Problem 33 - Calculus with Concepts in Calculus

Question

Find all extreme values (if any) of the given function on the given interval. Determine at which numbers in the interval these values occur. $$ f(x)=x-2 \ln x ;\left[\frac{1}{2}, 2\right] $$

Step-by-Step Solution

Answer

The maximum value of \( f(x) \) is at \( x = 2 \); the minimum is at \( x = \frac{1}{2} \).

Step 1: Find the Derivative

First, find the derivative of the function \( f(x) = x - 2 \ln x \). The derivative, \( f'(x) \), is obtained using the rules for differentiating polynomials and logarithmic functions: \( f'(x) = 1 - \frac{2}{x} \).

Step 2: Set the Derivative to Zero

Find the critical points by setting the derivative \( f'(x) = 1 - \frac{2}{x} = 0 \). Solving for \( x \) gives \( \frac{2}{x} = 1 \), so \( x = 2 \).

Step 3: Determine the Nature of Critical Points

Check the sign changes of \( f'(x) \) around \( x = 2 \) to determine if there's a minimum or maximum. For \( x < 2 \), \( f'(x) > 0 \); for \( x > 2 \), \( f'(x) < 0 \). This confirms \( x = 2 \) is a maximum.

Step 4: Evaluate the Function at Critical Points and Endpoints

Evaluate \( f(x) \) at the critical point and the endpoints of the interval. \( f\left(\frac{1}{2}\right) = \frac{1}{2} - 2 \ln \frac{1}{2} = \frac{1}{2} + 2 \ln 2 \), \( f(2) = 2 - 2 \ln 2 \), and \( f\left(2\right) = 2 - 2 \ln 2 \).

Step 5: Compare Function Values at Critical Points and Endpoints

From Step 4, \( f\left(\frac{1}{2}\right) \) is a minimum because \( f\left(\frac{1}{2}\right) > f(2) \). \( f(2) \) is a maximum value within the interval.

Key Concepts

Derivatives and How They Connect to Extreme Values

Derivatives play a crucial role in calculus especially when it comes to finding the extreme values of a function. A derivative, in simple terms, represents the rate at which a function is changing at any given point. In the context of extreme values, derivatives help us by identifying where these high or low values (maxima or minima) occur on a function.

• The process begins with finding the derivative of the function. For the given function, \( f(x) = x - 2 \ln x \), the derivative is \( f'(x) = 1 - \frac{2}{x} \).
• A critical step involves locating points where the derivative is zero or undefined, as these are potential points for extreme values, known as critical points.
• By finding where \( f'(x) = 0 \), such as when \( 1 - \frac{2}{x} = 0 \), we solve for \( x \) which leads to \( x = 2 \).

Knowing how to find and interpret derivatives is essential for understanding the behavior of the function, especially in finding where the function may reach its top or bottom within a specified interval.

Understanding Critical Points

Critical points are the values of \( x \) where the derivative of a function is either zero or undefined. These points are crucial in finding the peaks and valleys of a function.

• Once we find the derivative, the next step is setting \( f'(x) = 0 \) to identify potential critical points. For the function \( f(x) = x - 2 \ln x \), solving \( 1 - \frac{2}{x} = 0 \) gives us the critical point \( x = 2 \).
• Critical points tell us where the function's slope changes, which can indicate a maximum, minimum, or saddle point.
• Determining the nature of these critical points involves checking the derivative's sign change around these points. For example, if \( f'(x) \) changes from positive to negative, it indicates a maximum at that point because the function changes from increasing to decreasing.

Watching for how the sign of \( f'(x) \) changes provides a strong indication of where a function might have its extreme values.

Logarithmic Functions and Their Role in Calculus

Logarithmic functions are another vital component in calculus, intersecting frequently with concepts like derivatives and critical points. They have their distinct rules when it comes to differentiation, which are crucial for calculus problems like finding extreme values.

• Understanding the basics of logarithmic differentiation is key. For instance, the derivative of \( \ln x \) is \( \frac{1}{x} \), which is used in finding the derivative of complex functions like \( f(x) = x - 2 \ln x \).
• Logarithmic functions are particularly useful when dealing with growth and decay problems, and they often simplify the computation of derivatives.
• In the context of the exercise, understanding how the natural logarithm function \( \ln x \) behaves helps predict how the function \( x - 2 \ln x \) will act, especially when seeking its extreme values.

Grasping these concepts ensures that students not only solve calculus problems correctly but also develop a deeper understanding of how different functions behave within mathematical analyses.