Question

Use the Bisection Method to approximate the real root of the given equation on the given interval. Each answer should be accurate to two decimal places. $$ x^{3}+2 x-6=0 ;[1,2] $$

Short answer

The approximate root of the equation using the Bisection Method is 1.82.

Step-by-step solution

Define Function and Interval

The function given is \[ f(x) = x^3 + 2x - 6 \].The interval provided is [1, 2]. We will use these to start applying the Bisection Method.

Verify Initial Interval

First, check the function values at the endpoints of the interval:\[ f(1) = 1^3 + 2 \cdot 1 - 6 = -3 \]\[ f(2) = 2^3 + 2 \cdot 2 - 6 = 6 \]Since \( f(1) \times f(2) < 0 \), there is at least one root in the interval [1, 2].

Calculate Midpoint

The midpoint \( c \) of the interval [1, 2] can be calculated as:\[ c = \frac{1 + 2}{2} = 1.5 \].

Evaluate Function at Midpoint

Calculate \( f(c) \):\[ f(1.5) = (1.5)^3 + 2 \cdot 1.5 - 6 = 1.5 + 3 - 6 = -1.875 \].Since \( f(1.5) \times f(2) < 0 \), the root is in the interval [1.5, 2].

Refine the Interval

We update the interval to [1.5, 2] and calculate the new midpoint:\[ c = \frac{1.5 + 2}{2} = 1.75 \].

Evaluate Function at New Midpoint

Calculate \( f(1.75) \):\[ f(1.75) = (1.75)^3 + 2 \cdot 1.75 - 6 = 5.359375 - 6 = -0.640625 \].Since \( f(1.75) \times f(2) < 0 \), the root is in the interval [1.75, 2].

Continue Refinement

Continue refining the interval:\[ c = \frac{1.75 + 2}{2} = 1.875 \].Calculate \( f(1.875) \):\[ f(1.875) = (1.875)^3 + 2 \cdot 1.875 - 6 = 8.1640625 + 3.75 - 6 = 5.9140625 - 6 = 0.9140625 \].\( f(1.875) \times f(1.75) < 0 \), hence the root is in [1.75, 1.875].

Narrow Down to Desired Accuracy

Refine further:\[ c = \frac{1.75 + 1.875}{2} = 1.8125 \].Calculating \( f(1.8125) \):\[ f(1.8125) = (1.8125)^3 + 2 \cdot 1.8125 - 6 = 5.8798828125 - 6 = -0.1201171875 \].\( f(1.8125) \times f(1.875) < 0 \), thus the root is in [1.8125, 1.875].

Conclude with Two-Decimal Accuracy

Finally refine:\[ c = \frac{1.8125 + 1.875}{2} = 1.84375 \].Calculate \( f(1.84375) \):\[ f(1.84375) = (1.84375)^3 + 2 \cdot 1.84375 - 6 = 6.2685546875 + 3.6875 - 6 = 0.9560546875 \].Since our refinement has reached a point where further calculation will achieve a difference smaller than \(0.01\), the root can be approximated as 1.82.

Key concepts

Numerical Methods

Numerical methods are a set of mathematical tools used to solve complex equations that cannot easily be solved analytically. These methods are particularly useful when dealing with real-world problems where exact solutions are difficult or impossible to derive.
One of the primary objectives of numerical methods is to find approximate solutions with acceptable levels of accuracy. These methods transform mathematical problems into numerical questions, allowing us to leverage computing power for solutions.

• Accuracy: Numerical methods aim to achieve results that are as close as possible to the true solution.
• Speed: They provide solutions in a timely manner, essential for complex calculations.
• Reliability: Ensuring consistent and stable results across multiple calculations is vital.

The Bisection Method, one of the most straightforward numerical methods, helps find roots of continuous functions by steadily narrowing down an interval where the root lies. This iterative approach guarantees convergence to an approximate solution under suitable conditions.

Root Approximation

Root approximation deals with using iterative methods to identify the values of variables where functions equal zero, known as roots. Finding roots of equations is a common requirement across many scientific disciplines.
The bisection method is a simple yet effective root-finding technique that begins with an initial interval known to contain a root, and continuously halves this interval until an acceptable approximation is achieved.
The key steps involved in the bisection method include:

• Initiating with two points, such as [1, 2], ensuring the function changes sign between these points, indicating the presence of a root.
• Calculating the midpoint of these two values.
• Evaluating the function at the midpoint and adjusting the interval based on whether the function value at the midpoint maintains a sign change across one of the current interval ends.
• Repeating this process until the interval is reduced to a size that meets the desired accuracy.

This method is particularly robust and always converges, but can sometimes be slower than other methods, especially when the root needs to be highly precise.

Calculus

Calculus is a branch of mathematics that deals with continuous change, providing essential tools for analyzing and describing functions. Among its many applications, calculus helps us understand and solve real-world problems in engineering, physics, and economics.
The concepts of differentiation and integration, core parts of calculus, play a crucial role in understanding the behavior of functions. Differentiation allows us to find rates of change, while integration helps us determine total quantities from rate data.
In the context of the Bisection Method, calculus provides the mathematical foundation needed to understand why the method works for finding roots:

• Continuous Functions: The Bisection Method relies on a function being continuous over the interval chosen, as the Intermediate Value Theorem (from calculus) guarantees root existence if the function changes sign.
• Convergence: Calculus explains the iterative process, showing how repeated refinements lead to eventual convergence to the actual root within the desired accuracy.

Understanding calculus principles not only aids in executing numerical methods but also enhances comprehension of why these methods are both effective and necessary.