Problem 48 - Essential Calculus Early Transcendentals

Question

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(x)=x-\ln x, \quad\left[\frac{1}{2}, 2\right]$$

Step-by-Step Solution

Answer

The absolute maximum is 1.307 at x=2, and the absolute minimum is 1 at x=1.

Step 1: Find the derivative

First, we need to find the derivative of the given function, which is \( f(x) = x - \ln x \). The derivative is \( f'(x) = 1 - \frac{1}{x} \).

Step 2: Find critical points

Set the derivative equal to zero to find critical points: \( 1 - \frac{1}{x} = 0 \). Solving for \( x \), we get \( x = 1 \). This is our critical point within the interval \( \left[ \frac{1}{2}, 2 \right] \).

Step 3: Evaluate the function at endpoints and critical points

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

Step 4: Compare values to find the absolute maximum and minimum

From Step 3, we have:- \( f\left(\frac{1}{2}\right) = \frac{1}{2} + \ln 2 \approx 1.193 \)- \( f(2) = 2 - \ln 2 \approx 1.307 \)- \( f(1) = 1 \)The absolute maximum value is \( 1.307 \) at \( x = 2 \), and the absolute minimum value is \( 1 \) at \( x = 1 \).

Key Concepts

Critical Points

In calculus, finding critical points is essential when analyzing the behavior of a function within a certain interval. A critical point occurs where the derivative of the function is zero or undefined. For our function, critical points help identify where the function might shift from increasing to decreasing, or vice versa. These shifts can signify important changes such as maximums or minimums.

To locate critical points for the function \( f(x) = x - \ln x \), we first compute the derivative, resulting in \( f'(x) = 1 - \frac{1}{x} \). Once we have the derivative, setting \( f'(x) = 0 \) allows us to find the critical values. Solving \( 1 - \frac{1}{x} = 0 \) leads us to \( x = 1 \) within the interval \( \left[ \frac{1}{2}, 2 \right] \). This means \( x = 1 \) is a critical point to consider for further analysis.

Derivative

The concept of a derivative in calculus is a fundamental tool for understanding how functions change. Derivatives provide the rate at which a function value changes as its input changes. This is expressed through the slope of the tangent line to the curve of the function at any given point.

For the function \( f(x) = x - \ln x \), the derivative \( f'(x) = 1 - \frac{1}{x} \) was calculated. This derivative tells us how quickly the function \( f(x) \) is climbing or dropping at each point in its domain. A positive derivative indicates the function is increasing, while a negative value shows it's decreasing. When the derivative equals zero, it signals a potential peak or trough, known as a critical point.

Checking endpoints and critical points using the derivative helps us to fully grasp the behavior of the function, allowing us to find maximum and minimum values within a specified interval.

Absolute Maximum and Minimum

Finding absolute maximum and minimum values of a function over a defined interval involves comparing function values at critical points and endpoints. These values represent the highest and lowest points, respectively, of the function in that specific range.

For \( f(x) = x - \ln x \) evaluated over \( \left[ \frac{1}{2}, 2 \right] \), checking \( f(x) \) at the critical point \( x = 1 \), as well as at the endpoints \( x = \frac{1}{2} \) and \( x = 2 \), allows us to determine these extreme values.

• At \( x = \frac{1}{2} \), the calculation gives \( f\left(\frac{1}{2}\right) = \frac{1}{2} + \ln 2 \approx 1.193 \).
• At \( x = 1 \), the value is \( f(1) = 1 \).
• At \( x = 2 \), the result is \( f(2) = 2 - \ln 2 \approx 1.307 \).

Comparing these, the highest value, or absolute maximum, of approximately 1.307 occurs at \( x = 2 \), whereas the lowest, or absolute minimum, value of 1 occurs at the critical point \( x = 1 \). Understanding these concepts helps in sketching the function's graph and grasping its limits within a specific interval.