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.

Short answer

The positive root of \(x = g(x)\) is \(x = 0.5\).

Step-by-step solution

Analyze the Function

Consider the function given by the equation \(x = 2(x - x^2) = g(x)\). We can express this as \(g(x) = 2x - 2x^2\). The graph of this function and the line \(y = x\) need to be plotted to find the points where \(x = g(x)\).

Sketch the Graphs

1. The graph of \(y = x\) is a straight line passing through the origin with a slope of 1. 2. The graph of \(y = g(x) = 2x - 2x^2\) is a downward-opening parabola. 3. To sketch this parabola, identify key points such as when \(x = 0\), \(x = 0.5\), and \(x = 1\). The approximate shape of the graph can be plotted using these points. 4. The intersection of these graphs represents the solutions for \(x = g(x)\).

Initial Approximation of the Positive Root

From the graph, locate the intersection points of the line \(y = x\) and the parabola. The positive intersection point can be approximated around \(x = 0.6\) to \(x = 0.7\).

Apply the Fixed-Point Algorithm

Use the initial guess \(x_1 = 0.7\) and iterate with \(x_{n+1} = g(x_n) = 2x_n - 2x_n^2\). - Iteration 1: \(x_2 = g(0.7) = 2(0.7) - 2(0.7)^2 = 0.98\).- Iteration 2: \(x_3 = g(0.98) = 2(0.98) - 2(0.98)^2 = 0.9608\).- Continue iterating until \(x_n\) converges.

Algebraic Solution to the Equation

Solve \(x = 2x - 2x^2\) algebraically:1. Rearrange terms: \(2x^2 - x = 0\).2. Factor out \(x\): \(x(2x - 1) = 0\).3. Solve for \(x\): \(x = 0\) or \(2x - 1 = 0\).4. For \(2x - 1 = 0\), solve for \(x\): \(x = \frac{1}{2}\). The positive root is \(x = \frac{1}{2}\).

Key concepts

Graph of a function

A graph is a visual representation of a function, helping to quickly understand its behavior and key characteristics. For the equation from our exercise, we analyze the graphs of two functions:

• The line given by the equation \(y = x\).
• The parabola given by the equation \(y = g(x) = 2x - 2x^2\).

The straight line \(y = x\) passes through the origin and any point where the x and y coordinates are equal. This means the line has a slope of 1, creating a diagonal across the coordinate plane.
The parabola \(y = 2x - 2x^2\) opens downward because the coefficient of \(x^2\) is negative. We can find relevant points to sketch the parabola. For instance:

• At \(x = 0\), \(g(x) = 0\).
• At \(x = 0.5\), \(g(x) = 0.5\).
• At \(x = 1\), \(g(x) = 0\).

These points help sketch the parabola, and seeing both graphs together shows where they intersect.

Positive root

The positive root of an equation is the value of \(x\) that satisfies the equation and is greater than zero. In this exercise, we look for values where:
\[ x = g(x) = 2x - 2x^2 \]
From sketching the graphs of \(y = x\) and \(y = 2x - 2x^2\), we identify the realistic intersections. Of particular interest are intersections in the range where \(x\) is positive.
The graph gives us a visual approximation of where the positive root might be, suggesting that it can be around \(x = 0.6\) to \(x = 0.7\). This initial guess can be sharper by mathematical methods or further iterations.

Algebraic solution

To solve \(x = g(x)\) algebraically, we set
\[ x = 2x - 2x^2\]
Rearranging terms to zero gives us:
\[ 2x^2 - x = 0\]
We then factorize it:

• Factor out \(x\): \(x(2x - 1) = 0\).

This gives two potential solutions:

• \(x = 0\) and
• \(2x - 1 = 0\).

Solving \(2x - 1 = 0\),
we find:
\[ x = \frac{1}{2} \]
This indicates that among the factors, \(x = 0\) and \(x = \frac{1}{2}\), only \(x = \frac{1}{2}\) is a positive root, in line with the expected root found in graph approximations.

Intersection of graphs

The intersection of the graphs of \(y = x\) and \(y = g(x)\) (or \(y = 2x - 2x^2\)) tells us the solution(s) to the equation \(x = g(x)\).
Where they meet are points that satisfy both equations simultaneously. Graph intersections are crucial as they offer visual solutions even before formal computation.

• The line \(y = x\) intersects with the parabolic curve \(y = g(x)\), and this intersection can be visually traced.

For this particular graph, intersections occur at values such as \(x \approx 0.5\), the algebraic solution confirms \(x = \frac{1}{2}\).

• The intersection provides tangible answers to what would otherwise remain abstract algebraic terms.
• They are also the starting point for more precise numeric methods like fixed-point iteration.

Visually, this aids the intuition and understanding of what these values represent, making connections clearer.