Problem 8 - 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.) $$ f(x)=x \ln x, 1 \leq x \leq 2 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Function

We need to find the global extrema of the function \( f(x) = x \ln x \) on the closed interval \([1, 2]\). The function \( x \ln x \) is defined and continuous for \( x > 0 \), so it is defined on the given interval.

Step 2: Determine Critical Points

Find the derivative of the function to determine where it may have critical points. The derivative of \( f(x) = x \ln x \) is found using the product rule: \( f'(x) = \ln x + 1 \). To find critical points, solve the equation \( f'(x) = 0 \):\( \ln x + 1 = 0 \Rightarrow \ln x = -1 \Rightarrow x = \frac{1}{e} \). However, \( x = \frac{1}{e} \approx 0.3679 \) is not in the interval \([1, 2]\), so there are no critical points within the interval.

Step 3: Evaluate Endpoints

Since there are no critical points in the interval, the extrema will occur at the endpoints of the interval. Evaluate \( f(x) \) at the endpoints, \( x = 1 \) and \( x = 2 \):- \( f(1) = 1 \ln 1 = 0 \) (since \( \ln 1 = 0 \))- \( f(2) = 2 \ln 2 \).

Key Concepts

Continuous Function

Continuous functions are those that flow smoothly without any interruptions or jumps in their path. Imagine a curve that you can draw without lifting your pen from the paper. This characteristic is essential for many mathematical concepts and theorems, like the Extreme Value Theorem.

The function given in the exercise, defined as \( f(x) = x \ln x \), is continuous on the closed interval \([1, 2]\). A continuous function within a closed interval means the graph of the function behaves nicely, making it easier to analyze for extrema points – points where the function either hits its highest or lowest values.

Why does continuity matter? Simply put, it ensures that the function does not misbehave unexpectedly between endpoints, providing a standard and reliable behavior to work with. Next time you see a continuous function, you'll know it's predictable and smooth!

Global Extrema

Global extrema are the highest or lowest points (values) that a function can reach within a specified interval. It's like finding the tallest mountain or the deepest valley on a given terrain. This concept is crucial when analyzing the behavior of functions over specific intervals.

For the function \( f(x) = x \ln x \) over the interval \([1, 2]\), our goal is to identify these global extrema. This entails looking at both the endpoints of the interval and any potential critical points inside the interval, which are points where the derivative may be zero or undefined.

In this case, there are no critical points within the interval since \( x = \frac{1}{e} \) does not lie within \([1, 2]\). Therefore, we need to check the endpoints:

• At \( x = 1 \), \( f(1) = 0 \).
• At \( x = 2 \), we calculate \( f(2) = 2 \ln 2 \).

With this evaluation, \( f(x) = 0 \) at \( x = 1 \) is the global minimum, whereas \( f(x) = 2 \ln 2 \) at \( x = 2 \) is the global maximum point over the interval.

Derivative

The derivative of a function provides a wealth of information about its behavior. It's like a speedometer showing how fast or slow a function's value changes at any given point. In calculus, it allows us to pinpoint moments when a function might change from increasing to decreasing or vice versa.

For our function \( f(x) = x \ln x \), the product rule is used to find its derivative. The resulting derivative, \( f'(x) = \ln x + 1 \), unveils potential critical points when set equal to zero, leading to \( \ln x = -1 \) or \( x = \frac{1}{e} \). Critical points suggest possible local maxima, minima, or points of inflection.

However, in this scenario, \( x = \frac{1}{e} \approx 0.3679 \) is not in the segment \([1, 2]\) we are examining. This highlights how critical points outside our interval concern the function's behavior elsewhere but do not affect the current search for global extrema on \([1, 2]\).

Even if critical points weren't directly applicable here, derivatives remain an essential caliber for understanding and predicting functional landmarks.