/*! 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 4 Use the Bisection Method to appr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the Bisection Method to approximate the real root of the given equation on the given interval. Each answer should be accurate to two decimal places. $$ x-2+2 \ln x=0 ;[1,2] $$

Short Answer

Expert verified
The approximate root using the Bisection Method is 1.39.

Step by step solution

01

Understand the Equation

We have the equation \( x - 2 + 2\ln x = 0 \). Our task is to find the root of this equation using the Bisection Method in the interval \([1, 2]\).
02

Set Initial Points

Set \( a = 1 \) and \( b = 2 \). We want to ensure that \( f(a) \) and \( f(b) \) have opposite signs, indicating that there is a root between them.
03

Check Initial Function Values

Compute \( f(1) = 1 - 2 + 2\ln(1) = -1 \), and \( f(2) = 2 - 2 + 2\ln(2) \approx 1.386 \). Since \( f(1) < 0 \) and \( f(2) > 0 \), we have a root in \([1, 2]\).
04

Compute First Midpoint

Find the midpoint \( c = \frac{1 + 2}{2} = 1.5 \). Compute \( f(1.5) = 1.5 - 2 + 2\ln(1.5) \approx 0.194 \). Since \( f(1.5) > 0 \), the root is in \([1, 1.5]\).
05

Compute Second Midpoint

In the new interval \([1, 1.5]\), find the midpoint \( c = \frac{1 + 1.5}{2} = 1.25 \). Compute \( f(1.25) = 1.25 - 2 + 2\ln(1.25) \approx -0.215 \). Since \( f(1.25) < 0 \), the root is in \([1.25, 1.5]\).
06

Compute Third Midpoint

Now, in the interval \([1.25, 1.5]\), find the midpoint \( c = \frac{1.25 + 1.5}{2} = 1.375 \). Compute \( f(1.375) = 1.375 - 2 + 2\ln(1.375) \approx -0.019 \). Since \( f(1.375) < 0 \), the root is in \([1.375, 1.5]\).
07

Compute Fourth Midpoint

Find the new midpoint in \([1.375, 1.5]\), which is \( c = \frac{1.375 + 1.5}{2} = 1.4375 \). Compute \( f(1.4375) = 1.4375 - 2 + 2\ln(1.4375) \approx 0.084 \). Since \( f(1.4375) > 0 \), the root is in \([1.375, 1.4375]\).
08

Compute Fifth Midpoint

Find the midpoint \( c = \frac{1.375 + 1.4375}{2} = 1.40625 \). Compute \( f(1.40625) = 1.40625 - 2 + 2\ln(1.40625) \approx 0.0315 \). Since \( f(1.40625) > 0 \), the root is in \([1.375, 1.40625]\).
09

Compute Sixth Midpoint

Calculate the midpoint \( c = \frac{1.375 + 1.40625}{2} = 1.390625 \). Compute \( f(1.390625) = 1.390625 - 2 + 2\ln(1.390625) \approx 0.0060 \). Since \( f(1.390625) > 0 \), the root is in \([1.375, 1.390625]\).
10

Compute Seventh Midpoint

Calculate the midpoint \( c = \frac{1.375 + 1.390625}{2} = 1.3828125 \). Compute \( f(1.3828125) = 1.3828125 - 2 + 2\ln(1.3828125) \approx -0.0065 \). Since \( f(1.3828125) < 0 \), the root is in \([1.3828125, 1.390625]\).
11

Compute Eighth Midpoint

Find the midpoint \( c = \frac{1.3828125 + 1.390625}{2} = 1.38671875 \). Compute \( f(1.38671875) = 1.38671875 - 2 + 2\ln(1.38671875) \approx -0.0002 \), which is very close to zero.
12

Conclusion with Approximated Root

The Bisection Method approximates the root as \( x \approx 1.39 \), and since \( f(1.38671875) \approx 0 \), we conclude that the root is approximately 1.39 to two decimal places.

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 91Ó°ÊÓ!

