Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the formula for the amount \(E(x)\) by which a number \(x\) exceeds its square. Plot a graph of \(E(x)\) for \(0 \leq x \leq 1\). Use the graph to estimate the positive number less than or equal to 1 that exceeds its square by the maximum amount.

Step-by-Step Solution

Answer

The maximum amount by which a number \(x\) exceeds its square, for \(0 \leq x \leq 1\), is 0.25 at \(x = 0.5\).

Step 1: Set Up the Equation

To find the function for the amount by which a number \(x\) exceeds its square, we need to establish an equation. We know \(E(x) = x - x^2\). The expression \(x - x^2\) represents the difference between the number \(x\) and its square.

Step 2: Express the Function

We've identified the function \(E(x) = x - x^2\). This function shows how much a number \(x\) exceeds its square for any given \(x\).

Step 3: Plot the Graph

We now need to plot \(E(x) = x - x^2\) for \(0 \leq x \leq 1\). This will require calculating and plotting values within this range, such as \(E(0)\), \(E(0.5)\), etc. Generally, we can use graphing software or a graphing calculator for accuracy.

Step 4: Find the Turning Point

To find the point at which \(E(x)\) is maximized, we can calculate the derivative of \(E(x)\), which is \(E'(x) = 1 - 2x\). Setting \(E'(x) = 0\) gives us \(1 - 2x = 0\), leading to \(x = 0.5\). The function reaches its maximum when \(x = 0.5\).

Step 5: Verify the Maximum

To confirm that \(x = 0.5\) is indeed a maximum, check the values of \(E(x)\) around this point. For instance: \(E(0) = 0\), \(E(0.5) = 0.25\), and \(E(1) = 0\). The function values increase to 0.25 at \(x = 0.5\) and decrease thereafter.

Key Concepts

Function Analysis

Understanding the concept of a function is essential in calculus. In our exercise, the function is expressed as \(E(x) = x - x^2\). This function measures how much the number \(x\) exceeds its square. Function analysis allows us to break down this expression:

• \(x\) is a variable, representing any real number.
• \(x^2\) is the square of \(x\), a key term that reflects how numbers grow.
• \(E(x)\) symbolizes the difference between \(x\) and its square.

Function analysis often involves identifying the behavior of functions over set domains. In our case, the interval \(0 \leq x \leq 1\) is crucial. This means we are only interested in values of \(x\) between 0 and 1. The function transforms input values of \(x\) into output values of \(E(x)\), highlighting how \(x\) influences the difference it makes with its square.

Derivative

The derivative of a function provides insight into its rate of change. For \(E(x) = x - x^2\), we calculate its derivative, \(E'(x) = 1 - 2x\). In simple terms, the derivative tells us how the function \(E(x)\) changes with respect to \(x\).

• The term \(1\) shows the constant rate at which the function's base term \(x\) increases.
• The term \(-2x\) indicates how rapidly \(x^2\) reduces the amount that \(x\) exceeds its square.

By setting the derivative equal to zero, \(1 - 2x = 0\), we find critical points. These points provide potential maximum or minimum values. Solving for \(x\), we realize that the critical point \(x = 0.5\) is where \(E(x)\) attains a maximum, confirming that the function increases and then decreases around this point.

Graph Plotting

Plotting a graph is a visual representation of a function's behavior. For \(E(x) = x - x^2\), graph plotting within the interval \(0 \leq x \leq 1\) displays how \(E(x)\) changes.
Gathering some key values assists in plotting:

• \(E(0) = 0\)
• \(E(0.5) = 0.25\)
• \(E(1) = 0\)

When plotted, these reveal the graph is a downward opening parabola with a peak at \(x = 0.5\). This parabola represents the maximum point at \(x = 0.5\) where \(E(x) = 0.25\). As \(x\) moves from 0 to 0.5, the graph slopes upward to the maximum, then it declines back to zero as \(x\) approaches 1. Visualizing the function helps identify the maximum point effectively.

Optimization

Optimization is the process of finding the best solution, often a maximum or minimum value. In our exercise, \(E(x)\) needs to be maximized, meaning we want to know the point where \(x\) exceeds its square the most. This involves using the derivative \(E'(x) = 1 - 2x\) to find critical points by setting it equal to zero and solving for \(x\).

• Solving \(1 - 2x = 0\) gives \(x = 0.5\).
• Testing values around \(x = 0.5\), like \(E(0) = 0\), \(E(0.5) = 0.25\), and \(E(1) = 0\), confirms that \(E(0.5)\) is higher than others.

This optimal point at \(x = 0.5\) indicates the maximum amount, 0.25, by which \(x\) exceeds its square. Solutions like these are pivotal in calculus to help determine the most efficient point for any function's particular attributes within a given range.