Question

Use the Intermediate Value Theorem to show that \(\sqrt{x}-\cos x=0\) has a solution between 0 and \(\pi / 2 .\) Zoom in on the graph of \(y=\sqrt{x}-\cos x\) to find an interval having length \(0.1\) that contains this solution.

Short answer

A solution exists between 0 and \( \pi/2 \), with a specific interval of \( (0.6, 0.7) \).

Step-by-step solution

Define the Function

Let's consider the function \( f(x) = \sqrt{x} - \cos x \). Our task is to determine if this function equals zero within the interval \([0, \pi/2]\) using the Intermediate Value Theorem.

Evaluate Endpoints

Calculate the values of the function at the endpoints of the interval: \( f(0) = \sqrt{0} - \cos(0) = 0 - 1 = -1 \) and \( f(\pi/2) = \sqrt{\pi/2} - \cos(\pi/2) = \sqrt{\pi/2} - 0 = \sqrt{\pi/2} \). Since \( f(0) < 0 \) and \( f(\pi/2) > 0 \), the function changes sign between 0 and \( \pi/2 \).

Apply Intermediate Value Theorem

The Intermediate Value Theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\) and \( f(a) \) and \( f(b) \) have opposite signs, then there exists at least one \( c \) in \( (a, b) \) such that \( f(c) = 0 \). Since \( f(x) \) is continuous (being the combination of a continuous square root and cosine function), there exists a \( c \) in \((0, \pi/2)\) such that \( f(c) = 0 \).

Zoom into the Interval for Greater Precision

To zoom into an interval of length 0.1, test smaller intervals within \([0, \pi/2]\) to narrow down where the root exists. For instance, try evaluating the function at \( x = 0.5 \) and \( x = 0.1 \): \( f(0.1) = \sqrt{0.1} - \cos(0.1) \approx 0.3162 - 0.9950 = -0.6788 \) and \( f(0.5) = \sqrt{0.5} - \cos(0.5) \approx 0.7071 - 0.8776 = -0.1705 \). Notice that as \( x \) increases from \(0.1\) to \(0.5\), \( f(x) \) gets closer to zero.

Narrow Down the Interval

To find an interval with length 0.1 containing the root, continue evaluating \( f(x) \) in smaller subintervals. For example, check \( f(x) \) at \( x = 0.6 \): \( f(0.6) = \sqrt{0.6} - \cos(0.6) \approx 0.7746 - 0.8253 = -0.0507 \), and \( f(0.7) = \sqrt{0.7} - \cos(0.7) \approx 0.8367 - 0.7645 = 0.0722 \). Since \( f(0.6) < 0 \) and \( f(0.7) > 0 \), the interval \((0.6, 0.7)\) contains a zero of \( f(x) \) and is of length 0.1.

Key concepts

Continuous Functions

The concept of continuous functions is crucial in understanding the Intermediate Value Theorem. A function is said to be continuous if, intuitively, you can draw its graph without lifting your pencil from the paper. More formally, a function \( f(x) \) is continuous at a point \( c \) if the following three conditions are met:

• \( f(c) \) is defined.
• The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
• The limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).

In the case of the function \( f(x) = \sqrt{x} - \cos x \), it is continuous over the interval \([0, \pi/2]\) because it is composed of two continuous functions: the square root function (which is continuous for non-negative values) and the cosine function (which is continuous everywhere). Thus, by combining these, the resulting function is continuous on the closed interval \([0, \pi/2]\). This continuity is what allows us to apply the Intermediate Value Theorem to find where \( f(x) = 0 \).

Root Finding

Root finding is the process of identifying values of \( x \) for which an equation \( f(x) = 0 \) holds, known as the roots of the function. In the context of this exercise, root finding is accomplished through understanding that the function \( f(x) = \sqrt{x} - \cos x \) crosses the x-axis (i.e., equals zero) between certain points.

By evaluating the function at the endpoints of the interval \([0, \pi/2]\), we find that \( f(0) = -1 \) and \( f(\pi/2) \) is positive. Since the sign changes from negative to positive over this interval, there must be some point \( c \) within \([0, \pi/2]\) where \( f(c) = 0 \). This is a direct application of the Intermediate Value Theorem, which assists significantly in root finding when dealing with continuous functions. By further narrowing down subintervals with sign changes, at \( (0.6, 0.7) \), we discover a more precise location of the root.

Mathematical Proofs

Mathematical proofs are the logical sequences of statements to show the truth of a particular theorem or statement. The use of the Intermediate Value Theorem in this exercise provides not just a method, but also a proof that a solution exists in a given interval.

To prove that there is a root within \([0, \pi/2]\) using the Intermediate Value Theorem, the function must meet specific criteria: it must be continuous on the interval, and the function values at the endpoints must have opposite signs. By demonstrating both of these conditions for \( f(x) = \sqrt{x} - \cos x \), we are providing a mathematical proof of the existence of at least one root. These proofs are vital in making precise mathematical conclusions and developing confidence in the existence of solutions to complex problems.

Calculus

Calculus plays an integral role in understanding and applying the Intermediate Value Theorem, especially when working with continuous functions and in root-finding scenarios. Calculus provides the tools to analyze the behavior of functions across intervals, enabling us to engage in these types of mathematical verifications.

In this particular problem, calculus helps us by providing definitions and properties of functions like \( \sqrt{x} \) and \( \cos x \), ensuring they're continuous and thus suitable for application of the theorem. Calculus also guides us in evaluating limits and understanding the behavior of functions as they approach specific points from either side of the interval. As students progress in their studies of calculus, they can apply these principles to more complex functions and diverse scenarios, enhancing their mathematical capability and problem-solving skills.