < 0.02 \). Since \(x\) is between -2.01 and -1.99 and \(c = -2\), the distances from \(c\) to both of these bounds are 0.01. Thus, \( \delta = 0.01 \).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a formal way to define the concept of a limit in calculus. Limits help us understand the behavior of a function as it approaches a particular point. The epsilon-delta definition essentially provides a rigorous way of ensuring that as the input of the function gets closer to a specific number, the output gets closer to a pre-defined value. Here's how it works:

• Epsilon (\( \epsilon \)): This represents how close you want the function's output (\( f(x) \)) to be to a specific limit (\( L \)). It's a positive number that defines an acceptable range around the limit.
• Delta (\( \delta \)): This determines how close the input (\( x \)) of the function needs to be to a specific point (\( c \)) in order to keep the output within the epsilon range.

Simply put, the epsilon-delta definition helps ensure that if we choose a small delta, then the function value remains within the epsilon range of the limit. By applying this definition to our exercise, we were able to confirm a range of \( x \) values such that \( |f(x) - L| < \epsilon \). By using inequality transformations, we determined \( \delta \) using the endpoints derived from solving the inequalities relative to \( c \). This demonstrates the basis of limit calculation and helps ensure precision in calculus.

Absolute Value Inequality

Absolute value inequalities are essential in defining ranges for functions. They describe how far a number is from zero on the number line, regardless of direction. The inequality \( |a| < b \) indicates that \( a \) lies between \( -b \) and \( b \). Here's what this means:

• \( |a| < b \) is split into two separate inequalities: \( -b < a < b \)
• This transformation helps solve for variable \( a \).

In the context of our exercise, the absolute value inequality \( |2x + 4| < 0.02 \) was restated as \( -0.02 < 2x + 4 < 0.02 \). By breaking it down into two inequalities, we simplified and found the range of \( x \) that satisfies the condition.This approach demonstrates how managing absolute values allows us to handle ranges and provides the certainty needed for ensuring limits within specified boundaries. Absolute value inequalities are a cornerstone in limit problems and beyond.

Linear Functions

Linear functions are among the simplest types of functions in mathematics. They are characterized by being composed of terms that are either constant or have a single variable raised to the first power. The general form of a linear function is \( f(x) = mx + b \), where:

• \( m \) is the slope of the line. It describes the rate at which \( y \) changes with respect to \( x \).
• \( b \) is the y-intercept. It’s the value of \( y \) when \( x = 0 \).

In our specific exercise, the linear function provided was \( f(x) = 2x - 2 \). This function is linear because it fits the general structure of a linear equation. The slope \( m = 2 \) means for every unit increase in \( x \), \( f(x) \) increases by 2 units. The y-intercept \( b = -2 \) indicates that when \( x = 0 \), \( f(x) = -2 \).Understanding the properties of linear functions is essential, as they often serve as a foundation for exploring more complex functions in calculus. Linear functions provide a straightforward approach for evaluating expressions and offer predictability, which can be useful in calculating limits and solving inequalities efficiently.