Key Concepts

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

Numerical Analysis
Numerical analysis is a fascinating field dedicated to developing algorithms to approximate solutions of complex mathematical problems. These problems often can't be solved analytically with exact values. Bisection Method, used in our exercise, is an excellent example of numerical analysis in practice.
The goal of numerical analysis is to deliver solutions that are not only approximate but also efficient and practical for real-world applications. In the realm of root-finding problems, numerical analysis helps us devise strategies like bisection, which iteratively zero in on a potential root. The power of numerical analysis lies in its ability to handle large computations where manual calculations are impractical.
In general, various algorithms from this field control error margins carefully to ensure that calculations remain reliable. They are structured to execute functions rapidly while maintaining precision. As we saw in the exercise, a repeated sequence of straightforward operations enables the bisection method to converge on an answer quickly and accurately.
Root-Finding Algorithms
Root-finding algorithms are crucial in solving equations where typical algebraic methods falter. They assist in pinpointing values for which a given function equals zero. The Bisection Method is a straightforward and powerful root-finding algorithm employed in numerical analysis.
Here's a look at why it's effective:
  • Reliability: Utilizes a structured approach by narrowing down intervals where the root is contained. This eliminates guesswork, ensuring that the approximation steadily improves.
  • Guaranteed Convergence: As long as the initial interval is chosen correctly, this method is assured to converge towards a real root, due to its reliance on the Intermediate Value Theorem.
  • Simplicity: Its operation is basic, requiring only the computation of function values and their sign. This simplicity assures that the process is easy to understand and implement.
However, it should be noted that while bisection is simple and reliable, it can sometimes be slower than other methods, especially if one desires a highly precise root.
Logarithmic Functions
Logarithmic functions, denoted by \( \ln(x) \) for the natural logarithm, are an inverse operation of exponential functions. In our exercise, the equation involved a term \( 2 \ln(x) \) illustrating the role logarithms play in equations.
Logarithms have several properties that make them useful:
  • Transformative power: Complex expressions can be simplified using logarithms, making them easier to manipulate and solve.
  • Wide applicability: They appear in diverse fields such as science, engineering, and finance, aiding in modeling growth and decay phenomena.
  • Dimensions of Scale: As they scale exponentially growing values into more manageable numbers, they facilitate easy comprehension and analysis of substantial data magnitudes.
When employing root-finding algorithms, it's crucial to understand the characteristics of logarithmic functions, since they can influence the behavior of the function, especially regarding growth rates and asymptotic behavior. The interplay between this growth and the other terms in an equation can lead to intriguing challenges in finding roots.

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

A dead body is found at \(10 \mathrm{PM}\). to have temperature \(82^{\circ} \mathrm{F}\). One hour later the temperature was \(76^{\circ} \mathrm{F}\). The temperature of the room was a constant \(70^{\circ} \mathrm{F}\). Assuming that the temperature of the body was \(98.6^{\circ} \mathrm{F}\) when it was alive, estimate the time of death.

The population of the United States was \(3.9\) million in 1790 and 178 million in 1960 . If the rate of growth is assumed proportional to the number present, what estimate would you give for the population in \(2000 ?\) (Compare your answer with the actual 2000 population, which was 275 million.)

Brass is produced in long rolls of a thin sheet. To monitor the quality, inspectors select at random a piece of the sheet, measure its area, and count the number of surface imperfections on that piece. The area varies from piece to piece. The following table gives data on the area (in square feet) of the selected piece and the number of surface imperfections found on that piece. $$ \begin{array}{ccc} \hline \text { Piece } & \begin{array}{c} \text { Area in } \\ \text { Square Feet } \end{array} & \begin{array}{c} \text { Number of } \\ \text { Surface Imperfections } \end{array} \\ \hline 1 & 1.0 & 3 \\ 2 & 4.0 & 12 \\ 3 & 3.6 & 9 \\ 4 & 1.5 & 5 \\ 5 & 3.0 & 8 \\ \hline \end{array} $$ (a) Make a scatter plot with area on the horizontal axis and number of surface imperfections on the vertical axis. (b) Does it look like a line through the origin would be a good model for these data? Explain. (c) Find the equation of the least-squares line through the origin. (d) Use the result of part (c) to predict how many surface imperfections there would be on a sheet with area \(2.0\) square feet

GG 44. The Census Bureau estimates that the growth rate \(k\) of the world population will decrease by roughly \(0.0002\) per year for the next few decades. In \(2004, k\) was \(0.0132\). (a) Express \(k\) as a function of time \(t\), where \(t\) is measured in years since 2004 . (b) Find a differential equation that models the population \(y\) for this problem. (c) Solve the differential equation with the additional condition that the population in \(2004(t=0)\) was \(6.4\) billion. (d) Graph the population \(y\) for the next 300 years. (e) With this model, when will the population reach a maximum? When will the population drop below the 2004 level?

What constant acceleration will cause a car to increase its velocity from 45 to 60 miles per hour in 10 seconds?

See all solutions

Recommended explanations on Math Textbooks

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.