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. $$\sqrt{x^{4}+25 x^{3}+10}=5 ;(0,1)$$

Short answer

Short Answer: By applying the Intermediate Value Theorem (IVT), we showed that the equation has a solution on the given interval (0,1), since f(0) < 0 and f(1) > 0, and our function is continuous. Using a graphing utility, we found that the solution is x ≈ 0.484. A graph illustrating the function on the interval (0,1) confirms this result, showing the function crossing the x-axis at only one point, x ≈ 0.484.

Step-by-step solution

Part a: Apply Intermediate Value Theorem

First, we need to find the continuous function which we will use the IVT on. We can rewrite the given equation as $$f(x) = x^{4}+25 x^{3}+10-25$$ Since every term in the equation is a polynomial, f(x) is continuous for all x. Now we need to show that there is a solution to the equation in the interval (0,1) by applying the IVT. First, let's find f(0) and f(1): $$f(0) = 0^{4}+25(0)^{3}+10-25 = -15$$ $$f(1) = 1^{4}+25(1)^{3}+10-25 = 11$$ Since f(0) < 0 and f(1) > 0, and our function is continuous on the interval (0, 1), we can apply the Intermediate Value Theorem. The theorem states that there is a number c in the interval (0,1) such that f(c) = 0. Therefore, the equation has a solution on the given interval.

Part b: Use a Graphing Utility to Find Solutions

Using a graphing utility (such as Desmos or WolframAlpha) to graph the function f(x) = \(0\). We can observe that there is one point where the function crosses the x-axis, meaning there's one solution for x in the interval (0,1). Using a more precise algorithm or the utility's root-finding functionality, we can estimate that this solution is x ≈ 0.484.

Part c: Illustrate with a Graph

To illustrate our answers with a graph, plot the equation f(x) = \(x^{4}+25 x^{3}+10-25\) starting from the interval (0,1). Our graph should show the function's behavior in this interval, including the point where f(x) = 0. You should see that the function is continuous on the interval (0,1), and that it crosses the x-axis at only one point, x ≈ 0.484, which is the solution we found using the graphing utility.

Key concepts

Continuous Functions

Continuous functions are the backbone of many concepts in calculus, including the Intermediate Value Theorem. Simply put, a function is continuous over an interval if you can draw the function's graph over the interval without lifting your pencil. Such functions have no breaks, jumps, or holes within that interval.
The function given in the exercise is \(f(x) = x^4 + 25x^3 + 10 - 25\), which is a polynomial function. Polynomial functions are continuous everywhere on the real number line. Therefore, the function \(f(x)\) is continuous over the interval (0, 1).
Understanding that \(f(x)\) is continuous allows us to apply the Intermediate Value Theorem to find solutions in our interval of interest.

Polynomial Functions

Polynomial functions are a specific type of mathematical expression that consist of variables raised to whole number powers, multiplied by coefficients, and summed together. They are among the smoothest types of functions because they are continuous and differentiable across the entire real number line.
In the exercise, you encounter \(f(x) = x^4 + 25x^3 + 10 - 25\). This is a quartic polynomial, as it includes a term raised to the fourth power. The characteristics of polynomial functions make them easy to analyze in calculus. Their continuous nature means that they do not introduce unexpected behavior such as asymptotes or discontinuities within the domain.
Analyzing polynomial functions with the Intermediate Value Theorem becomes straightforward because their continuity is guaranteed, enabling us to find roots confidently within the specified intervals.

Root-Finding

Root-finding involves determining the points where a function equals zero—these are known as the roots of the function. In the context of the Intermediate Value Theorem, root-finding leverages the theorem's assurance that if a continuous function changes sign over an interval, it must have a root somewhere within that interval.
For our function \(f(x) = x^4 + 25x^3 + 10 - 25\), we see that \(f(0) = -15\) and \(f(1) = 11\). A sign change from negative to positive indicates that a root exists in between. By applying a graphing utility, we find that the root in the interval (0, 1) is approximately \(x \approx 0.484\).
Root-finding is central in solving equations beyond simple manual calculations, especially for more complex polynomials where algebraic solutions aren't readily available.

Graphical Solution

A graphical solution provides a visual confirmation of the solutions we've derived analytically. By graphing the function \(f(x) = x^4 + 25x^3 + 10 - 25\), we gain insights into the function's behavior and verify the points where it crosses the x-axis.
Using a graphing tool, such as Desmos, allows you to plot the function over the interval (0, 1). You will see the curve smoothly crossing the x-axis, confirming the function's continuity and reinforcing the findings of the Intermediate Value Theorem that the function must have a root. The precise point where the function touches the x-axis—our sought-after root—can be identified as approximately \(x \approx 0.484\).
Graphing is not only instrumental in confirming analytical findings but also in building an intuitive understanding of function behavior, which is essential for mastering the subject.