Problem 49 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ f(x)=x-\ln x $$

Step-by-Step Solution

Answer

The function is decreasing on \((0, 1)\) and increasing on \((1, \infty)\).

Step 1: Find the derivative of the function

We start by finding the derivative of the given function \( f(x) = x - \ln x \). This will help us determine the intervals of increase or decrease. The derivative \( f'(x) \) is given by the following formula:\[ f'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\ln x) = 1 - \frac{1}{x} \]

Step 2: Set the derivative equal to zero

Next, we find the critical points by setting the derivative equal to zero:\[ f'(x) = 1 - \frac{1}{x} = 0 \]Solving for \( x \), we get:\[ \frac{1}{x} = 1 \]\[ x = 1 \]

Step 3: Determine the sign of the derivative

Now, we determine the sign of \( f'(x) = 1 - \frac{1}{x} \) in the intervals determined by the critical point. We'll test the intervals \((0, 1)\) and \((1, \infty)\):- For \( x \) in \((0, 1)\), choose \( x = 0.5 \): \[ f'(0.5) = 1 - \frac{1}{0.5} = 1 - 2 = -1 \] (negative)- For \( x \) in \((1, \infty)\), choose \( x = 2 \): \[ f'(2) = 1 - \frac{1}{2} = 1 - 0.5 = 0.5 \] (positive)

Step 4: Conclude the intervals of increase and decrease

Based on the signs of the derivative in different intervals:- \( f(x) \) is **decreasing** on \((0, 1)\) because \( f'(x) < 0 \).- \( f(x) \) is **increasing** on \((1, \infty)\) because \( f'(x) > 0 \).

Key Concepts

Critical Points

Critical points in calculus are where the function's derivative equals zero or is undefined. These points are essential as they help us understand where the function might change behavior, such as switching from increasing to decreasing or vice versa.
To find the critical points, first find the derivative of the function. For the function \( f(x) = x - \ln x \), the derivative is \( f'(x) = 1 - \frac{1}{x} \).
Set this derivative equal to zero to find the x-values where the function has potential critical points:

• \( 1 - \frac{1}{x} = 0 \)

Solving for \( x \) gives \( x = 1 \). This is a critical point because it is where the derivative changes sign. Identifying and understanding these points provide valuable insights into the behavior of the function across various intervals.

Intervals of Increase and Decrease

Determing the intervals where a function increases or decreases involves analyzing the sign of the derivative over specified regions along the x-axis. Once critical points are found, they serve as boundaries for these intervals.
To find where the function \( f(x) = x - \ln x \) is increasing or decreasing, examine \( f'(x) = 1 - \frac{1}{x} \) within the intervals created by the critical point \( x = 1 \).

• In the interval \((0, 1)\), choose a test point such as \( x = 0.5 \). Calculate \( f'(0.5) = 1 - \frac{1}{0.5} = -1 \), indicating the function is decreasing on this interval.
• In the interval \((1, \infty)\), choose \( x = 2 \) and find that \( f'(2) = 1 - \frac{1}{2} = 0.5 \), showing the function increases.

Analyzing these intervals helps provide a clear picture of where the function grows or declines, an invaluable skill in studying the behavior of functions.

Derivative of Natural Logarithm

The derivative of the natural logarithm function plays a key role in calculus. It often appears in problems involving growth and decay, economics, and other applied sciences.
The natural logarithm function \( \ln x \) has a simple yet extremely useful derivative:

• \( \frac{d}{dx} (\ln x) = \frac{1}{x} \)

The derivative \( \frac{1}{x} \) is pivotal in solving problems involving differentiation of logarithmic expressions. Knowing this derivative by heart assists in quickly identifying changes in growth models or decay processes. In our problem, it was used to find the derivative \( f'(x) = 1 - \frac{1}{x} \), critical for determining where the function increases or decreases. Such understanding greatly enhances the ability to tackle a range of problems in calculus and beyond.