Problem 30 - Applied Calculus

Question

Find the exact global maximum and minimum values of the function. The domain is all real numbers unless otherwise specified. $$f(x)=x-\ln x \text { for } x>0$$

Step-by-Step Solution

Answer

Global minimum at \( x = 1 \) with value 1; no finite global maximum.

Step 1: Find the derivative of the function

To find the maximum and minimum values of the function, we first need to find the derivative. Start by rewriting the function: \( f(x) = x - \ln x \). Now, differentiate with respect to \( x \): \( f'(x) = 1 - \frac{1}{x} \).

Step 2: Set the derivative to zero

Set the derivative equal to zero to find the critical points: \( 1 - \frac{1}{x} = 0 \). Solve for \( x \): \( \frac{1}{x} = 1 \) which leads to \( x = 1 \).

Step 3: Second derivative test

Find the second derivative to determine the concavity and identify maximum or minimum: The first derivative is \( f'(x) = 1 - \frac{1}{x} \). Differentiate again to get \( f''(x) = \frac{1}{x^2} \). Since \( f''(x) > 0 \) for all \( x > 0 \), the function is concave up, indicating a local minimum at \( x=1 \).

Step 4: Evaluate the function at critical points

Calculate \( f(x) \) at the critical point: \( f(1) = 1 - \ln 1 = 1 \). Since \( \ln 1 = 0\), \( f(1) = 1 \). This is the minimum value because the function is concave up around \( x = 1 \).

Step 5: Analyze endpoint behavior and limits

As \( x \to 0^+ \), \( f(x) = x - \ln x \to +\infty \) because the logarithm approaches \(-\infty\). As \( x \to \infty \), \( f(x) = x - \ln x \to \infty \). Thus, there is no finite global maximum on the domain \( x > 0 \).

Key Concepts

Derivative

In calculus, a derivative helps us understand how a function changes as its input changes. It's like a rate of change or the slope of the function at any given point. For the function given in the exercise, \( f(x) = x - \ln x \), the derivative is a key step in finding maximum and minimum values.
The derivative of \( f(x) \) is \( f'(x) = 1 - \frac{1}{x} \). This expression comes from taking the derivative of both \( x \) and \( \ln x \). Remember:

• The derivative of \( x \) is \( 1 \).
• The derivative of \( \ln x \) is \( \frac{1}{x} \).

By subtracting these, we get the derived expression. Understanding how to find and interpret derivatives is essential in solving problems involving rates of change.

Critical Points

Critical points of a function are where its derivative is zero or undefined. They are potential spots for finding local maximums or minimums. In our case, we set the derivative to zero:
\( f'(x) = 1 - \frac{1}{x} = 0 \)
Solving this equation helps us find the critical points, which in this solution's context comes out to \( x = 1 \).

Critical points are crucial because they are where the function can change direction, moving from increasing to decreasing, or vice versa. It's akin to finding the peaks and troughs in a landscape, and for optimization problems, this is where you'll want to focus your attention.

Second Derivative Test

Once we have the critical points, we use the second derivative test to determine if these points are local minimums or maximums.
Here, the second derivative of \( f(x) \) is \( f''(x) = \frac{1}{x^2} \). Since \( f''(x) > 0 \) for all \( x > 0 \), the function is concave up around \( x = 1 \).
A positive second derivative indicates a local minimum, as the curve is shaped like a U. Conversely, if the second derivative were negative, the curve would be shaped like an upside-down U, suggesting a local maximum. This test is a handy tool to quickly verify the nature of critical points.

Concavity Analysis

Concavity tells us about the "bend" of the function. If the function is concave up, it resembles a U-shape, indicating that it's opening upwards. Conversely, if it's concave down, it resembles an upside-down U.
To analyze concavity, we look at the second derivative. For the function \( f(x) = x - \ln x \), the second derivative is \( f''(x) = \frac{1}{x^2} \) which is always greater than zero for \( x > 0 \).
This positive value tells us that the entire function is concave up for all specified \( x \). Because of the concavity analysis, we can confidently identify \( x = 1 \) as a local minimum.
Understanding concavity can provide deeper insights into the behavior of a function and clarify why certain points act as maximums or minimums.