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Chapter 4: Problem 4

Use Newton's method to estimate the two zeros of the function \(f(x)=2 x-x^{2}+1 .\) Start with \(x_{0}=0\) for the left-hand zero and with \(x_{0}=2\) for the zero on the right. Then, in each case, find \(x_{2}\).

Short Answer

Expert verified
The second iterations are \( x_2 = -0.583 \) for the left-hand zero, and \( x_2 = 2.417 \) for the right-hand zero.

Step by step solution

01

Find the Derivative of the Function

To apply Newton's method, we need to find the first derivative of the function. Given the function \[ f(x) = 2x - x^2 + 1, \]we differentiate it to get:\[ f'(x) = 2 - 2x. \]
02

Apply Newton's Method Formula

Newton's method is given by the iterative formula:\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}. \] We'll use this formula to find approximations for the zeros of the function on each interval.
03

Find the First Iteration for the Left-Hand Zero

Start with the initial guess \( x_0 = 0 \). Calculate \( x_1 \) using the formula:\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0 - \frac{f(0)}{f'(0)}.\]Substitute \( f(0) = 2(0) - 0^2 + 1 = 1 \) and \( f'(0) = 2 - 2(0) = 2 \), giving:\[ x_1 = 0 - \frac{1}{2} = -0.5. \]
04

Find the Second Iteration for the Left-Hand Zero

Now with \( x_1 = -0.5 \), find \( x_2 \):\[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = -0.5 - \frac{f(-0.5)}{f'(-0.5)}.\]Calculate \( f(-0.5) = 2(-0.5) - (-0.5)^2 + 1 = 0.25 \) and \( f'(-0.5) = 2 + 1 = 3 \):\[ x_2 = -0.5 - \frac{0.25}{3} \approx -0.583. \]
05

Find the First Iteration for the Right-Hand Zero

Start with the initial guess \( x_0 = 2 \). Calculate \( x_1 \):\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{f(2)}{f'(2)}.\]Substitute \( f(2) = 2(2) - 2^2 + 1 = 1 \) and \( f'(2) = 2 - 4 = -2 \), giving:\[ x_1 = 2 - \frac{1}{-2} = 2.5.\]
06

Find the Second Iteration for the Right-Hand Zero

Now with \( x_1 = 2.5 \), find \( x_2 \):\[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2.5 - \frac{f(2.5)}{f'(2.5)}.\]Calculate \( f(2.5) = 2(2.5) - (2.5)^2 + 1 = -0.25 \) and \( f'(2.5) = 2 - 5 = -3 \):\[ x_2 = 2.5 - \frac{-0.25}{-3} = 2.5 + 0.083 \approx 2.417.\]

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Key Concepts

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

Derivative of Function
Understanding derivatives is essential for many calculus applications, including Newton's Method. A derivative represents the rate at which a function is changing at any given point. It's essentially the slope of the tangent line to the curve at a particular location. When handling the function \( f(x) = 2x - x^2 + 1 \), the process of differentiation gives us \( f'(x) = 2 - 2x \). This derivative is key to Newton's Method because it allows us to determine how sharply the function is climbing or descending at a point. By calculating \( f'(x) \), we can effectively make predictions about how to adjust our estimates in Newton's iterative process.
Iterative Formula
Newton's Method relies on a specific iterative formula. This formula is beautifully simple yet powerful for numerical approximations. Defined as:
  • \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \)
Each term in the formula has a unique role:
  • \( x_n \) is our current estimate of the root.
  • \( f(x_n) \) is the value of the function at \( x_n \).
  • \( f'(x_n) \) provides the slope at that point.
  • The fraction \( \frac{f(x_n)}{f'(x_n)} \) tells us how much to correct our current estimate.
This correction brings us closer to the actual root with each iteration, making the estimation process both efficient and insightful.
Approximating Zeros
Approximating zeros means finding values of \( x \) where the function \( f(x) \) equals zero. These are the points where the graph of the function touches or crosses the x-axis. For the function \( f(x) = 2x - x^2 + 1 \), we employed Newton's Method starting with initial guesses at 0 and 2. Through consecutive iterations, our approximations for the zeros became more precise:
  • Beginning with \( x_0 = 0 \), the next estimation \( x_1 \) was \(-0.5\), and further refined to \( x_2 = -0.583\).
  • For \( x_0 = 2 \), the first adjustment provided \( x_1 = 2.5 \), followed by \( x_2 = 2.417\).
This systematic approach helps in incrementally zooming in on the actual root locations.
Calculus Problem Solving
Calculus offers various methodologies for solving complex problems, one of which is using Newton's Method to find function zeros. The process involves several calculus concepts:
  • First, compute the derivative to understand the function's behavior.
  • Next, apply Newton's iterative formula to find a zero.
  • Each iteration refines the approximation, gradually converging to a solution.
This problem-solving technique is invaluable because it transforms the abstract calculus problem into a concrete step-by-step process of prediction and correction. By breaking down problems into manageable parts, students can systematically approach daunting calculus challenges and achieve a deeper understanding through practice.

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Most popular questions from this chapter

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