Problem 13 - Thomas Calculus

Question

In Exercises \(9-28 :\) a. Find the intervals on which the function is increasing and decreasing. b. Then identify the function's local extreme values, if any, saying where they are taken on. c. Which, if any, of the extreme values are absolute? d. Support your findings with a graphing calculator or computer grapher. $$ f(\theta)=3 \theta^{2}-4 \theta^{3} $$

Step-by-Step Solution

Answer

The function is increasing on \((0, \frac{1}{2})\) and decreasing on \(( -\infty, 0)\) and \((\frac{1}{2}, \infty)\). It has a local minimum at \(0\) and an absolute maximum at \(\frac{1}{2}\).

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, begin by finding the first derivative, which tells us the slope of the tangent line to the graph at any point. Differentiate \( f(\theta) = 3\theta^2 - 4\theta^3 \). The derivative is calculated as follows:\[ f'(\theta) = \frac{d}{d\theta}(3\theta^2 - 4\theta^3) = 6\theta - 12\theta^2. \]

Step 2: Solve for Critical Points

Critical points occur where the derivative is zero or undefined. Set the derivative equal to zero to find these points:\[ 6\theta - 12\theta^2 = 0. \]Factor the expression:\[ 6\theta(1 - 2\theta) = 0. \]Solve for \( \theta \):\[ \theta = 0 \quad \text{or} \quad \theta = \frac{1}{2}. \]

Step 3: Determine Intervals of Increase and Decrease

To determine whether the function is increasing or decreasing, test the intervals defined by the critical points \( \theta = 0 \) and \( \theta = \frac{1}{2} \). Choose test points in each interval: For \( \theta < 0 \), use \( \theta = -1 \).For \( 0 < \theta < \frac{1}{2} \), use \( \theta = \frac{1}{4} \).For \( \theta > \frac{1}{2} \), use \( \theta = 1 \).Plug these into \( f'(\theta) = 6\theta - 12\theta^2 \): 1. \( \theta = -1: 6(-1) - 12(-1)^2 = -6 - 12 = -18 \) (decreasing)2. \( \theta = \frac{1}{4}: 6\left(\frac{1}{4}\right) - 12\left(\frac{1}{4}\right)^2 = \frac{3}{2} - \frac{3}{4} = \frac{3}{4} \) (increasing)3. \( \theta = 1: 6(1) - 12(1)^2 = 6 - 12 = -6 \) (decreasing)Thus, the function is decreasing on \(( -\infty, 0 )\) and \(( \frac{1}{2}, \infty)\), and increasing on \(( 0, \frac{1}{2} ) \).

Step 4: Identify Local Extremes

Next, evaluate the behavior of the function at the critical points \( \theta = 0 \) and \( \theta = \frac{1}{2} \):- At \( \theta = 0 \), the derivative changes from negative to positive, indicating a local minimum.- At \( \theta = \frac{1}{2} \), the derivative changes from positive to negative, indicating a local maximum.Calculate the function values at these points:- For \( \theta = 0 \), \( f(0) = 3(0)^2 - 4(0)^3 = 0 \).- For \( \theta = \frac{1}{2} \), \( f\left(\frac{1}{2}\right) = 3\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right)^3 = \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \).

Step 5: Determine Absolute Extremes

To determine if any local extremes are absolute, consider the end behavior of the function. As \( \theta \to \pm \infty \), the term \( -4\theta^3 \) will dominate, causing \( f(\theta) \to -\infty \).Thus, the local maximum at \( \theta = \frac{1}{2} \), with value \( f\left(\frac{1}{2}\right) = \frac{1}{4} \), is an absolute maximum for the function on its entire domain \( (-\infty, \infty) \).

Step 6: Graphical Verification

Finally, use a graphing calculator or computer graphing software to plot \( f(\theta) = 3\theta^2 - 4\theta^3 \). The graph will confirm that the function has a local maximum at \( \theta = \frac{1}{2} \) and a local minimum at \( \theta = 0 \). The function decreases as \( \theta \to \infty \) or \( \theta \to -\infty \), which matches our derivatives and calculations.

Key Concepts

Critical Points

Critical points are where the first derivative of a function equals zero or becomes undefined. These points are crucial because they can indicate potential locations for local maximums or minimums in a function. To find these points for the function \( f(\theta) = 3\theta^2 - 4\theta^3 \), we first calculate the derivative as \( f'(\theta) = 6\theta - 12\theta^2 \).
Setting the derivative to zero gives us the equation \( 6\theta(1 - 2\theta) = 0 \), which results in critical points at \( \theta = 0 \) and \( \theta = \frac{1}{2} \).
These points are where the slope of the function's graph is zero, meaning the graph has a horizontal tangent line at these locations.

Increasing and Decreasing Intervals

A function is increasing where its derivative is positive and decreasing where its derivative is negative. By examining the sign of \( f'(\theta) = 6\theta - 12\theta^2 \) in different intervals, we can determine where the function is increasing or decreasing.
Using test points between and around critical points:

• For \( \theta < 0 \), such as \( \theta = -1 \), \( f'(\theta) = -18 \). The function is decreasing.
• For \( 0 < \theta < \frac{1}{2} \), such as \( \theta = \frac{1}{4} \), \( f'(\theta) = \frac{3}{4} \). The function is increasing.
• For \( \theta > \frac{1}{2} \), such as \( \theta = 1 \), \( f'(\theta) = -6 \). The function is decreasing.

The intervals of increase and decrease are therefore:

• Decreasing on \( (-\infty, 0) \) and \( \left(\frac{1}{2}, \infty\right) \)
• Increasing on \( (0, \frac{1}{2}) \)

Local Extremes

Local extremes, or extremum points, are points where a function reaches a local maximum or a local minimum. They occur at critical points where the derivative changes sign. Evaluating the function at these critical points helps determine the nature of the extremes.
For \( f(\theta) = 3\theta^2 - 4\theta^3 \):

• At \( \theta = 0 \), the derivative changes from negative to positive, indicating a local minimum, with the function value \( f(0) = 0 \).
• At \( \theta = \frac{1}{2} \), the derivative changes from positive to negative, indicating a local maximum, with the function value \( f\left(\frac{1}{2}\right) = \frac{1}{4} \).

These local extremes help in sketching the behavior of the function around these points.

Absolute Extremes

Absolute extremes are the largest or smallest values of a function over its entire domain. They include absolute maximums and minimums. By comparing local extremes within the context of the entire function, we can identify any absolute extreme values.
For this function, \( f(\theta) = 3\theta^2 - 4\theta^3 \), the behavior at the ends of the domain is considered. As \( \theta \to \pm\infty \), \( -4\theta^3 \) dominates, making \( f(\theta) \to -\infty \).

Hence, the local maximum at \( \theta = \frac{1}{2} \) with a value of \( \frac{1}{4} \) is an absolute maximum, as no higher value exists on the entire real line. There is no absolute minimum since the function decreases without bound in both directions.