/*! 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 7 Use Horner's method to calculate... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use Horner's method to calculate \(g(-2)\) and \(g^{\prime}(-2)\) where \(g(x)=4 x^{4}-5 x^{3}+6 x-7\). Do not use a computer.

Short Answer

Expert verified
\(g(-2) = -19\) and \(g'(-2) = 16\).

Step by step solution

01

Setup the Polynomial Coefficients

Write down the coefficients of the polynomial \(g(x) = 4x^4 - 5x^3 + 0x^2 + 6x - 7\). These are \([4, -5, 0, 6, -7]\). Notice that the coefficient of \(x^2\) is 0.
02

Apply Horner's Method to Evaluate \(g(-2)\)

Start with the first coefficient (4) and bring it down. Then multiply it by \(-2\) (the x-value we are evaluating at), and add this result to the next coefficient. Continue this process.\[\begin{array}{cccccc}-2 & | & 4 & -5 & 0 & 6 & -7 \ & & & \,8 & -6 & 12 & -12\\hline & & 4 & 3 & -6 & 18 & -19 \\end{array}\]The final number \(-19\) is \(g(-2)\).
03

Setup the Derivative Calculation using Horner's Method

To find \(g'(-2)\), use the same coefficients and method but modify for the derivative. This starts the same, but each row represents the polynomial degree descending by one.
04

Calculate the Derivative \(g^{ ext{'} ext{'}}(-2)\)

Use the same table with a little variation. Discard the last number and multiply the intermediate calculations by \(-2\) again, but now use \(b_{k+1} = c_k\).\[\begin{array}{cccc}-2 & | & 4 & 3 & -6 & 18 & -19\ & & & \, -8 & -2 & 16\\hline & & 4 & -5 & -2 & 16 \\end{array}\]\(g'(-2) = 16\).
05

Verify Results

Verify your calculations: \(g(-2) = -19\) and \(g'(-2) = 16\). Double-check the coefficients and multiplication to ensure the calculations are correct.

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.

Horner's Method
Horner's Method is a clever way to simplify polynomial computations manually. It can be used for both evaluating polynomials and their derivatives at a given point. The process involves a nested sequence of multiplications and additions, dramatically reducing the workload compared to the traditional approach of substituting values into the polynomial.
To use Horner's Method, you begin by writing down the coefficients of the polynomial in descending order of power. For example, for \(g(x) = 4x^4 - 5x^3 + 6x - 7\), the coefficients are [4, -5, 0, 6, -7]. Notice the zero coefficient for \(x^2\), ensuring all terms are accounted for.
Next, select the x-value (let's say \(x = -2\)) at which you want to evaluate the polynomial. You start with the leading coefficient, then repeatedly multiply by the x-value and add the next coefficient. This continues until you reach the final term, yielding the polynomial's evaluated result. This method is advantageous because it requires fewer steps and calculations, thus making manual computations easier.
Polynomial Evaluation
Evaluating a polynomial means finding its value for a specific input. With Horner's Method, this process becomes much simpler. Traditionally, you would substitute the input value into each term, calculate each power, and compute the result.
  • Simplicity: Horner's Method streamlines this by focusing on coefficients and avoiding repetitive calculations of powers.
  • Efficiency: Instead of evaluating powers and multiplying separately, you perform a sequence of add-multiply steps, greatly reducing computation.
  • No Extra Powers: The process only requires basic arithmetic operations without needing to calculate high powers again.
For the polynomial \(g(x) = 4x^4 - 5x^3 + 6x - 7\), evaluating it manually at \(x = -2\), we find that the final result is \(-19\). This is done by starting with the leading coefficient, and sequentially processing the rest as described in the Horner's process.
Derivative Calculation
Calculating a derivative manually can be time-consuming, but Horner's Method offers a streamlined alternative. The method for derivative takes advantage of the same framework used to evaluate the polynomial.
To find the derivative at a specific point, the process involves adjusting the polynomial's order by one less degree each time you move to the next row in calculations. This adjustment accounts for the decrease in power that's inherent in differentiation.
For example, when finding \(g'(-2)\) for the polynomial \(g(x) = 4x^4 - 5x^3 + 6x - 7\), start as if evaluating for \(g(-2)\), but use the results of each multiplication as if it dropped a degree in power. By doing this and leaving out the calculation involving the constant term, you can find that the derivative at \(x = -2\) is 16.
Manual Calculations
In an era where computers and calculators are widespread, understanding how to perform manual calculations is still significant, especially for educational purposes. Manual calculations enhance your understanding of algebraic operations and the behavior of polynomials.
While computers automate the process, using methods such as Horner’s for manual calculations helps develop insight and an intuitive grasp of polynomials and derivatives. You learn the importance of each coefficient and incrementally how changes affect the polynomial's outcome.
  • Practice: Doing manual calculations strengthens your mathematical understanding.
  • Foundation: It reinforces the principles underlying algorithms and prepares you for more complex problem-solving tasks.
  • Check Understanding: By checking computer-based work against manual methods, you gain confidence in your skills.
Thus, although modern tools exist, honing your proficiency in techniques like Horner's Method is invaluable for reinforcing core mathematical concepts.

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

Write an Octave function that implements Horner's method. The first line of your function should be function [p,pprime] \(=\) horner \((\mathrm{x} 0, \mathrm{c})\) where \(c\) is an array containing the coefficients of the polynomial, \(x 0\) is the number at which to evaluate it, \(\mathrm{p}\) is the value of the polynomial at \(\mathrm{x} 0,\) and pprime is the value of the derivative of the polynomial at \(\mathrm{x} 0 .\) This way, the values \(\mathrm{p}\) and pprime are returned by the function in an array. The function is called like this: $$ [\mathrm{y}, \mathrm{yy}]=\text { horner }(-2,[5,4,3,2,1]), $$ setting y to the value of the polynomial and yy to the value of its derivative. Test your code well by comparing outputs of your function to hand/calculator computations.

Compute \(a_{0}, a_{1},\) and \(a_{2}\) of Aitken's delta-squared method for the sequence in problem 2 on page \(29 .\) Since the sequence has an undefined term at \(n=1\), start the sequence \(\left\langle\frac{n+1}{n-1}\right\rangle\) with \(n=2 .\) In other words, consider the sequence in problem 2 on page 29 to be \(3,2, \frac{5}{3}, \frac{3}{2}, \frac{7}{5} \ldots\) so \(p_{0}=3, p_{1}=2, p_{2}=\frac{5}{3},\) and so on.

Using Horner's method as an aid, and not using your calculator, find the first iteration of Newton's method for the function \(f(x)=2 x^{3}-10 x+1\) using \(x_{0}=2\).

Suppose you are using the secant method with \(x_{0}=1\) and \(x_{1}=1.1\) to find a root of \(f(x)\). (a) Find \(x_{2}\) given that \(f(1)=0.3\) and \(f(1.1)=0.23\). (b) Create a sketch (graph) that illustrates the calculation. HINT: \(x_{2}\) will be located where the line through \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(x_{1}, f\left(x_{1}\right)\right)\) crosses the \(x\) axis.

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.