Problem 73 - 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. \(g(x)=-x^{3}+2 ; a=0 ; b=2 ; \epsilon=0.1,0.01,0.001\)

Step-by-Step Solution

Answer

For \( \epsilon = 0.1, 0.01, 0.001 \), \( \delta \approx 1.03, 0.998, 0.999 \) respectively.

Step 1: Understand the Problem

We are given a function \( g(x) = -x^3 + 2 \) and need to find a value \( \delta \) for given \( \epsilon \) values: 0.1, 0.01, and 0.001. The requirement is that the graph exits the strip \( a-\delta < x < a+\delta \) through the sides within \( b-\epsilon < y < b+\epsilon \) range.

Step 2: Evaluate the Function at the Center Point

The function is centered at \( x = a = 0 \). At this point, \( g(a) = g(0) = -0^3 + 2 = 2. \) Therefore, the point on the graph at this center is \( (0, 2) \).

Step 3: Define the Range for y Values

Given \( b = 2 \), the target range for \( y \) is \( b-\epsilon < y < b+\epsilon \). For \( \epsilon = 0.1 \), \( 1.9 < y < 2.1 \). Similarly, for \( \epsilon = 0.01 \), \( 1.99 < y < 2.01 \), and for \( \epsilon = 0.001 \), \( 1.999 < y < 2.001 \).

Step 4: Solve for \( \delta \) with \( \epsilon = 0.1 \)

The inequalities become \( 1.9 < -x^3 + 2 < 2.1 \). Rearranging gives \( 0.9 < -x^3 < 1.1 \), simplifying to \( -1.1 < x^3 < -0.9 \). Solving these inequalities for \( x \), we have \( \sqrt[3]{-1.1} < x < \sqrt[3]{-0.9} \). Thus, the corresponding \( \delta \) is \( \delta \approx \min(|\sqrt[3]{-1.1}|, |\sqrt[3]{-0.9}|) \approx 1.03. \)

Step 5: Solve for \( \delta \) with \( \epsilon = 0.01 \)

The inequalities become \( 1.99 < -x^3 + 2 < 2.01 \). Rearranging gives \( 0.99 < -x^3 < 1.01 \), simplifying to \( -1.01 < x^3 < -0.99 \). Solving these inequalities for \( x \), we have \( \sqrt[3]{-1.01} < x < \sqrt[3]{-0.99} \). Thus, the corresponding \( \delta \) is \( \delta \approx \min(|\sqrt[3]{-1.01}|, |\sqrt[3]{-0.99}|) \approx 0.998. \)

Step 6: Solve for \( \delta \) with \( \epsilon = 0.001 \)

The inequalities become \( 1.999 < -x^3 + 2 < 2.001 \). Rearranging gives \( 0.999 < -x^3 < 1.001 \), simplifying to \( -1.001 < x^3 < -0.999 \). Solving these inequalities for \( x \), we have \( \sqrt[3]{-1.001} < x < \sqrt[3]{-0.999} \). Thus, the corresponding \( \delta \) is \( \delta \approx \min(|\sqrt[3]{-1.001}|, |\sqrt[3]{-0.999}|) \approx 0.999. \)

Key Concepts

Limits

The concept of limits is fundamental in calculus. It helps us understand the behavior of a function as the input approaches a particular value. Imagine driving a car towards a traffic signal - the idea of limits is like predicting what speed your car's speedometer will show as you get closer to stopping. Limits let us see what value a function approaches, even if it never actually reaches that exact point. To find the limit of a function as it approaches a certain point, we observe the values the function gets close to from both the left and right sides of that point.
In the epsilon-delta definition of limits, \( \epsilon \) represents how close the function's output needs to be to its limit, and \( \delta \) ensures the inputs are close enough to the point of interest. This creates a window or range within which the function behaves predictably. The challenge with limits is formalizing them with these epsilon and delta values - this helps us rigorously define closeness and continuity in math.

Continuous Functions

Continuous functions are smooth without breaks or jumps - imagine gliding over a surface without abruptly falling. For a function to be continuous at a point, its limit must exist at that point, and the function value must match that limit. Using our exercise, the function \( g(x) = -x^3 + 2 \) is continuous, which is why we can apply the epsilon-delta method effectively.
If you plot the graph of a continuous function, you can draw it without lifting your pen. Mathematically, this means, for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( |x - a| < \delta \, \, then \, |g(x) - b| < \epsilon \). This principle ensures that within a very small interval of inputs (determined by \( \delta \)), the outputs remain close to a target value by no more than \( \epsilon \). Understanding this concept is crucial for solving calculus problems involving continuity and limits.

Calculus Problem Solving

Solving calculus problems effectively requires mastering certain strategies, especially when dealing with limits and continuous functions. Here's how you can tackle these problems:

• Understand the Problem: Start by carefully reading the question and identifying what's being asked. In our exercise, it's essential to find how small \( \delta \) must be for various \( \epsilon \) values.
• Evaluate Key Points: Calculate the function at specific points to better understand its behavior. Recognizing where it's continuous gives insight into how the function's values change.
• Apply Mathematical Tools: Use algebraic manipulation for solving inequalities involving \( \delta \) and \( \epsilon \). This includes solving cubic equations as we saw with determining \( \delta \).
• Deductive Reasoning: Visualize how the epsilon-delta window confines the solution. This acts as a guideline, allowing you to determine necessary conditions for the function's output.

With practice, tackling calculus problems becomes more intuitive, and the epsilon-delta approach is a great way to see these concepts in action. Remember, patience and a solid grasp of these foundational ideas are key to mastering limits and continuity in calculus.