Problem 69 - 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. $$x^{3}-5 x^{2}+2 x=-1 ;(-1,5)$$

Step-by-Step Solution

Answer

Based on the given equation $$x^3 - 5x^2 + 2x = -1$$ and the interval $$(-1, 5)$$, we have successfully used the Intermediate Value Theorem (IVT) to show that the equation has a solution in the given interval. Additionally, we have used a graphing utility to approximate the solution as $$x \approx 3.49$$. The graph of the function $$f(x) = x^3 - 5x^2 + 2x + 1$$ illustrates that it intersects the x-axis at the point corresponding to the solution we found.

Step 1: Check the continuity of the function

We have the function $$f(x) = x^3 - 5x^2 + 2x + 1$$. This is a polynomial function, and polynomial functions are continuous everywhere. Thus, this function is continuous on the given interval $$(-1, 5)$$.

Step 2: Apply the Intermediate Value Theorem

Since the function $$f(x)$$ is continuous on $$(-1, 5)$$, we can apply the Intermediate Value Theorem (IVT). We first have to compute $$f(-1)$$ and $$f(5)$$: $$f(-1) = (-1)^3 - 5(-1)^2 + 2(-1) + 1 = -1 - 5 - 2 + 1 = -7$$ $$f(5) = (5)^3 - 5(5)^2 + 2(5) + 1 = 125 - 125 + 10 + 1 = 11$$ Since $$f(x)$$ is continuous on $$(-1, 5)$$ and $$f(-1) = -7 < 0 < 11 = f(5)$$, by the Intermediate Value Theorem, there must exist a solution $$c$$ in the interval $$(-1, 5)$$ such that $$f(c) = 0$$. So, the given equation has a solution on this interval.

Step 3: Use a graphing utility to find the solutions

We can use an online graphing calculator, such as Desmos, to graph the function $$f(x) = x^3 - 5x^2 + 2x + 1$$ on the interval $$(-1, 5)$$. From the graph, we can see that the function intersects the x-axis at one point in the given interval. This means there is only one solution to the given equation on this interval, which is approximately $$x \approx 3.49$$.

Step 4: Illustrate the answer with a graph

To illustrate our answer, we can show the graph of the function $$f(x) = x^3 - 5x^2 + 2x + 1$$ with the x-axis intersection point marked. [Here you would insert a graph of the function with the x-axis intersection point highlighted on the given interval $$(-1, 5)$$] We have now shown that the given equation has a solution in the interval $$(-1, 5)$$ using the Intermediate Value Theorem and found the approximate solution of $$x \approx 3.49$$ using a graphing utility.

Key Concepts

Continuity of Polynomial Functions

Understanding the concept of continuity, especially for polynomial functions, is essential in calculus. Continuity simply means that you can draw the graph of a function without lifting your pencil. Polynomial functions, which include terms like x^3, x^2, x, and constants, are known to be continuous at every point in their domain — that is, for all real numbers.

Now, why is this important? In our exercise, we have a polynomial function f(x) = x^3 - 5x^2 + 2x + 1. Due to the continuity of polynomials, this particular function doesn't have any breaks, jumps, or holes for any value of x. This means we can confidently use tools like the Intermediate Value Theorem to analyze its behavior over an interval, such as (-1, 5).

In practice, the continuity ensures that if you're searching for where a polynomial crosses the x-axis (its roots), as we are in our exercise, you know that if it starts above the axis on one end of an interval and ends below on the other, it must have crossed the axis somewhere in between. That's the power of understanding the continuity of polynomial functions!

Graphing Polynomials

Having a visual representation of a polynomial function can greatly enhance our understanding of its behavior. Graphing polynomials provides us with a visible shape of the equation, revealing important features such as intercepts, turns, and the overall trend as x approaches positive or negative infinity.

When graphing the polynomial function from our exercise, f(x) = x^3 - 5x^2 + 2x + 1, we would expect to see a smooth, continuous curve without any gaps. The graph would allow us to pinpoint exactly where the function crosses the x-axis, which signifies the roots or solutions of the equation f(x)=0.

With the help of graphing utilities like Desmos or graphing calculators, students can easily input their polynomial function and see where it intersects with the x-axis. This visual approach is not only helpful for finding the approximate value of the roots but also for confirming the existence of roots within a given interval — just as the Intermediate Value Theorem suggests.

Finding Roots of Equations

Finding the roots of an equation involves determining where the curve of the polynomial function intersects the x-axis, which are the points where the function reaches a value of zero. Roots are essential for understanding the behavior of polynomial functions, as they represent the points at which the output of the function changes sign.

In our exercise, we use the Intermediate Value Theorem (IVT) to guarantee the presence of at least one root within the interval (-1, 5) for the function f(x) = x^3 - 5x^2 + 2x + 1. The IVT states that for any continuous function that produces two values of opposite sign in an interval, there must be at least one root between them.

By calculating f(-1) and f(5), we have shown that the function moves from a negative value to a positive value, confirming the existence of a root. Finally, a graphing utility was used to find the approximate location of this root. Tools such as the bisection method, Newton's method, or simply utilizing a graphing calculator are typically employed to find root values more precisely, which is a critical skill in calculus and has numerous applications in science, engineering, and beyond.