Problem 9 - University Calculus: Early Transcendentals

Question

Answer the following questions about the functions whose derivatives are given. a. What are the critical points of \(f ?\) b. On what open intervals is \(f\) increasing or decreasing? c. At what points, if any, does \(f\) assume local maximum and minimum values? $$f^{\prime}(x)=1-\frac{4}{x^{2}}, \quad x \neq 0$$

Step-by-Step Solution

Answer

Critical points: \(x = 2, -2\). Increasing: \((-\infty, -2)\) and \((2, \infty)\). Local max at \(-2\), local min at \(2\).

Step 1: Find Critical Points

First, we find the critical points of the function by setting the derivative equal to zero: \[ f'(x) = 1 - \frac{4}{x^2} = 0 \]Solve for \(x\):\[ 1 = \frac{4}{x^2} \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = \pm 2 \]Thus, the critical points are \(x = 2\) and \(x = -2\).

Step 2: Determine Increasing or Decreasing Intervals

To determine where \(f(x)\) is increasing or decreasing, consider the number line and test intervals between the critical points:1. Interval \((-\infty, -2)\): Choose \(x = -3\). \[f'(-3) = 1 - \frac{4}{(-3)^2} = 1 - \frac{4}{9} > 0\] So, \(f(x)\) is increasing on \((-\infty, -2)\).2. Interval \((-2, 2)\): Choose \(x = 0\) (only check \(x\) close to it but not \(0\)). \[ f'(0.1) \approx 1 - \frac{4}{0.01} < 0 \] So, \(f(x)\) is decreasing on \((-2, 2)\).3. Interval \((2, \infty)\): Choose \(x = 3\). \[f'(3) = 1 - \frac{4}{9} > 0\] So, \(f(x)\) is increasing on \((2, \infty)\).

Step 3: Identify Local Maxima and Minima

Using the first derivative test at the critical points:1. At \(x = -2\): \(f'(x)\) changes from positive to negative. Thus, \(x = -2\) is a local maximum.2. At \(x = 2\): \(f'(x)\) changes from negative to positive. Thus, \(x = 2\) is a local minimum.

Key Concepts

Understanding Derivatives

Derivatives measure how a function changes as its input changes. By taking the derivative of a function, we can understand how quickly or slowly the function value is increasing or decreasing.
For example, when we have a derivative like \( f'(x) = 1 - \frac{4}{x^2} \), it tells us how the function \( f(x) \) behaves around different points \( x \).

At critical points, the derivative is zero or undefined, which indicates a potential change in the function's behavior.
By setting \( f'(x) = 0 \), we solve to find these points, leading us to discover any turning points on the graph.

• Critical points are found where the derivative equals zero or does not exist.
• Here we calculated \( x = \pm 2 \) as critical points.

Understanding these points is crucial for sketching graphs and optimizing functions.

Local Maximum and Minimum Values

Local maxima and minima refer to the highest or lowest points in a specific section of a graph. When we examine how the derivative changes around critical points, it helps determine the local behavior.

Specifically for this function,

• We used the first derivative test to see how \( f'(x) \) changes sign around \( x = -2 \) and \( x = 2 \).

For \( f'(x) \) changing from positive to negative, we find a local maximum at \( x = -2 \). This indicates a peak in the graph.
Conversely, when \( f'(x) \) goes from negative to positive, as it does at \( x = 2 \), a local minimum is observed. Here, the graph reaches a trough before ascending.

Determining Increasing and Decreasing Intervals

An open interval can be identified as increasing or decreasing based on the sign of the derivative.

To determine where a function is increasing or decreasing, we select test points in the intervals defined by critical points.

• If \( f'(x) > 0 \), the function is increasing in that interval.
• If \( f'(x) < 0 \), the function is decreasing.

When assessing \( f'(x) = 1 - \frac{4}{x^2} \):
- The function increases on \((-\infty, -2)\) and \((2, \infty)\), where the derivative is positive.
- It decreases within \((-2, 2)\), where the derivative is negative.

These insights help us understand the overall shape and behavior of the graph.