/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Find all fixed points of the fol... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 1

Find all fixed points of the following \(g(x)\). (a) \(\frac{3}{x}\) (b) \(x^{2}-2 x+2\) (c) \(x^{2}-4 x+2\)

Short Answer

Expert verified
Question: List all the fixed points of the following functions: (a) \(g(x) = \frac{3}{x}\) (b) \(g(x) = x^{2}-2 x+2\) (c) \(g(x) = x^{2}-4 x+2\) Answer: (a) \(x = \sqrt{3}\), \(x = -\sqrt{3}\) (b) \(x = 1\), \(x = 2\) (c) \(x = \frac{5 + \sqrt{17}}{2}\), \(x = \frac{5 - \sqrt{17}}{2}\)

Step by step solution

01

(a) Find fixed points of \(g(x) = \frac{3}{x}\)

To find the fixed points, we need to solve the equation \(g(x) = x\) for x: \[\frac{3}{x} = x\] To solve this equation, multiply both sides by x: \[3 = x^2\] Now, we can find the square root of both sides: \[x = \pm\sqrt{3}\] So, (a) has two fixed points, \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).
02

(b) Find fixed points of \(g(x) = x^{2}-2 x+2\)

To find the fixed points, we need to solve the equation \(g(x) = x\) for x: \[x^2 - 2x + 2 = x\] Subtract x from both sides: \[x^2 - 3x + 2 = 0\] Now let's find the roots using the quadratic formula: \[x = \frac{-\text{b} \pm \sqrt{\text{b}^2 - 4\text{ac}}}{2\text{a}}\] In this case, \(a = 1\), \(b = -3\), and \(c = 2\). Plug in the values: \[x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)}\] \[x = \frac{3 \pm \sqrt{1}}{2}\] \[x = \frac{3 \pm 1}{2}\] The two fixed points are: \[x = \frac{3 + 1}{2} = 2\] \[x = \frac{3 - 1}{2} = 1\] So, (b) has two fixed points, \(x = 1\) and \(x = 2\).
03

(c) Find fixed points of \(g(x) = x^{2}-4 x+2\)

To find the fixed points, we need to solve the equation \(g(x) = x\) for x: \[x^2 - 4x + 2 = x\] Subtract x from both sides: \[x^2 - 5x + 2 = 0\] Now let's find the roots using the quadratic formula: \[x = \frac{-\text{b} \pm \sqrt{\text{b}^2 - 4\text{ac}}}{2\text{a}}\] In this case, \(a = 1\), \(b = -5\), and \(c = 2\). Plug in the values: \[x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(2)}}{2(1)}\] \[x = \frac{5 \pm \sqrt{17}}{2}\] The two fixed points are: \[x = \frac{5 + \sqrt{17}}{2}\] \[x = \frac{5 - \sqrt{17}}{2}\] So, (c) has two fixed points, \(x = \frac{5 + \sqrt{17}}{2}\) and \(x = \frac{5 - \sqrt{17}}{2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Numerical Analysis
Numerical analysis is a field of mathematics that deals with the development of algorithms and methods to obtain numerical solutions to various mathematical problems. These can range from simple algebraic equations to complex differential equations.

In the context of finding fixed points for a function, numerical analysis would involve iterative methods which provide successively better approximations to the actual solutions. This can be especially handy when dealing with more complicated functions where analytical solutions are difficult to find or when an exact solution is not necessary, and a close approximation suffices.

Examples of numerical methods include the Newton-Raphson method, the bisection method, and the fixed-point iteration method. These methods are iterative in nature, meaning they repeatedly apply a calculation to move closer to the desired fixed point.
  • The Newton-Raphson method uses the derivative of the function to find its roots.
  • The bisection method divides an interval into halves to locate the root.
  • Fixed-point iteration involves rearranging the equation to make x the subject and then using iteration to approach the fixed point.
For the simpler problems in our exercise, such as a linear or quadratic equation, numerical analysis might not be necessary, as analytical solutions are readily obtainable via algebraic manipulations.
Quadratic Formula
The quadratic formula is a critical tool in algebra for solving quadratic equations, which are of the form \(ax^2 + bx + c = 0\). This formula provides the roots, or solutions, to the quadratic equation:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Using the quadratic formula is a straightforward process:
  • Identify the coefficients \(a\), \(b\), and \(c\) from the equation.
  • Plug these values into the formula
  • Simplify to find the values of \(x\), which are the roots.
The quadratic formula is essential for finding fixed points in quadratic functions when set equal to \(x\). As seen in our solved examples, it simplifies what could otherwise be a complex algebraic process.
Root Finding
Root finding in mathematics is the process of determining the values for which a given function equals zero. These values are known as the roots of the function. Root finding is a fundamental aspect of algebra and forms the basis for solving equations.

There are various methods available for root finding, from graphical interpretations to analytic methods like factoring, using the quadratic formula, or applying the aforementioned numerical analysis techniques.

In the case of finding fixed points, we are specifically looking for the values of \(x\) where the function \(g(x)\) intersects the line \(y=x\). Solving \(g(x) = x\) is essentially a root-finding problem where we want to find the roots of the function \(g(x) - x\). The methods we choose may vary depending on the complexity of the function \(g(x)\) and the precision required for the roots. Our examples demonstrate how fixed points for various functions are found by using algebraic manipulation and applying the quadratic formula, which is an analytical root-finding technique suitable for well-defined polynomials.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If the Secant Method converges to \(r, f^{\prime}(r) \neq 0\), and \(f^{\prime \prime}(r) \neq 0\), then the approximate error relationship \(e_{i+1} \approx\left|f^{\prime \prime}(r) /\left(2 f^{\prime}(r)\right)\right| e_{i} e_{i-1}\) can be shown to hold. Prove that if in addition \(\lim _{i \rightarrow \infty} e_{i+1} / e_{i}^{\alpha}\) exists and is nonzero for some \(\alpha>0\), then \(\alpha=(1+\sqrt{5}) / 2\) and \(e_{i+1} \approx\left|\left(f^{\prime \prime}(r) / 2 f^{\prime}(r)\right)\right|^{\alpha-1} e_{i}^{\alpha} .\)

Consider Fixed-Point Iteration with \(g(x)=x-x^{3}\). (a) Show that \(x=0\) is the only fixed point. (b) Show that if \(0x_{1}>x_{2} \ldots>0\). (c) Show that FPI converges to \(r=0\), while \(g^{\prime}(0)=1\). (Hint: Use the fact that every bounded monotonic sequence converges to a limit.)

Define Fixed-Point Iteration on the equation \(x^{2}+x=5 / 16\) by isolating the \(x\) term. Find both fixed points, and determine which initial guesses lead to each fixed point under iteration.

Let \(n\) be a positive integer. The equation defining the \(n\)th root of a positive number \(A\) is \(x^{n}-A=0\). (a) Find the multiplicity of the root. (b) Show that, for an approximate \(n\)th root with small forward error, the backward error is approximately \(n A^{(n-1) / n}\) times the forward error.

Consider the equation \(8 x^{4}-12 x^{3}+6 x^{2}-x=0\). For each of the two solutions \(x=0\) and \(x=1 / 2\), decide which will converge faster (say, to eight-place accuracy), the Bisection Method or Newton's Method, without running the calculation.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.