Problem 69 - Thomas Calculus

Question

Use the Intermediate Value Theorem in Exercises \(69-76\) to prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations. $$ x^{3}-3 x-1=0 $$

Step-by-Step Solution

Answer

There is a root on the interval \([-2, 2]\) where the function changes sign.

Step 1: Identify Function and Interval

Consider the function \( f(x) = x^3 - 3x - 1 \). To use the Intermediate Value Theorem (IVT), first select an interval \([a, b]\) where \(f(a)\) and \(f(b)\) have opposite signs. Let's choose \([-2, 2]\).

Step 2: Evaluate Function at Interval Bounds

Calculate \( f(-2) \) and \( f(2) \):- \( f(-2) = (-2)^3 - 3(-2) - 1 = -8 + 6 - 1 = -3 \).- \( f(2) = (2)^3 - 3(2) - 1 = 8 - 6 - 1 = 1 \).

Step 3: Apply Intermediate Value Theorem

The Intermediate Value Theorem states if a continuous function changes signs over an interval, there is at least one root in that interval. Since \( f(-2) = -3 \) and \( f(2) = 1 \) change signs from negative to positive, there must be a root in \([-2, 2]\).

Step 4: Graph the Function

Use a graphing calculator to graph \( f(x) = x^3 - 3x - 1 \). The graph should show that the function crosses the x-axis between \(-2\) and \(2\). The precise point of intersection will give the approximate solution.

Key Concepts

Continuous Function

A continuous function is one where there are no breaks, gaps, or jumps when you plot it on a graph. This means you can draw the function without lifting your pencil from the paper. Think of it like a smooth curve that keeps going without interruption.
Continuous functions are important in calculus because they allow us to apply the Intermediate Value Theorem (IVT). The IVT relies on a function being continuous over a given interval to ensure that, if the function takes on different signs at either end of the interval, it must cross zero somewhere in between.
For the function in our exercise, \( f(x) = x^3 - 3x - 1 \), it is continuous because it is a polynomial, and polynomials are always continuous on their domains. This property is key for finding a root in a specified interval.

Root of a Function

The root of a function refers to the value of \(x\) where the function equals zero, or in simpler terms, where the graph of the function crosses the x-axis. In the context of equations, finding the root often means solving the equation to find what value of \(x\) makes the expression equal to zero.
For example, in the equation \(x^3 - 3x - 1 = 0\), we are looking for the point(s) on the graph where this function crosses the x-axis. According to the Intermediate Value Theorem, if a continuous function changes from a negative to a positive value over an interval, it confirms that there is at least one root (or zero) in that interval.
Thus, after evaluating \(f(-2) = -3\) and \(f(2) = 1\), we can be sure there's a root somewhere between \(-2\) and \(2\).

Graphing Calculator

A graphing calculator is a powerful tool that allows students and mathematicians to visually interpret and analyze the behavior of functions by plotting their graphs.

• Graphical representation helps in better understanding the function's properties.
• It quickly shows where the function crosses the x-axis, helping to identify roots.
• They support checking the solutions predicted by theoretical methods such as the Intermediate Value Theorem by visual confirmation.

For the function \( f(x) = x^3 - 3x - 1 \), using a graphing calculator would show that the function crosses the x-axis in the interval from \(-2\) to \(2\). This visual cross-check is an effective way to verify the existence and estimate the position of the root found theoretically through the IVT.