/*! 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 23 The graphs of \(y=x^{2}(x+1)\) a... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 23

The graphs of \(y=x^{2}(x+1)\) and \(y=1 / x(x>0)\) intersect at one point \(x=r .\) Use Newton's method to estimate the value of \(r\) to four decimal places.

Short Answer

Expert verified
The value of \( r \) is approximately 0.9055.

Step by step solution

01

Set up the equation

To find the intersection points of the graphs, set the equations equal to each other:\[ x^2(x+1) = \frac{1}{x} \] Simplify and rearrange the equation:\[ x^3 + x^2 - \frac{1}{x} = 0 \] Multiply through by \( x \) to eliminate the fraction:\[ x^4 + x^3 - 1 = 0 \] This is the function \( f(x) = x^4 + x^3 - 1 \) that we want to solve using Newton's method.
02

Derivative of the function

Find the derivative of the function \( f(x) = x^4 + x^3 - 1 \):\[ f'(x) = 4x^3 + 3x^2 \] We will use this derivative in Newton's method to find the root.
03

Apply Newton's Method

Newton's method formula is:\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Start with an initial guess \( x_0 \). Since \( x > 0 \) and we need the graphs to intersect, a reasonable initial guess is around \( x_0 = 1 \).
04

Iterations

Perform iterations using Newton's method until the desired precision is achieved.1. **Iteration 1**: \( x_0 = 1 \) \[ f(1) = 1^4 + 1^3 - 1 = 1 \] \[ f'(1) = 4 \times 1^3 + 3 \times 1^2 = 7 \] \[ x_1 = 1 - \frac{1}{7} \approx 0.8571 \]2. **Iteration 2**: \( x_1 = 0.8571 \) \[ f(0.8571) = 0.8571^4 + 0.8571^3 - 1 \approx -0.2013 \] \[ f'(0.8571) = 4 \times 0.8571^3 + 3 \times 0.8571^2 \approx 4.655 \] \[ x_2 = 0.8571 - \frac{-0.2013}{4.655} \approx 0.9003 \]3. **Iteration 3**: \( x_2 = 0.9003 \) \[ f(0.9003) = 0.9003^4 + 0.9003^3 - 1 \approx -0.0257 \] \[ f'(0.9003) = 4 \times 0.9003^3 + 3 \times 0.9003^2 \approx 5.173 \] \[ x_3 = 0.9003 - \frac{-0.0257}{5.173} \approx 0.9053 \]4. **Iteration 4**: \( x_3 = 0.9053 \) \[ f(0.9053) = 0.9053^4 + 0.9053^3 - 1 \approx -0.0014 \] \[ f'(0.9053) = 4 \times 0.9053^3 + 3 \times 0.9053^2 \approx 5.250 \] \[ x_4 = 0.9053 - \frac{-0.0014}{5.250} \approx 0.9055 \]5. **Iteration 5**: \( x_4 = 0.9055 \) \[ f(0.9055) = 0.9055^4 + 0.9055^3 - 1 \approx -0.000005 \] \[ f'(0.9055) = 4 \times 0.9055^3 + 3 \times 0.9055^2 \approx 5.253 \] \[ x_5 = 0.9055 - \frac{-0.000005}{5.253} \approx 0.9055 \]Since \( |x_5 - x_4| \) is very small and within our desired precision, we can stop.
05

Conclude result

The intersection point's value of \( x = r \) is approximately \( 0.9055 \), which solves the original problem to four decimal places.

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Key Concepts

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

Intersection of Graphs
Understanding the intersection of graphs is essential in finding where two or more functions collide or meet. When we have functions like \(y = x^2(x+1)\) and \(y = \frac{1}{x} (x > 0)\), the intersection point reveals the \(x\)-value where their outputs (\(y\)-values) are the same.
Finding these intersection points involves setting the functions equal to each other and solving for \(x\). By equating these, you're looking for the values of \(x\) that satisfy both equations simultaneously. This is akin to finding a common solution or root for both equations.
This exercise involves dynamic graphical analysis as it doesn't check just the visual point of intersection but calculates it using precise mathematical techniques like those involved in Newton's Method.
Derivative Calculation
The derivative of a function provides critical information about the function's rate of change. In the context of Newton's Method, the derivative helps in adjusting our guesses to approximate better the roots of an equation.
For the function \(f(x) = x^4 + x^3 - 1\), its derivative is calculated as \(f'(x) = 4x^3 + 3x^2\). This derivative essentially provides the slope of the tangent line at any given point on \(f(x)\).
  • The slope tells us how steep the curve is at a particular point.
  • It indicates whether the function is increasing or decreasing.
  • It's used in Newton's Method to refine our estimation of the root.
In Newton's Method, the derivative plays a crucial role in repositioning our estimate closer to where the graph actually intersects the x-axis.
Polynomial Equation
Polynomial equations, like our function \(x^4 + x^3 - 1 = 0\), are expressions containing variables with whole number exponents. The degree of this polynomial is four because the highest power of \(x\) is 4.
Working with polynomial equations requires finding the roots or zeros, which are the \(x\)-values leading the polynomial to equal zero. These roots are where the polynomial crosses or touches the \(x\)-axis.
  • High-degree polynomials can have multiple roots.
  • The Fundamental Theorem of Algebra tells us that this polynomial has at least one root for every degree it possesses.
  • Advanced methods like Newton's Method are often required to efficiently find real roots.
In the intersection problem, rearranging the initial equation helps derive a solvable polynomial, which we then tackle using iterative methods for precise solutions.
Root Finding
Root finding involves determining where a function equals zero, essentially where the graph of the function intersects the \(x\)-axis. Methods for root finding can vary from simple algebraic manipulations to sophisticated numerical approaches.
Newton's Method is a popular technique used for root finding, especially for its speed and simplicity.
  • An initial guess for the root \(x_0\) is required.
  • The method uses an iterative formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), continually refining the guess.
  • This iterative process stops when the changes in \(x\) become negligible, reaching the desired precision.
In our problem, Newton's method began with a guess of \(x = 1\) and quickly narrowed down to an accurate root of approximately 0.9055, showcasing its efficiency in root approximation.

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