/*! 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 63 The equation \(\frac{1}{3} \sqrt... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 63

The equation \(\frac{1}{3} \sqrt[3]{x}-x+2=0\) has a root near \(2 .\) To approximate this root, rewrite the equation as \(x=\frac{1}{3} \sqrt[3]{x}+2\) Let \(x_{1}=2\) and find successive approximations \(x_{2}, x_{3}, \ldots\) by using the formulas $$ x_{2}=\frac{1}{3} \sqrt[3]{x_{1}}+2, \quad x_{3}=\frac{1}{3} \sqrt[3]{x_{2}}+2, \quad \ldots $$ until four-decimal-place accuracy is obtained.

Short Answer

Expert verified
The root is approximately 2.0944 with four-decimal-place accuracy.

Step by step solution

01

Identify Initial Condition

We are given the initial approximation for the root, which is \(x_1 = 2\). This will be our starting point for the iterative process.
02

Apply the Iterative Formula for x_2

Substitute \(x_1 = 2\) into the iterative formula to find \(x_2\):\[x_2 = \frac{1}{3} \sqrt[3]{2} + 2\]Calculate \(\sqrt[3]{2} \approx 1.2599\). Then, \(x_2 = \frac{1}{3}\times 1.2599 + 2 \approx 2.0866\).
03

Calculate Next Approximation x_3

Using the value for \(x_2\), calculate \(x_3\):\[x_3 = \frac{1}{3} \sqrt[3]{x_2} + 2\]Find \(\sqrt[3]{2.0866} \approx 1.2808\). Then, \(x_3 = \frac{1}{3} \times 1.2808 + 2 \approx 2.0936\).
04

Continue Iterative Process for x_4

Compute \(x_4\) using \(x_3\):\[x_4 = \frac{1}{3} \sqrt[3]{x_3} + 2\]Calculate \(\sqrt[3]{2.0936} \approx 1.2830\). Therefore, \(x_4 = \frac{1}{3} \times 1.2830 + 2 \approx 2.0943\).
05

Determine x_5 for More Accuracy

Proceed to calculate \(x_5\) from \(x_4\):\[x_5 = \frac{1}{3} \sqrt[3]{x_4} + 2\]Find \(\sqrt[3]{2.0943} \approx 1.2832\). Thus, \(x_5 = \frac{1}{3} \times 1.2832 + 2 \approx 2.0944\).
06

Check for Desired Accuracy

Comparing the values of \(x_4\) and \(x_5\), both are \(2.0943\) and \(2.0944\) respectively. The values agree up to four decimal places. Thus, we ensure that the desired accuracy has been achieved.

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Key Concepts

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

Cubic Roots
Cubic roots are essential when dealing with equations that require solving for the root of a variable raised to the third power. The cubic root of a number, say \( n \), is the value that when multiplied by itself three times equals \( n \). In math, this is written as \( \sqrt[3]{n} \). For instance, \( \sqrt[3]{8} = 2 \), because \( 2 \times 2 \times 2 = 8 \). Understanding cubic roots is fundamental when transforming complex equations into simpler forms for iterative solutions. In this exercise, cubic roots are directly applied to isolate \( x \), making it possible to find its approximate value using a sequential approach.
Equation Solving
Solving equations is the process of finding unknown values that make the equation true. Equations can often be rearranged to isolate the variable of interest. In this specific problem, the goal was to find the value of \( x \) that satisfies the equation \( \frac{1}{3} \sqrt[3]{x} - x + 2 = 0 \). By rearranging this, isolating \( x \) gives the equation \( x = \frac{1}{3} \sqrt[3]{x} + 2 \). Such rearrangements simplify the process to solve equations iteratively. This involves repeatedly applying a formula to zero in on more accurate approximations.
Approximation Techniques
Approximation techniques are handy when exact solutions are difficult due to computational complexity or when only a sufficiently accurate result is required. In this exercise, the iterative method is used to approximate a root. Iteration begins with an initial guess, \( x_1 \), and successively applies a formula to improve this estimate. For example, using the equation \( x = \frac{1}{3} \sqrt[3]{x} + 2 \), the iterative process employs calculations like \( x_2 = \frac{1}{3} \sqrt[3]{x_1} + 2 \), and so on. This continues until the results converge to a value accurate to the required number of decimal places.
Precalculus
Precalculus serves as the bridge between algebra, trigonometry, and calculus. It encompasses a variety of mathematical concepts, including functions and their transformations, which are vital for understanding more advanced topics. In this exercise, understanding functions and transformations is crucial. By isolating \( x \), a function is formed that can be iteratively solved to get an approximate root. Techniques from precalculus allow for recognizing patterns and applying suitable changes to an equation, setting the stage for deeper exploration in calculus. Such foundational knowledge in equation manipulation is indispensable for higher-level mathematics.

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