Problem 59 - Calculus Early Transcendentals

Question

a. Use the Intermediate Value Theorem to show that the following equations have a solution on the given interval. b. Use a graphing utility to find all the solutions to the equation on the given interval. c. Illustrate your answers with an appropriate graph. $$2 x^{3}+x-2=0 ;(-1,1)$$

Step-by-Step Solution

Answer

Answer: The Intermediate Value Theorem shows that there must be at least one solution within the interval $$(-1,1)$$ for the equation $$2x^3 + x - 2 = 0$$, since there is a sign change in the interval. The approximate solution within this interval is $$x \approx 0.66$$.

Step 1: Evaluate the function at the interval endpoints

Calculate \(f(-1)\) and \(f(1)\) to check for a sign change in the interval $$(-1,1)$$: $$f(-1) = 2(-1)^3 + (-1) - 2 = -5$$ $$f(1) = 2(1)^3 + (1) - 2 = 1$$

Step 2: Use the Intermediate Value Theorem

Since \(f(-1) < 0\) and \(f(1) > 0\), we can conclude that there must be a value \(c\) in the interval \((-1, 1)\) such that \(f(c) = 0\). This is because the function is continuous, and there is a sign change in the interval. Thus, the equation $$2x^3 + x - 2 = 0$$ has at least one solution in the given interval.

Step 3: Use a graphing utility

To find the exact solution of the equation, we can use a graphing utility. Once the graph is plotted, we can observe the root (where the graph intersects the x-axis) within the interval $$(-1,1)$$. Based on the graph, there is one root in the interval, and this root is approximately \(x \approx 0.66\).

Step 4: Illustrate answers with a graph

We can illustrate our answers by drawing a graph of the function $$f(x) = 2x^3 + x - 2$$, showing the interval $$(-1,1)$$, and marking the root that we have found using the graphing utility. This way, we can visualize how the Intermediate Value Theorem was used.