Problem 63 - Single Variable Calculus

Question

In Exercises 63-66, use Theorem 8 to show that there is at least one root of the equation in the given interval. \(x^{3}-2 x-1=0 ; \quad(0,2)\)

Step-by-Step Solution

Answer

Using the Intermediate Value Theorem (Theorem 8), we evaluate the function \(f(x) = x^3 - 2x - 1\) at the endpoints of the interval \((0, 2)\). We find that \(f(0) = -1\) and \(f(2) = 5\), which have opposite signs. Therefore, there is at least one root of the equation \(x^3 - 2x - 1 = 0\) in the interval \((0, 2)\).

Step 1: Evaluate at Endpoint: 0

Plug in the left endpoint, 0, into the equation: \(f(0) = (0)^3 - 2(0) - 1 = -1\)

Step 2: Evaluate at Endpoint: 2

Plug in the right endpoint, 2, into the equation: \(f(2) = (2)^3 - 2(2) - 1 = 5\)

Step 3: Apply Intermediate Value Theorem

Since the values at the endpoints are of opposite signs (f(0) = -1 and f(2) = 5), we can apply the Intermediate Value Theorem. It states that there must be at least one root of the equation within the interval (0, 2). This approach shows that there is at least one root of the equation \(x^3 - 2x - 1 = 0\) in the interval \((0, 2)\) using Theorem 8 (Intermediate Value Theorem).

Key Concepts

Root of the Equation

Understanding the concept of the 'root of an equation' is fundamental in algebra and calculus. It refers to the value of the variable that makes the equation true, or in other words, results in a zero value when substituted into the equation. In the context of the exercise, finding a root for the polynomial equation x^3 - 2x - 1 = 0 means we are looking for a value of x that when plugged into the equation will give the result of zero.

When we evaluate the given equation at the endpoints of the interval (0, 2), we get f(0) = -1 and f(2) = 5. These evaluations are crucial because they indicate a change in the sign of the value of f(x) across the interval. This suggests that a root lies between these two endpoints, an insight that is formalized by the Intermediate Value Theorem.

Continuity in Calculus

Continuity in calculus is a quality that ensures a function has no breaks, jumps, or holes in its graph within a given interval. A function f(x) is continuous at a point x = c if f(c) exists, the limit as x approaches c exists, and the limit equals f(c). Continuity over an interval means the function is continuous at every point in that interval.

For the Intermediate Value Theorem (Theorem 8) to hold, the function must be continuous on the closed interval [a, b]. The theorem implies that if you have a function that is continuous on [a, b] and takes on different signs at a and b, there must be at least one root in the interval (a, b). In our exercise, because polynomial functions are inherently continuous everywhere, this condition is satisfied.

Polynomial Equations

Polynomial equations are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. The simplest example is a linear equation ax + b = 0, and they extend to higher degrees like the cubic equation in our exercise, x^3 - 2x - 1 = 0. Solving polynomial equations is about finding the values of x that satisfy the equation. Importantly, all polynomial equations are continuous and differentiable, and their roots can be real or complex numbers.

Polynomial equations play a pivotal role in numerous areas of mathematics and applied sciences. They can model a wide range of phenomena, and finding their roots is essential for understanding the behavior of the system they represent. The cubic equation in our example has at least one real root in the interval (0,2), as shown by the application of the Intermediate Value Theorem.