Problem 25 - Thomas Calculus

Question

Use the Intermediate Value Theorem from Section 2.5 to show that \(f(x)=x^{3}+2 x-4\) has a root between \(x=1\) and \(x=2 .\) Then find the root to five decimal places.

Step-by-Step Solution

Answer

The root of \(f(x)=x^{3}+2x-4\) between \(x=1\) and \(x=2\) is approximately \(x=1.17969\).

Step 1: Understand the Intermediate Value Theorem

The Intermediate Value Theorem states that if \( f \) is continuous on \([a, b]\) and \( N \) is any number between \( f(a) \) and \( f(b) \), then there is a \( c \) in the interval \([a, b]\) such that \( f(c) = N \). In this case, we need to demonstrate that \( f(x) = x^3 + 2x - 4 \) has a root, so we have to check if the function changes sign between \( x = 1 \) and \( x = 2 \).

Step 2: Calculate Values at the Endpoints

Compute \( f(1) \) and \( f(2) \). \( f(1) = 1^3 + 2 \times 1 - 4 = 1 + 2 - 4 = -1 \). \( f(2) = 2^3 + 2 \times 2 - 4 = 8 + 4 - 4 = 8 \). Since \( f(1) = -1 \) and \( f(2) = 8 \), the function changes sign, confirming the existence of a root between \( x = 1 \) and \( x = 2 \).

Step 3: Use Bisection Method to Approximate the Root

Use the bisection method, which involves evaluating the function at the midpoint and adjusting the interval based on where the sign change occurs. 1. Start with the interval \([1, 2]\). 2. Compute \( f(1.5) = (1.5)^3 + 2 \times 1.5 - 4 = 3.375 + 3 - 4 = 2.375 \).3. The root lies between \(1\) and \(1.5\) since \( f(1) = -1 \) and \( f(1.5) = 2.375\).4. Compute \( f(1.25) = (1.25)^3 + 2 \times 1.25 - 4 = 1.953125 + 2.5 - 4 = 0.453125 \).5. The root lies between \(1\) and \(1.25\).6. Repeat more bisections (\([1, 1.125], [1, 1.0625], [1, 1.03125], etc.\)) until \( b - a \) is less than or equal to \(0.00001\), obtaining a root between \(1.179685\) and \(1.179695\).

Step 4: Confirm the Root by Further Calculation

Once you have narrowed down the interval to a sufficiently small size, try evaluating \( f(x) \) around this interval to obtain more accuracy. For example, check between \( 1.17968 \) and \( 1.17969 \) to ensure the estimated root to five decimal places lies in this interval. Use a smaller interval if needed to tighten the approximation.

Key Concepts

Root Finding

Root finding is a fundamental concept in mathematics where the aim is to determine the values of a variable, say \( x \), such that a given function \( f(x) \) equals zero. These values are known as the roots of the function. The process is crucial in various fields because it helps find solutions to equations that model real-world situations.
When searching for these roots, the Intermediate Value Theorem becomes quite handy. It assures us that if a continuous function changes signs over an interval, there must be at least one root in that interval. This theorem ties directly into the concept of continuous functions, which we'll explore later.
Root finding methods like the bisection method provide systematic approaches to isolate and approximate these roots accurately.

Bisection Method

The bisection method is a straightforward and reliable algorithm used for root finding, particularly applicable when dealing with continuous functions. It stands out for its simplicity, making it an excellent choice for approximating roots when precision is needed. This method leverages the Intermediate Value Theorem, ensuring a sign change over an interval implies a contained root.
Here's how the method works:

• Select two initial points, \( a \) and \( b \), such that \( f(a) \) and \( f(b) \) have opposite signs, indicating a root is present between them.
• Calculate the midpoint \( m = \frac{a + b}{2} \) and evaluate \( f(m) \).
• If \( f(m) \) is zero, \( m \) is the root. Otherwise, use the sign of \( f(m) \) to determine which half-interval contains the root.
• Repeat the process on the new interval until the desired precision is reached.

This method is especially useful because it generates results that converge to the true root, albeit sometimes requiring many iterations for high precision.

Continuous Functions

A function is called continuous if it is, in simple terms, free from any breaks, holes, or jumps. Mathematically, this means that small changes in \( x \) lead to small changes in \( f(x) \). Continuity is crucial in guaranteeing the existence of roots within a closed interval when paired with certain conditions like sign changes, as described in the Intermediate Value Theorem.
In the context of root finding, continuous functions play a vital role because they ensure predictable behavior across intervals and enable specific methods like the bisection method to work effectively. For example, when dealing with a polynomial function like \( x^3 + 2x - 4 \), which is inherently continuous, we can confidently use methods relying on this property to approximate roots successfully. Understanding continuity helps in analyzing and solving problems involving real-world data represented by mathematical functions.

Polynomial Functions

Polynomial functions are a special class of continuous functions characterized by their simple algebraic form. Typically expressed as \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \, ... \, + a_1 x + a_0 \), they consist of variables raised to whole-number powers with constant coefficients. This layout makes them smooth and predictable across their domain.
Because they are both continuous and differentiable, polynomial functions are excellent candidates for root-finding techniques like the bisection method. They avoid irregularities that would complicate computations or violate the assumptions required by the Intermediate Value Theorem.
In practical applications, solving polynomial equations to find roots is crucial in engineering, physics, economics, and beyond, as these functions often model complex systems and patterns in a manageable format.