Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider the equation \(x=2\left(x-x^{2}\right)=g(x)\). (a) Sketch the graph of \(y=x\) and \(y=g(x)\) using the same coordinate system, and thereby approximately locate the positive root of \(x=g(x)\). (b) Try solving the equation by the Fixed-Point Algorithm starting with \(x_{1}=0.7\). (c) Solve the equation algebraically.

Step-by-Step Solution

Answer

The positive root of the equation is approximately \(x = 0.5\).

Step 1: Understand the Functions

We have two functions to analyze. The first function is the line \(y = x\). The second is \(y = g(x) = 2(x - x^2)\), which simplifies to \(y = 2x - 2x^2\). We will plot both functions to find their point of intersection, which gives us the root of \(x = g(x)\).

Step 2: Sketch the Graphs

Plot the graph of the linear function \(y = x\), which is a straight line passing through the origin with slope 1. Next, plot the graph of the quadratic function \(y = 2x - 2x^2\), which is a downward-facing parabola with roots at \(x = 0\) and \(x = 1\) and a vertex at \(x = 0.5\). The intersection of these graphs approximates the positive root of the equation.

Step 3: Approximate Root by Graphs

By inspecting the graph, identify the approximate positive root where \(y = x\) intersects \(y = g(x)\). Upon close observation, the intersection appears around \(x = 0.6\).

Step 4: Set Up Fixed-Point Iteration

To find the root using the Fixed-Point Algorithm, start with an initial guess \(x_1 = 0.7\). Use the iteration formula \(x_{n+1} = g(x_n) = 2x_n - 2x_n^2\).

Step 5: Perform Iterations

1. Calculate \(x_2 = g(0.7) = 2(0.7) - 2(0.7)^2 = 0.84\). 2. Calculate \(x_3 = g(0.84) = 2(0.84) - 2(0.84)^2 = 0.5376\). 3. Calculate \(x_4 = g(0.5376) = 2(0.5376) - 2(0.5376)^2 \approx 0.9974\). Continue iterating until \(x_n\) stabilizes.

Step 6: Algebraic Solution

Set \(x = 2x - 2x^2\) and simplify to \(0 = x - 2x^2\). Factor the equation to get \(0 = x(1 - 2x)\). This yields the solutions \(x = 0\) and \(x = 0.5\). The positive root we're interested in is \(x = 0.5\).

Key Concepts

Fixed-Point Algorithm

The Fixed-Point Algorithm is a simple iterative method used to find solutions to equations of the form \(x = g(x)\). The basic idea is to choose an initial guess, denoted as \(x_1\), and then apply the function iteratively to get closer to the solution.

Here's how it works:

• Start with an initial guess \(x_1\).
• Compute \(x_2 = g(x_1)\).
• Continue this process to get \(x_3 = g(x_2)\), and so on.
• Iterate until the values converge, meaning \(x_{n+1} \approx x_n\).

In our exercise, starting with \(x_1 = 0.7\), we apply the function \(2x - 2x^2\) iteratively. As a result, the values start to converge towards the solution. This method is particularly useful when the equation is difficult to solve algebraically. Always ensure to choose a good initial guess close to the actual solution for faster convergence.

Function Intersection

Function intersection involves finding points where two or more graphs intersect. This is a powerful concept because it helps determine solutions common to multiple equations. For instance, in the problem, we need to find where the graphs of \(y = x\) and \(y = g(x)\) intersect. This gives us the roots of the equation

Here are the steps:

• Sketch the graphs of both functions on the same coordinate plane.
• Identify the coordinates where they meet. These points are the solutions.

For the problem above, plotting \(y = x\) (a straight line) and \(y = 2x - 2x^2\) (a parabola) reveals that they intersect around \(x = 0.6\). Visually inspecting graphs helps to approximate the solutions before solving them numerically or algebraically.

Quadratic Functions

Quadratic functions form one of the fundamental concepts in algebra and calculus. They are polynomial functions of degree 2, expressed as \(y = ax^2 + bx + c\). A key feature of quadratic functions is their parabolic shape.

In the exercise, we have the quadratic function \(y = 2x - 2x^2\), which can be rewritten as \(y = -2x^2 + 2x\). Important characteristics include:

• The vertex of this parabola is located at \(x = 0.5\), the midpoint of the roots.
• The graph is downward facing because the coefficient of \(x^2\) is negative.
• The roots are found by setting \(y = 0\), giving us solutions \(x = 0\) and \(x = 1\).

Understanding these properties helps in both graphing and solving quadratics, whether visually or using algebraic techniques.

Graph Sketching

Graph sketching is a vital skill for analyzing functions. By sketching graphs, we can observe the overall behavior of functions, especially how they intersect. This can often give insights into approximate solutions with simple visual inspection.

When sketching graphs:

• Identify key points like intercepts, roots, and vertex for quadratic functions.
• Mark these points with precision on your graph.
• Connect the points smoothly, adhering to the nature of the graph (e.g., linear or parabolic).
• Consider the slope and the direction (upward or downward) for the parabolas.

For the problem at hand, sketching both the line \(y = x\) and the parabola \(y = 2x - 2x^2\) allows us to visually approximate intersections near \(x = 0.6\). This technique is not only useful for solving equations but also for gaining a deeper understanding of function behaviors.