Problem 39 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 39 through 42 , find the largest and small values of the given function over the prescribed closi bounded interval. 39\. \(f(x)=\ln \left(4 x-x^{2}\right) \quad\) for \(1 \leq x \leq 3\)

Step-by-Step Solution

Answer

Largest value: \( \ln(4) \) at \( x = 2 \); Smallest value: \( \ln(3) \) at \( x = 1 \) and \( x = 3 \)

Step 1 - Determine Function Domain

Determine the domain of the function. The expression inside the logarithm must be positive: \[ 4x - x^2 > 0 \]Solve the inequality: \[ x(4 - x) > 0 \]This inequality is satisfied for \(0 < x < 4\), so the valid range within the given interval is \(1 \leq x \leq 3\).

Step 2 - Find Critical Points

Take the first derivative of the function to find critical points: \[ f'(x) = \frac{d}{dx} [\ln(4x - x^2)] = \frac{4 - 2x}{4x - x^2} \]Set the derivative equal to zero to find critical points: \[ 4 - 2x = 0 \]\[ x = 2 \]Check if \(x = 2\) is within the interval [1, 3]. Since it is, we will use it.

Step 3 - Evaluate Function at Critical Points and Endpoints

Evaluate the function at the critical points and endpoints of the interval:\[ f(1) = \ln(4(1) - 1^2) = \ln(3) \]\[ f(3) = \ln(4(3) - 3^2) = \ln(3) \]\[ f(2) = \ln(4(2) - 2^2) = \ln(4) \]

Step 4 - Compare Values

Compare the values obtained in the previous step:\[ f(1) = \ln(3) \]\[ f(3) = \ln(3) \]\[ f(2) = \ln(4) \]Since \(\ln(4) > \ln(3)\), the largest value is at \(x = 2\) and smallest values are at the endpoints \(x = 1\) and \(x = 3\).

Key Concepts

Critical Points

Critical points are the values of x where the first derivative of a function is zero or undefined. These points are essential because they can indicate local maxima, minima, or horizontal inflection points. In our example, we have the function \(f(x) = \ln(4x - x^2)\). To find the critical points, we take the first derivative of this function and set it to zero. The resulting critical point is \(x = 2\). We then ensure that this point lies within our given interval, \([1, 3]\). Since it does, we will evaluate the function at this critical point.

First Derivative

The first derivative of a function describes the rate at which the function's value changes. For the logarithmic function \(f(x) = \ln(4x - x^2)\), the first derivative is: \[f'(x) = \frac{4 - 2x}{4x - x^2} \]. Derivatives are fundamental in calculus because they help us to find slopes of tangent lines, velocities, and to identify critical points. For this problem, we set the first derivative equal to zero to find our critical points: \[4 - 2x = 0 \]. Solving this equation gives us the critical point \(x = 2\).

Domain of a Function

The domain of a function refers to the set of all possible input values (x-values) for which the function is defined. For logarithmic functions, the argument inside the logarithmic function must be greater than zero. In our example, \(f(x) = \ln(4x - x^2)\), we solve the inequality \[4x - x^2 > 0 \] to find the domain. This inequality is satisfied for \(0 < x < 4\). Given the prescribed interval \(1 \leq x \leq 3\), we further constrain the domain to \[1 \leq x \leq 3 \]. This means the function is defined and positive within this interval, ensuring all calculations will be valid.

Evaluating Function at Endpoints

To find the maximum and minimum values of a function over a closed interval, it's essential to evaluate the function at the endpoints and the critical points within the interval. For our problem, we evaluated the function \(f(x) = \ln(4x - x^2)\) at the critical point \(x = 2\) and at the endpoints \(x = 1\) and \(x = 3\). The evaluations are as follows:
\[ f(1) = \ln(3) \], \[ f(3) = \ln(3) \], and \[ f(2) = \ln(4) \].
Comparing these values, \( \ln(4) \) is the largest, and \( \ln(3) \) is the smallest. Therefore, the largest value is at \(x = 2\) and the smallest values are at the endpoints \(x = 1\) and \(x = 3\).