Problem 34 - 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. $$ f(x)=\ln \frac{x}{x^{2}+1} ;\left[\frac{1}{2}, 3\right] $$

Step-by-Step Solution

Answer

The maximum value occurs at \( x = 1 \), and the minimum value occurs at \( x = 3 \).

Step 1: Find the Derivative

To find the critical points where the extreme values might occur, we start by finding the derivative of the function. Let's apply the quotient rule for derivatives. If \( u = \ln x \) and \( v = \ln(x^2 + 1) \), then \( \frac{d}{dx} \left( \ln \frac{x}{x^2+1} \right) = \frac{u'v - uv'}{v^2} = \frac{1/x \cdot (x^2+1) - \ln x \cdot \frac{d}{dx}(x^2+1)}{(x^2+1)^2} \). Simplifying, we find \( f'(x) = \frac{1 - x^2}{x(x^2+1)} \).

Step 2: Find Critical Points

To find critical points, solve the equation \( f'(x) = 0 \). This implies \( \frac{1 - x^2}{x(x^2+1)} = 0 \), which gives \( 1 - x^2 = 0 \). Solving, we find \( x = 1 \) as the critical point within the interval \([\frac{1}{2}, 3]\).

Step 3: Evaluate the Function at Critical and Endpoint Values

Calculate the value of the function at the critical point and at the endpoints of the interval. We have three points to evaluate: \( f(\frac{1}{2}) = \ln \frac{\frac{1}{2}}{\frac{1}{2}^2 + 1} \), \( f(1) = \ln \frac{1}{1^2 + 1} \), and \( f(3) = \ln \frac{3}{3^2 + 1} \). Simplifying these, we get:- \( f(\frac{1}{2}) = \ln \frac{\frac{1}{2}}{\frac{5}{4}} = \ln \frac{2}{5} \).- \( f(1) = \ln \frac{1}{2} \).- \( f(3) = \ln \frac{3}{10} \).

Step 4: Compare Values to Determine Extremes

Now compare the calculated function values: \( f(\frac{1}{2}) = \ln \frac{2}{5} \), \( f(1) = \ln \frac{1}{2} \), and \( f(3) = \ln \frac{3}{10} \). Since \( \ln x \) is increasing with respect to \( x \), order these by checking numerically: \( \frac{1}{2} \approx 0.5 \), \( \frac{3}{10} \approx 0.3 \), and \( \frac{2}{5} \approx 0.4 \). The maximum value occurs at \( x = 1 \) (\( \ln 0.5 \)), and the minimum occurs at \( x = 3 \) (\( \ln 0.3 \)).

Key Concepts

Derivative Calculation

Calculating derivatives is a key step in finding the extreme values of a function. The derivative provides information about the rate of change of the function, revealing points where the slope is zero. This is where extreme values often occur.

To find the derivative of the given function, \( f(x) = \ln \frac{x}{x^2 + 1} \), we use the quotient rule. When using the quotient rule, remember the formula for the derivative of a function \( \frac{u}{v} \) is given by \( \frac{u'v - uv'}{v^2} \).

Let \( u = \ln x \) and \( v = \ln(x^2 + 1) \). By applying the quotient rule, we find that the derivative \( f'(x) \) is:

• \( u' = \frac{1}{x} \)
• \( v' = \frac{d}{dx}(x^2 + 1) = 2x \)

After applying, simplifying gives: \[ f'(x) = \frac{1 - x^2}{x(x^2 + 1)} \] This derivative will help find where the function has extreme values.

Critical Points

Critical points are where a function's derivative is zero or undefined. These points are candidates for local maxima or minima, helping identify potential extreme values.

Let's solve \( f'(x) = 0 \) for \( f(x) = \ln \frac{x}{x^2+1} \). Since the derivative is \( \frac{1 - x^2}{x(x^2+1)} \), setting \( 1 - x^2 = 0 \) results in \( x^2 = 1 \). Solving gives \( x = 1 \).

Checking the interval \([\frac{1}{2}, 3]\), we find \( x = 1 \) lies within. This is our critical point.

Interval Analysis

Interval analysis helps find extreme values by comparing function values at critical points and the interval's endpoints. This analysis ensures we don't miss any extremum outside critical points.

We plug the found critical point \( x = 1 \) and the interval endpoints into the function \( f(x) \). Calculate:

• \( f(\frac{1}{2}) = \ln \frac{\frac{1}{2}}{\frac{1}{2}^2 + 1} = \ln \frac{2}{5} \)
• \( f(1) = \ln \frac{1}{2} \)
• \( f(3) = \ln \frac{3}{10} \)

This setup helps compare and order values seen in the interval, aiding in determining the maximum and minimum points.

Logarithmic Functions

Understanding logarithmic functions is key for analyzing expressions like \( f(x) = \ln \frac{x}{x^2 + 1} \). The natural logarithm \( \ln \) of \( x \) is the power to which \( e \) (Euler's number: approximately 2.718) must be raised to get \( x \). Logarithms simplify multiplication and division into addition and subtraction, making complex relationships easier to handle.

With this exercise, note that logarithmic functions are continuously increasing for positive values of \( x \). Thus, numerical comparisons (like \( \ln \frac{1}{2} \) or \( \ln \frac{3}{10} \)) use increasing property to help identify the minimum and maximum values by comparing the numerators (or data). Simply put, larger inputs to a logarithmic function yield larger outputs, guiding us in identifying the extreme values.