Problem 72 - Calculus

Question

For each value of \(\epsilon\) in Problems \(72-73,\) find a positive value of \(\delta\) such that the graph of the function leaves the window \(a-\delta < x < a+\delta, b-\epsilon < y < b+\epsilon\) by the sides and not through the top or bottom. \(f(x)=-2 x+3 ; a=0 ; b=3 ; \epsilon=0.2,0.1,0.02\) 0.01,0.002,0.001

Step-by-Step Solution

Answer

For each \(\epsilon\), \(\delta = \frac{\epsilon}{2}\). Calculate \(\delta\) values accordingly.

Step 1: Identify the function and determine limits

The function is given as \( f(x) = -2x + 3 \). The problem asks us to analyze this function's behavior around the point \( (a,b) = (0,3) \). We want to find a \(\delta\) such that within \( a - \delta < x < a + \delta \), the function's values satisfy \( b - \epsilon < f(x) < b + \epsilon \).

Step 2: Set up the epsilon condition

For each given \( \epsilon \), express the condition as \( 3 - \epsilon < -2x + 3 < 3 + \epsilon \). This requires solving this inequality for \( x \).

Step 3: Solve the inequality

Starting with the inequality \( 3 - \epsilon < -2x + 3 < 3 + \epsilon \), subtract 3 from each part to get: \[ -\epsilon < -2x < \epsilon \] Divide each part by \(-2\) and reverse the inequality signs to solve for \(x\):\[ \frac{\epsilon}{2} > x > -\frac{\epsilon}{2} \].

Step 4: Determine \(\delta\) for each \(\epsilon\)

Since the problem requests the function to leave through the sides, this means that the safe range for \(x\) is \(-\frac{\epsilon}{2} < x < \frac{\epsilon}{2} \). Here, \(\delta \) is the greater of the two absolute values, namely \( \delta = \frac{\epsilon}{2} \).

Step 5: Calculate \(\delta\) for each specific \(\epsilon\)

For \(\epsilon = 0.2\), \(\delta = \frac{0.2}{2} = 0.1\).For \(\epsilon = 0.1\), \(\delta = \frac{0.1}{2} = 0.05\).For \(\epsilon = 0.02\), \(\delta = \frac{0.02}{2} = 0.01\).For \(\epsilon = 0.01\), \(\delta = \frac{0.01}{2} = 0.005\).For \(\epsilon = 0.002\), \(\delta = \frac{0.002}{2} = 0.001\).For \(\epsilon = 0.001\), \(\delta = \frac{0.001}{2} = 0.0005\).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a crucial concept in calculus, serving as the foundation of the formal definition of limits. It provides a rigorous mathematical way to define when a function approaches a particular limit. Here's how it works:

• For a given function \( f(x) \) to approach a limit \( L \) as \( x \) approaches \( a \), we want the function values to be as close to \( L \) as desired.
• This closeness is measured using \( \epsilon \), which represents a small positive distance from \( L \).
• The challenge is to find a corresponding \( \delta \) for every \( \epsilon \), such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \epsilon \).

In simpler terms, our task is to ensure that as \( x \) comes close to \( a \) within a \( \delta \) neighborhood, the function \( f(x) \) stays within an \( \epsilon \) band around \( L \).This approach allows us to prove and understand limits without ambiguity, critically supporting concepts such as continuity and derivative.

Linear Functions

Linear functions are among the simplest types of functions you will encounter in calculus. A linear function has the form \( f(x) = mx + c \), where:

• \( m \) is the slope that determines the steepness of the line.
• \( c \) is the y-intercept, which is where the line crosses the y-axis.

In the provided exercise, the function is \( f(x) = -2x + 3 \):

• The slope \(-2\) tells us that for each unit increase in \( x \), \( f(x) \) decreases by 2 units.
• The y-intercept \(+3\) indicates that the line crosses the y-axis at \( y = 3 \).

Understanding these parameters aids in predicting how changes in \( x \) affect \( f(x) \), which is essential for calculating limits and examining function behavior in local neighborhoods.

Inequality Solving

Solving inequalities is a vital skill when dealing with calculus problems, particularly those involving the epsilon-delta definition. Here's how to approach them:1. **Set Up the Inequality**: Start by placing your function within the desired bounds. For the exercise, you have \( 3 - \epsilon < -2x + 3 < 3 + \epsilon \).2. **Manipulate the Inequality**: Adjust the inequality to isolate \( x \), keeping track of changes such as reversing signs when multiplying or dividing by negative values. Here, you first subtract 3 to simplify.3. **Divide and Reverse**: If needed, divide through the inequality by a coefficient, ensuring to reverse the inequality sign when dealing with negative factors, ending with \( \frac{\epsilon}{2} > x > -\frac{\epsilon}{2} \).4. **Determine Valid \( x \) Ranges**: Use the solution to find the valid range for \( x \) where the function acts as desired. This strategic approach helps you identify critical cutoff values for \( x \), which directly provides the necessary \( \delta \) for the epsilon condition.

Calculus Problems

Calculus is full of fascinating problems, many driven by limits, derivatives, and integrals. Let's explore the corridor of thought required to solve such problems.

• **Identifying the Function**: Clearly understand how your function is structured, such as linear, quadratic, etc., since it significantly influences the mathematical tactics you'll employ.
• **Setting Initial Conditions**: Before calculating, note any points or limits you're examining. Utilize these to frame your approach, as they define the boundaries within which you work.
• **Applying Strategic Solutions**: Break down each calculus problem into manageable parts, applying strategies like substitution, integration, or differentiation as needed.
• **Iterate and Refine**: Calculus problems may require iterations and refinements, especially when variables and operations grow more complex.

By grasping these strategies, students can enhance their problem-solving versatility, allowing them to transition smoothly from fundamental definitions, like epsilon-delta, to tackling advanced calculus computations efficiently.