Problem 4 - Calculus for Biology and Medicine

Question

\mathrm{\\{} I n ~ P r o b l e m s ~ , each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extreme- value theorem. With the help of a graphing calculator or spreadsheet, graph each function and locate its global extrema. (Note that a function may have more than one global minimum or maximum point.) $$ f(x)=\sin \frac{x}{2}, 0 \leq x \leq 2 \pi $$

Step-by-Step Solution

Answer

The global maximum is 1 at \( x = \pi \), and the global minimum is 0 at both \( x = 0 \) and \( x = 2\pi \).

Step 1: Identify the Function and Interval

The function given is \( f(x) = \sin \left( \frac{x}{2} \right) \), and it is defined over the interval \( 0 \leq x \leq 2\pi \). We need to find the global extrema within this interval.

Step 2: Analyze Critical Points

Find the derivative of the function: \( f'(x) = \frac{1}{2} \cos \left( \frac{x}{2} \right) \). Set \( f'(x) = 0 \) to find the critical points: \( \cos \left( \frac{x}{2} \right) = 0 \), which gives \( \frac{x}{2} = \frac{\pi}{2} + n\pi \) for integers \( n \). Solving this gives potential critical points \( x = \pi, 3\pi \). Since \( x = 3\pi \) is outside the interval, we only consider \( x = \pi \).

Step 3: Evaluate at Critical and Boundary Points

Evaluate the function at the endpoints of the interval and the critical point. At \( x = 0 \), \( f(0) = \sin(0) = 0 \). At \( x = \pi \), \( f(\pi) = \sin(\frac{\pi}{2}) = 1 \). At \( x = 2\pi \), \( f(2\pi) = \sin(\pi) = 0 \).

Step 4: Determine Global Extrema

Compare function values at these points: \( f(0) = 0 \), \( f(\pi) = 1 \), and \( f(2\pi) = 0 \). The largest value is 1 at \( x = \pi \), and the smallest value is 0 at both \( x = 0 \) and \( x = 2\pi \).

Step 5: Conclude with Extrema Locations

The global maximum of the function \( f(x) = \sin \left( \frac{x}{2} \right) \) on the interval \( 0 \leq x \leq 2\pi \) is 1 at \( x = \pi \), and the global minimum is 0 at \( x = 0 \) and \( x = 2\pi \).

Key Concepts

Global Extrema

When we talk about global extrema, we are referring to the highest or lowest points of a function within a specific interval. These points are known as the global maximum and global minimum. Identifying these points is important as they represent the peak and trough values that a function can reach over its given range.

In our original problem, the function is given as \( f(x) = \sin \left( \frac{x}{2} \right) \) and is defined on the interval \( 0 \leq x \leq 2\pi \). By finding the values of \( f(x) \) at the critical points and the endpoints of the interval, we determine the global extrema.

Our analysis found that the global maximum of this function occurs at \( x = \pi \) with a value of 1, and the global minimum occurs at both endpoints \( x = 0 \) and \( x = 2\pi \), where the value is 0.

Critical Points

Critical points are where the derivative of a function is zero or undefined. They are vital in determining where the extrema of a function might be.

For our function \( f(x) = \sin \left( \frac{x}{2} \right) \), the derivative is calculated as \( f'(x) = \frac{1}{2} \cos \left( \frac{x}{2} \right) \). Setting this equal to zero helps us find the critical points. In our example, \( \cos \left( \frac{x}{2} \right) = 0 \) leads us to potential critical points at \( x = \pi \) and \( x = 3\pi \).

However, since our interval is \( 0 \leq x \leq 2\pi \), we only consider \( x = \pi \) as a critical point because \( x = 3\pi \) falls outside our interval. Finding these critical points helps pinpoint where potential maxima and minima may occur, aiding in identifying the global extrema.

Continuous Function

A continuous function, as the name suggests, is one where, intuitively speaking, you can draw its graph without lifting your pencil off the paper. Mathematically, a function is continuous at a point if the limit of the function as it approaches the point is equal to the function's value at that point.

For our problem, the function \( f(x) = \sin \left( \frac{x}{2} \right) \) is continuous over the interval \( 0 \leq x \leq 2\pi \). This continuity is crucial because it allows us to apply the extreme-value theorem, which states that a continuous function on a closed interval will always have a global maximum and minimum.

The assurance of the function's continuity is why we can confidently locate its extrema by simply checking the critical points and endpoints, knowing there are no sudden jumps or holes in its graph.