Problem 3 - University Calculus: Early Transcendentals

Question

Use Newton's method to estimate the two zeros of the function \(f(x)=x^{4}+x-3 .\) Start with \(x_{0}=-1\) for the left-hand zero and with \(x_{0}=1\) for the zero on the right. Then, in each case, find \(x_{2}\).

Step-by-Step Solution

Answer

Left-hand zero: \(x_2 \approx -1.645\); Right-hand zero: \(x_2 \approx 1.1655\).

Step 1: Define the Function and Its Derivative

First, we need to find the derivative of the function. We start with \( f(x) = x^4 + x - 3 \). The derivative, \( f'(x) \), is calculated as follows: \ \[ f'(x) = \frac{d}{dx}(x^4 + x - 3) = 4x^3 + 1. \] \"},{

Step 2: Set Up Newton's Method Formula

Newton's method uses the formula: \ \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] \for the next approximation \( x_{n+1} \) based on the current approximation \( x_n \). We will use this formula to find the zeros of \( f(x) \).

Step 3: Calculate \(x_1\) for \(x_0 = -1\)

Start with \( x_0 = -1 \) to find the left-hand zero. Plug \( x_0 = -1 \) into the function and its derivative: \\[ f(-1) = (-1)^4 + (-1) - 3 = 1 - 1 - 3 = -3 \] \\[ f'(-1) = 4(-1)^3 + 1 = -4 + 1 = -3 \] \Use Newton's method formula: \\[ x_1 = -1 - \frac{-3}{-3} = -1 - 1 = -2. \]

Step 4: Calculate \(x_2\) for Initial \(x_0 = -1\)

With \( x_1 = -2 \), calculate \( x_2 \): \\[ f(-2) = (-2)^4 + (-2) - 3 = 16 - 2 - 3 = 11 \] \\[ f'(-2) = 4(-2)^3 + 1 = -32 + 1 = -31 \] \Use Newton's method: \\[ x_2 = -2 - \frac{11}{-31} = -2 + \frac{11}{31} \approx -1.645. \]

Step 5: Calculate \(x_1\) for \(x_0 = 1\)

Now calculate \( x_1 \) for the right-hand zero with \( x_0 = 1 \). Evaluate \( f(x) \) and \( f'(x) \) at \( x_0 = 1 \): \\[ f(1) = (1)^4 + 1 - 3 = 1 + 1 - 3 = -1 \] \\[ f'(1) = 4(1)^3 + 1 = 4 + 1 = 5 \] \Use Newton's method: \\[ x_1 = 1 - \frac{-1}{5} = 1 + \frac{1}{5} = 1.2. \]

Step 6: Calculate \(x_2\) for Initial \(x_0 = 1\)

With \( x_1 = 1.2 \), calculate \( x_2 \): \\[ f(1.2) = (1.2)^4 + 1.2 - 3 = 2.0736 + 1.2 - 3 = 0.2736 \] \\[ f'(1.2) = 4(1.2)^3 + 1 = 6.928 + 1 = 7.928 \] \Use Newton's method: \\[ x_2 = 1.2 - \frac{0.2736}{7.928} \approx 1.1655. \]

Key Concepts

Derivative Calculation

Calculating the derivative of a function is a crucial step in understanding how that function behaves rather than just what it looks like. When we talk about Newton's Method, the derivative helps us find the slope of the tangent, which we use to estimate the zero of a function.
To get started with a function like \( f(x) = x^4 + x - 3 \), first take its derivative. This involves applying the power rule. For \( x^4 \), the derivative is \( 4x^3 \). For \( x \), it’s simply 1. Constant terms, like -3, have a derivative of 0, since they don’t change.

• The function's derivative is \( f'(x) = 4x^3 + 1 \).

The derivative \( f'(x) \) gives us the rate of change of the function \( f(x) \) at any point \( x \). Knowing the derivative allows you to apply Newton's Method, which relies on calculating slopes to find zeros.

Zero Estimation

Zero estimation is the process of finding solutions where the function equals zero, often referred to as 'roots'. Newton's Method is particularly useful in estimating these zeros, requiring both a function and its derivative.
Begin with an initial guess \( x_0 \), which is ideally close to where the zero is suspected to be. The goal is to refine this guess iteratively to get closer to the actual zero.
In practice, starting with \( x_0 = -1 \) for the left-hand zero involves plugging \( x_0 \) into your function \( f(x) = x^4 + x - 3 \) and its derivative \( f'(x) = 4x^3 + 1 \). Calculating these values gives insights into how far off the initial guess is.

• For example, \( f(-1) = -3 \) and \( f'(-1) = -3 \).
• An initial guess \( x_0 = 1 \) allows you to calculate \( f(1) = -1 \) and \( f'(1) = 5 \).

These initial computations set the stage for using the Newton's Method formula to adjust and improve your guesses.

Iterative Approximation

Iterative approximation is a repeating process that gradually hones in on a specific target. In the context of Newton's Method, it’s about making successive improvements on zero estimates.
The method uses the formula: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). This means each new estimate, \( x_{n+1} \), is computed using the current estimate, \( x_n \), the function value \( f(x_n) \), and its derivative \( f'(x_n) \).

• For \( x_0 = -1 \), the next step is \( x_1 = -2 \), improving upon \( x_0 \) by applying the formula.
• Continue iterating: \( x_2 \approx -1.645 \). Each step moves you closer to the actual zero.
• For \( x_0 = 1 \), calculate \( x_1 = 1.2 \), and the iteration for \( x_2 \approx 1.1655 \).

This method requires a few rounds of calculation, but with each iteration, you edge closer to finding the function’s zero, improving precision with each step.