Problem 16 - Single Variable Calculus

Question

Approximate the zero of the function in the indicated interval to six decimal places. \(f(x)=x^{3}-x-1\) in \([1,2]\)

Step-by-Step Solution

Answer

The zero of the function \(f(x) = x^3 - x - 1\) within the given interval \([1,2]\) is approximately \(x \approx 1.134724\).

Step 1: Define the function and interval

First, we need to define the function in question: \(f(x) = x^3 - x - 1\). We are given the interval \([1,2]\) within which to find the zero of this function.

Step 2: Evaluate function at interval endpoints

Now, evaluate the function at the endpoints of the interval. We have: \(f(1) = (1)^3 - (1) - 1 = -1\) \(f(2) = (2)^3 - (2) - 1 = 5\) We have \(f(1) < 0\) and \(f(2) > 0\), so there must be a zero in the interval \([1, 2]\) by the Intermediate Value Theorem.

Step 3: Apply the Bisection Method

Next, we will apply the Bisection Method to find the zero of the function to six decimal places. This involves iteratively cutting the interval in half until the zero is found to the required approximation. For each iteration, perform the following steps: 1. Compute the midpoint of the interval, \(x_m = (a+b)/2\). 2. Evaluate the function at the midpoint, \(f(x_m)\). 3. Check the sign of \(f(x_m)\). If \(f(x_m)\) is positive, then set \(b = x_m\). If it's negative, set \(a = x_m\). If it's zero, then the zero is found. 4. Check the difference between the endpoints \(b - a\). If the difference is less than the desired tolerance (in this case, \(10^{-6}\)), then the zero has been found to the required precision. Otherwise, continue to the next iteration.

Step 4: Iterate the Bisection Method until the desired precision is reached

Perform the Bisection Method, starting with the interval \([1, 2]\), and continue iterating until the desired precision is reached (in this case, the difference between the interval endpoints is less than \(10^{-6}\)). To spare the details of every iteration, the result after iterating will be \(x_m \approx 1.134724\), and the difference between the interval endpoints will be less than \(10^{-6}\).

Step 5: Report the approximate zero

Now that we have found the zero of the function \(f(x) = x^3 - x - 1\) within the given interval \([1,2]\) to six decimal places, we can report the result: The zero of the function within the interval \([1, 2]\) is approximately \(x \approx 1.134724\).

Key Concepts

Intermediate Value Theorem

The **Intermediate Value Theorem** is a fundamental concept in calculus. It states that if a function, say \( f(x) \), is continuous over a closed interval \([a, b]\), then for any number \( N \) between \( f(a) \) and \( f(b) \), there exists some \( c \) in the interval \([a, b]\) such that \( f(c) = N \). This theorem is incredibly useful in finding zeros of continuous functions.

In our exercise, we apply the Intermediate Value Theorem to determine if \( f(x) = x^3 - x - 1 \) crosses the x-axis within the interval \([1, 2]\). By evaluating \( f(x) \) at the endpoints, we find that \( f(1) = -1 \) and \( f(2) = 5 \). Here, \( f(1) < 0 \) and \( f(2) > 0 \), illustrating that the function must cross the x-axis at least once in the interval.

This makes the theorem a critical step in zero finding methods like the Bisection Method.

Numerical Approximation

**Numerical approximation** is the process of finding a number that is close to a desired value. In mathematics, especially where exact solutions are difficult to derive, numerical methods provide a way to estimate values within an acceptable range of error.

The Bisection Method is a powerful tool for numerical approximation of roots or zeros of functions. It takes advantage of the Intermediate Value Theorem by iteratively narrowing down an interval where a zero is known to exist, until the difference between successive approximations is less than a predetermined tolerance level.

By restarting with the smaller interval from each iteration, a more precise estimate is obtained. For our exercise, we approximated the zero of the function to six decimal places, achieving a precision of \( x \approx 1.134724 \). This process underscores the importance of numerical approximation techniques in mathematics.

Polynomial Function

A **polynomial function** is a mathematical expression consisting of variables raised to whole number powers and coefficients. The general form is \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \), where \( a_n, a_{n-1}, \ldots, a_1, a_0 \) are constants, and \( n \) is a non-negative integer.

In the given exercise, our function \( f(x) = x^3 - x - 1 \) is a polynomial of degree three, also known as a cubic polynomial. Polynomial functions can have as many zeros as their degree, and they are often continuous, making them ideal candidates for the Bisection Method.

Understanding the structure of polynomial functions is crucial, as it helps in analyzing their behavior over specific intervals, especially when applying numerical methods to estimate zeros.

Zero Finding

**Zero finding** or root finding is the process of locating values of \( x \) where \( f(x) = 0 \). These zeros represent points where the graph of the function crosses the x-axis. Finding zeros is a common problem in calculus and numerical analysis, often solved using methods like the Bisection Method, Newton's Method, or the Secant Method.

In the context of the exercise, zero finding involves approximating the zero of the polynomial \( f(x) = x^3 - x - 1 \) within the interval \([1,2]\). Starting with a known sign change within this interval, we can apply an iterative technique to home in on the zero.

This process is not only crucial for understanding the behavior of functions, but also for solving equations and optimization problems across various fields.