Problem 3 - Calculus for Biology and Medicine

Question

Each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extremevalue theorem. With the help of a graphing calculator, graph each function and locate its global extrema. (Note that a function may assume a global extremum at more than one point.) \(f(x)=\sin (2 x), 0 \leq x \leq \pi\)

Step-by-Step Solution

Answer

Global maximum is 1 at \(x=\frac{\pi}{4}\); global minimum is -1 at \(x=\frac{3\pi}{4}\).

Step 1: Understand the Function

The given function is \(f(x) = \sin(2x)\). It is a trigonometric function with argument \(2x\), which affects the frequency of the sine wave, causing it to complete its period from 0 to \(\pi\). This means the function will complete its full sine wave within this interval.

Step 2: Determine Critical Points

To find critical points, we first differentiate the function. The derivative of \(f(x) = \sin(2x)\) is \(f'(x) = 2\cos(2x)\). Setting the derivative to zero gives the critical points: \(2\cos(2x) = 0\). Solving for \(x\), we find \(x = \frac{\pi}{4}, \frac{3\pi}{4}\), which are within the interval.

Step 3: Evaluate Function at Critical Points and Endpoints

Evaluate \(f(x)\) at the critical points and the endpoints of the interval.\- At \(x = 0: f(0) = \sin(0) = 0\).\- At \(x = \frac{\pi}{4}: f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1\).\- At \(x = \frac{3\pi}{4}: f\left(\frac{3\pi}{4}\right) = \sin\left(\frac{3\pi}{2}\right) = -1\).\- At \(x = \pi: f(\pi) = \sin(2\pi) = 0\).

Step 4: Identify the Global Extrema

Comparing the values:\- \(f(0) = 0\)\- \(f\left(\frac{\pi}{4}\right) = 1\) (this is a maximum)\- \(f\left(\frac{3\pi}{4}\right) = -1\) (this is a minimum)\- \(f(\pi) = 0\)\Therefore, the global maximum is 1 at \(x = \frac{\pi}{4}\), and the global minimum is -1 at \(x = \frac{3\pi}{4}\).

Step 5: Verify with a Graphing Calculator

Input the function \(f(x) = \sin(2x)\) into a graphing calculator and observe its behavior over the interval \([0, \pi]\). Confirm visually that the function reaches its highest point at \(x = \frac{\pi}{4}\) with a value of 1, and its lowest point at \(x = \frac{3\pi}{4}\) with a value of -1.

Key Concepts

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are fundamental in mathematics. They relate to angles and are often used to describe periodic phenomena. The sine function, in particular, comes up in various real-world contexts. In this exercise, we look at the function \(f(x) = \sin(2x)\). The argument \(2x\) indicates that the frequency of the sine wave is increased compared to \(\sin(x)\). This means the function completes its cycle faster. Specifically, \(\sin(x)\) completes its period in \(2\pi\), while \(\sin(2x)\) completes it in \(\pi\). This behavior is crucial when analyzing periodic functions and their critical points across given intervals.
Understanding how these functions behave is essential for solving calculus problems, especially when working with phenomena that repeat over time or space.

Critical Points

Finding critical points is a key step in determining where a function might achieve its highest or lowest values, known as extrema. Critical points occur where the derivative of the function is zero or undefined. For our function \(f(x) = \sin(2x)\), we first find its derivative, \(f'(x) = 2\cos(2x)\). By setting this derivative to zero, \(2\cos(2x) = 0\), we identify potential points where the function’s behavior changes.

• The solutions \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\) within the interval \([0, \pi]\) are the critical points.

Remember that not all critical points are points of maximum or minimum values. To determine whether they are, you need to evaluate the function at these points, as well as at the interval's endpoints.

Global Extrema

Global extrema refer to the maximum and minimum values that a function attains on the entire interval. In this case, we examine \(f(x) = \sin(2x)\) over \([0, \pi]\). We evaluate the function at critical points and endpoints:

• \(f(0) = 0\)
• \(f\left(\frac{\pi}{4}\right) = 1\) — Global Maximum
• \(f\left(\frac{3\pi}{4}\right) = -1\) — Global Minimum
• \(f(\pi) = 0\)

These calculations tell us that the highest point, the global maximum, is 1, occurring at \(x = \frac{\pi}{4}\). The lowest point, the global minimum, is -1, at \(x = \frac{3\pi}{4}\). Understanding how to find these extrema helps solve problems involving optimization and ensures that predictions based on these functions are accurate.

Graphing Calculators

Graphing calculators are invaluable tools for visually analyzing mathematical functions. They offer a way to verify calculations and see the bigger picture by graphically representing the function along a certain interval. When working with \(f(x) = \sin(2x)\), a graphing calculator helps confirm results by plotting the function and highlighting where it achieves its extrema.
Input the function into the calculator, set the viewing window to the appropriate interval \([0, \pi]\), and observe the curve:

• Check that the highest point of the function is at \(x = \frac{\pi}{4}\).
• Confirm the lowest point at \(x = \frac{3\pi}{4}\).

This visual approach provides an intuitive understanding of the function’s behavior and complements analytical solutions. Utilizing technology alongside mathematical methods leverages the best of both worlds, ensuring accuracy and providing a clearer understanding.