Question

Proof (a) Let \(f_{1}(x)\) and \(f_{2}(x)\) be continuous on the closed interval \([a, b] .\) If \(f_{1}(a)f_{2}(b),\) prove that there exists \(c\) between \(a\) and \(b\) such that \(f_{1}(c)=f_{2}(c)\) (b) Show that there exists \(c\) in \(\left[0, \frac{\pi}{2}\right]\) such that \(\cos x=x\) . Use a graphing utility to approximate \(c\) to three decimal places.

Short answer

For part (a), by applying the Intermediate Value Theorem to the function \(g(x) = f_{1}(x) - f_{2}(x)\), we proved that there exists some \(c\) in the interval \([a, b]\) where \(f_{1}(c) = f_{2}(c)\). As for part (b), using the same theorem, we determined that there exists a \(c\) in the interval \([0, \frac{\pi}{2}]\) such that \(cos x = x\). We can approximate \(c\) using a graphing utility to three decimal places.

Step-by-step solution

Understand the Problem

We have two tasks to accomplish, which can be solved using the Intermediate Value Theorem (IVT). In the first task we are given two functions \(f_{1}\) and \(f_{2}\) that are continuous on the interval \([a, b]\). We know that \(f_{1}(a)f_{2}(b)\). Using the IVT, we need to show there exists some \(c\) in \([a, b]\) where \(f_{1}(c)=f_{2}(c)\). In the second task, we are looking for a value \(c\) in \([0, \frac{\pi}{2}]\) where \(cos x = x\). Again, we will use the IVT to solve this.

Apply the Intermediate Value Theorem - part(a)

Let \(g(x) = f_{1}(x) - f_{2}(x)\). Since \(f_{1}\) and \(f_{2}\) are continuous in \([a, b]\), \(g\) must also be continuous in \([a, b]\). By the conditions we know that \(g(a) = f_{1}(a) - f_{2}(a) < 0\) and \(g(b) = f_{1}(b) - f_{2}(b) > 0\). Thus, by the Intermediate Value Theorem, there must be some real number \(c\) in the interval \((a, b)\) such that \(g(c) = 0\), which means \(f_{1}(c) = f_{2}(c)\).

Apply the Intermediate Value Theorem - part(b)

Now we find the \(c\) in the interval \([0, \frac{\pi}{2}]\) such that \(cos(x) = x\). Let \(g(x) = cos(x) - x\). By nature, \(g\) is continuous. We also know that \(g(0) = cos(0) - 0 = 1\) and \(g(\frac{\pi}{2}) = cos(\frac{\pi}{2}) - \frac{\pi}{2} = -\frac{\pi}{2}\). So \(g(0) > 0\) and \(g(\frac{\pi}{2}) < 0\). According to the Intermediate Value Theorem, there must exist a \(c\) in the interval so that \(g(c) = 0\). That is, \(cos(c) = c\) in \([0, \frac{\pi}{2}]\). The approximate value can be computed with a graphing utility to three decimal places.

Key concepts

Understanding Continuous Functions

Continuous functions play a crucial role in calculus and real-world applications—they smoothly transition without any gaps or jumps. This continuity means that, in a continuous function, small changes to the input result in small changes to the output, much like smoothly varying terrain follows the undulations of a hillside, rather than the abruptness of stairs.

For instance, if you consider the temperature over the course of a day, the reading transitions gradually over time, never suddenly leaping from one value to another—it graphically represents a continuous function over that period. This is important because the Intermediate Value Theorem (IVT) specifically applies to functions that are continuous on a closed interval \[a, b\]. The functions \(f_1\) and \(f_2\) from the given problem highlight this by being continuous between the points \(a\) and \(b\).

When we describe a function as continuous on an interval, it's like saying you can draw its graph without lifting your pencil from the start to the end points of that interval, ensuring a smooth, unbroken line. This inherent characteristic is essential for the IVT to assure us that the function hits all values between its endpoints on the graph, which sets the stage for proving the existence of solutions within intervals.

Existence of Solutions in Intervals

The Intermediate Value Theorem guarantees that if you've got a continuous function spanning an interval from \(a\) to \(b\), and the function's value at \(a\) is less than some number, while at \(b\) it's greater than that number, the function must cross that particular value at some point between \(a\) and \(b\). In layman's terms, if you're tunneling from one side of a mountain to the other and you start below sea level and end up above it, you must have passed through sea level at some point during your journey.

In the task at hand, you're playing mathematical detective, ascertaining that there must exist a \(c\) within \(a, b\) such that \(f_1(c) = f_2(c)\). Considering the conditions, \(f_1(a) < f_2(a)\) and \(f_1(b) > f_2(b)\), this theorem becomes an invaluable tool. By constructing a new function, \(g(x) = f_1(x) - f_2(x)\), it leverages the continuity of \(f_1\) and \(f_2\) to facilitate the use of IVT. This approach illustrates the theorem's power to pinpoint the existence of a specific value within an interval, where previously it was only likely.

Graphical Approximation of Solutions

Some problems in math can be like guessing the number of jellybeans in a jar—it helps to have a visual. For complex equations, graphing is a similar strategy; it's a powerful means to estimate solutions where algebra alone might not suffice. This is vividly demonstrated in the second part of the exercise by graphing \(g(x) = cos(x) - x\) over the interval \[0, \frac{\pi}{2}\].

Using a graphing utility offers a graphical approximation to identify where the curve intersects the x-axis, a point we denote as \(c\), where the equation \(cos(x) = x\) holds true. This intersection is evident because graphical representations allow us to observe function behaviors, such as the intersection points, that equations define abstractly.

A keen focus on the graphical method not only serves as an approximate solution but also encourages a deeper understanding of the function's behavior within the domain. A visual check sometimes reveals patterns or anomalies that purely numerical methods may overlook, making it an essential step in the mathematical toolkit for understanding and approximating solutions.