Problem 24 - Thomas Calculus

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$ f(\theta)=6 \theta-\theta^{3} $$

Step-by-Step Solution

Answer

The function increases on \((-\sqrt{2}, \sqrt{2})\), decreases on \((-\infty, -\sqrt{2})\) and \((\sqrt{2}, \infty)\), with a local min at \(\theta = -\sqrt{2}\) and a local max at \(\theta = \sqrt{2}\).

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first find the first derivative of the function. The given function is \( f(\theta) = 6\theta - \theta^3 \). Differentiating with respect to \( \theta \), we get: \[ f'(\theta) = 6 - 3\theta^2 \].

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. Set the derivative to zero and solve for \( \theta \): \[ 6 - 3\theta^2 = 0 \] Solving, we find: \[ \theta^2 = 2 \] \[ \theta = \pm\sqrt{2} \] These are our critical points.

Step 3: Test Intervals Around Critical Points

To determine where the function is increasing or decreasing, test intervals around the critical points \( \theta = -\sqrt{2} \) and \( \theta = \sqrt{2} \). - Choose \( \theta < -\sqrt{2} \), for example \( \theta = -2 \): \[ f'(-2) = 6 - 3(-2)^2 = -6 \] (negative, function decreasing)- Choose \( -\sqrt{2} < \theta < \sqrt{2} \), for example \( \theta = 0 \): \[ f'(0) = 6 - 0 = 6 \] (positive, function increasing)- Choose \( \theta > \sqrt{2} \), for example \( \theta = 2 \): \[ f'(2) = 6 - 3(2)^2 = -6 \] (negative, function decreasing).

Step 4: Determine Intervals

Based on testing, the function is increasing on the interval \((-\sqrt{2}, \sqrt{2})\) and decreasing on the intervals \((-\infty, -\sqrt{2})\) and \((\sqrt{2}, \infty)\).

Step 5: Find Local Extremes Using the First Derivative

From the first derivative test, the function changes from decreasing to increasing at \( \theta = -\sqrt{2} \) (local minimum) and from increasing to decreasing at \( \theta = \sqrt{2} \) (local maximum).

Step 6: Calculate Function Values at Critical Points

Evaluate the function at the critical points to determine the extreme values: \[ f(-\sqrt{2}) = 6(-\sqrt{2}) - (-\sqrt{2})^3 = -6\sqrt{2} + 2\sqrt{2} = -4\sqrt{2} \] (l ext{ocal minimum at } \theta = -\sqrt{2}) \[ f(\sqrt{2}) = 6(\sqrt{2}) - (\sqrt{2})^3 = 6\sqrt{2} - 2\sqrt{2} = 4\sqrt{2} \] (l ext{ocal maximum at } \theta = \sqrt{2}).

Key Concepts

Critical Points

In calculus, critical points are crucial for understanding the behavior of a function. They are found where the derivative of the function is either zero or undefined. These points indicate where there might be a change in the direction of the graph, such as from increasing to decreasing or vice versa.

To find the critical points, we set the first derivative equal to zero and solve for the variable. In our example function, \[ f(\theta) = 6\theta - \theta^3 \]the derivative is \[ f'(\theta) = 6 - 3\theta^2. \] Setting this equal to zero yields:\[ 6 - 3\theta^2 = 0 \] which simplifies to \[ \theta^2 = 2. \] Taking the square root, we obtain two critical points, \[ \theta = +\sqrt{2} \] and \[ \theta = -\sqrt{2}. \]

These points are important as they may indicate possible locations of local minimums or maximums, depending on the function's behavior before and after these points.

First Derivative Test

The First Derivative Test is a handy tool to determine if the critical points are local maxima or minima. It involves analyzing the sign of the derivative before and after each critical point.

Here's how it works in practice:

• If the derivative changes from negative to positive as you pass through a critical point, the function has a local minimum at that point.
• If the derivative changes from positive to negative, the function has a local maximum at that point.
• If the sign does not change, the critical point is neither a maximum nor a minimum.

Using the first derivative \[ f'(\theta) = 6 - 3\theta^2 \] and the critical points \( \theta = \pm\sqrt{2} \), interval testing assists in deciding the nature of these points.

We found:

• At \( \theta = -\sqrt{2} \), \( f' \) changes from negative to positive, indicating a local minimum.
• At \( \theta = \sqrt{2} \), \( f' \) changes from positive to negative, indicating a local maximum.

These results help us understand where our function has turning points.

Local Extremes

Local extremes refer to the highest or lowest points in a particular section of a function's graph. Local maxima and minima are these points where a function's direction changes. This happens specifically at the critical points where \( f'(\theta) \) either equals zero or is undefined.

By applying the First Derivative Test, we found the local extreme values at the critical points. For the function \( f(\theta) = 6\theta - \theta^3 \):

• The local minimum occurs at \( \theta = -\sqrt{2} \) with a value of \( -4\sqrt{2} \).
• The local maximum occurs at \( \theta = \sqrt{2} \) with a value of \( 4\sqrt{2} \).

These local extremes signify turning points, where the function transitions from decreasing to increasing or vice versa.

Interval Testing

Interval testing involves selecting and checking test points in the intervals defined by the critical points. It's a significant step for understanding the behavior of the function between these points.

For the function \( f(\theta) = 6\theta - \theta^3 \), the critical points divide the number line into sections: \(( -\infty, -\sqrt{2} )\), \(( -\sqrt{2}, \sqrt{2} )\), and \(( \sqrt{2}, \infty )\).

To identify whether the function is increasing or decreasing in these intervals:

• For \( \theta < -\sqrt{2} \), choose a test point such as \( \theta = -2 \) and find \( f'(-2) = -6 \). This negative value indicates the function decreases on \(( -\infty, -\sqrt{2} )\).
• For \( -\sqrt{2} < \theta < \sqrt{2} \), choose a test point like \( \theta = 0 \) where \( f'(0) = 6 \). A positive value means the function increases on \(( -\sqrt{2}, \sqrt{2} )\).
• For \( \theta > \sqrt{2} \), pick \( \theta = 2 \) and \( f'(2) = -6 \). Again, a negative value shows the function decreases on \(( \sqrt{2}, \infty )\).

Through interval testing, we've clearly identified where the function is increasing or decreasing, assisting in sketching the behavior of the entire function around its critical points.