Question

In calculus you will learn that if $$ p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} $$ is a polynomial function, then the derivative of \(p(x)\) is $$ p^{\prime}(x)=n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\cdots+2 a_{2} x+a_{1} $$ Newton's Method is an efficient method for approximating the \(x\) -intercepts (or real zeros) of a function, such as \(p(x)\). The following steps outline Newton's Method. STEP 1: Select an initial value \(x_{0}\) that is somewhat close to the \(x\) -intercept being sought. STEP 2: Find values for \(x\) using the relation $$ x_{n+1}=x_{n}-\frac{p\left(x_{n}\right)}{p^{\prime}\left(x_{n}\right)} \quad n=0,1,2, \ldots $$ until you get two consecutive values \(x_{n}\) and \(x_{n+1}\) that agree to whatever decimal place accuracy you desire. STEP 3: The approximate zero will be \(x_{n+1}\). Consider the polynomial \(p(x)=x^{3}-7 x-40\) (a) Evaluate \(p(3)\) and \(p(5)\). (b) What might we conclude about a zero of \(p ?\) (c) Use Newton's Method to approximate an \(x\) -intercept, \(r\), \(3

Short answer

The approximate zero of p(x) in the interval 3 < r < 5 is 4.0933.

Step-by-step solution

Evaluate p(x) at x = 3 and x = 5

First, substitute 3 and 5 into the polynomial function to find their values. For x = 3: \[ p(3) = 3^3 - 7 \times 3 - 40 = 27 - 21 - 40 = -34 \] For x = 5: \[ p(5) = 5^3 - 7 \times 5 - 40 = 125 - 35 - 40 = 50 \]

Analyze the possible zero

Since p(3) is -34 (negative) and p(5) is 50 (positive), by the Intermediate Value Theorem, there must be at least one real zero between x = 3 and x = 5, because the function changes signs in this interval.

Calculate the derivative of p(x)

Find the derivative of the polynomial function \( p(x) = x^3 - 7x - 40 \). Using the power rule:\[ p'(x) = 3x^2 - 7 \]

Apply Newton's Method

Use the iterative formula \[ x_{n+1} = x_n - \frac{p(x_n)}{p'(x_n)} \] to approximate the x-intercept. Start with an initial guess close to the zero, say \( x_0 = 4 \). Then:1. Calculate \( x_1 = x_0 - \frac{p(x_0)}{p'(x_0)} \). Substitute \( x_0 = 4 \): \[ p(4) = 64 - 28 - 40 = -4 \] \[ p'(4) = 3(16) - 7 = 41 \] \[ x_1 = 4 - \frac{-4}{41} = 4 + \frac{4}{41} = 4.0976 \] (rounded to four decimal places)2. Repeat the process with \( x_1 \): \[ p(4.0976) \approx 4.0976^3 - 7\cdot4.0976 - 40 = 0.1816 \] \[ p'(4.0976) \approx 3\cdot4.0976^2 - 7 = 42.346 \] \[ x_2 = 4.0976 - \frac{0.1816}{42.346} \approx 4.0933 \]3. Repeat the process again:\[ p(4.0933) \approx 0.0001, \]\[ p'(4.0933) = 42.239, \]\[ x_3 = 4.0933 - \frac{0.0001}{42.239} \approx 4.0933 \](agreement to four decimal places).

Conclude the result

After a few iterations, two consecutive values \( x_i \) and \( x_{i+1} \) agree to four decimal places, hence the approximate zero is 4.0933.

Key concepts

Calculus

Calculus is a branch of mathematics that studies change. It is divided mainly into two parts: differential calculus and integral calculus. Differential calculus focuses on the concept of a derivative, which represents the rate of change of a function. It helps in understanding how a function changes at any given point. Integral calculus, on the other hand, deals with the accumulation of quantities and areas under or between curves.
The concepts of limits, derivatives, integrals, and infinite series are essential in calculus. In the context of Newton's Method, calculus, especially differential calculus, plays a crucial role in finding roots of functions effectively. We use derivatives to construct a sequence of points that converge to a zero (root) of the function. By analyzing how the function behaves, specifically how it changes, we can predict values that inch us closer to the roots we seek.

Polynomial Functions

A polynomial function is represented as an expression of the form p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\text{...} +a_{1} x+a_{0} where each coefficient \(a_i\) is a constant and \(n\) is a non-negative integer that defines the degree of the polynomial. Polynomial functions are smooth and continuous.
You can find polynomial functions everywhere, like quadratic and cubic functions. In our exercise, the function given is \(p(x) = x^3 - 7x - 40\), which is a cubic polynomial because the highest power of \(x\) is 3.

• These polynomial functions have various characteristics such as intercepts, turning points, and end behavior.
• To solve polynomial equations, we look for values of \(x\) that satisfy the equation \(p(x) = 0\), called the roots or zeros of the function.

Derivatives

The derivative of a function measures how the function's value changes as its input changes. Mathematically, the derivative of a function \(p(x)\) is represented as \(p'(x)\). For a polynomial function like \(p(x) = x^3 - 7x - 40\), its derivative can be found using power rules.
Using the power rule, the derivative of \(x^n\) is \(nx^{n-1}\). Thus, the derivative of our polynomial \(p(x)\) is:

• \(p'(x) = 3x^2 - 7\)
  Derivatives are essential in Newton's Method because we need the derivative \(p'(x)\) to apply the iterative formula and find the zero of the polynomial efficiently. Without derivatives, we wouldn't understand the slope or rate of change of our function at specific points, making it challenging to achieve accurate approximations.

Root-Finding Algorithm

Newton's Method is a root-finding algorithm used to approximate the zeros or roots of a real-valued function. To start, it requires an initial guess \(x_0\) that is somewhat close to the actual root. The iterative formula used in Newton's Method is
{x_{n+1} = x_n - \frac{p(x_n)}{p'(x_n)}\}
Here's how it works step-by-step:

• Select Initial Guess: Choose an initial guess \(x_0\) close to where you believe the root lies.The closer it is to the actual root, the fewer iterations you'll need.
• Iterate: Use the formula to calculate new values \(x_{n+1}\) until consecutive values converge to an acceptable degree of accuracy. In our example, starting with \(x_0 = 4\), subsequent iterations bring us closer to the root.
• Converge: Continue the iterations until the values of \(x_n\) and \(x_{n+1}\) agree to a specified decimal precision; in our case, four decimal places.

The key to Newton's Method is the use of both the function \(p(x)\) and its derivative \(p'(x)\) in the iterative formula, making it a powerful and efficient tool for finding roots of nonlinear functions.