Problem 23 - 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. $$ f(\theta)=3 \theta^{2}-4 \theta^{3} $$

Step-by-Step Solution

Answer

The function is increasing on \((0, \frac{1}{2})\) and decreasing on \((-\infty, 0)\) and \((\frac{1}{2}, \infty)\). Local max at \(\theta = \frac{1}{2}\), local min at \(\theta = 0\).

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first find the derivative. Given the function \( f(\theta) = 3\theta^2 - 4\theta^3 \), let's find the first derivative \( f'(\theta) \). Differentiate term by term:\[f'(\theta) = \frac{d}{d\theta}(3\theta^2) - \frac{d}{d\theta}(4\theta^3)\]\[f'(\theta) = 6\theta - 12\theta^2\]This derivative will help us determine the intervals of increase and decrease.

Step 2: Find Critical Points

Set the first derivative equal to zero to find critical points:\[6\theta - 12\theta^2 = 0\]Factor out the common term:\[6\theta(1 - 2\theta) = 0\]The solutions to this equation are:\[ \theta = 0 \quad \text{and} \quad \theta = \frac{1}{2} \]These \(\theta\)-values are potential locations where the function changes from increasing to decreasing or vice versa.

Step 3: Test Intervals Around Critical Points

Select test points from the intervals determined by these critical points: \((-\infty, 0)\), \((0, \frac{1}{2})\), and \((\frac{1}{2}, \infty)\). Substitute these test \(\theta\)-values into \(f'(\theta)\):- For \(\theta < 0\), choose \(\theta = -1\): \[f'(-1) = 6(-1) - 12(-1)^2 = -6 - 12 = -18 \text{ (negative)}\]- For \(0 < \theta < \frac{1}{2}\), choose \(\theta = \frac{1}{4}\): \[f'(\frac{1}{4}) = 6(\frac{1}{4}) - 12(\frac{1}{4})^2 = \frac{3}{2} - \frac{3}{2} = 0 \text{ or positive, depending on precise values}\]- For \(\theta > \frac{1}{2}\), choose \(\theta = 1\): \[f'(1) = 6(1) - 12(1)^2 = 6 - 12 = -6 \text{ (negative)}\]This tells you the function is decreasing on \((-\infty, 0)\) and \((\frac{1}{2}, \infty)\), and increasing on \((0, \frac{1}{2})\).

Step 4: Analyze Function at Critical Points for Extrema

The critical points \(\theta = 0\) and \(\theta = \frac{1}{2}\) need to be tested in \(f(\theta)\) to determine if they are local extrema:- \(f(0) = 3(0)^2 - 4(0)^3 = 0\)- \(f(\frac{1}{2}) = 3\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right)^3 = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}\)The function changes from increasing at \(\theta = \frac{1}{2}\) to decreasing thereafter, suggesting a local maximum at \(\theta = \frac{1}{2}\). At \(\theta = 0\), the function decreases ahead and increases thereafter, indicating a local minimum.

Key Concepts

Derivative

The derivative of a function is a powerful tool in calculus. It helps us understand how a function changes by giving us the slope of the function at any point. To find the derivative of any function, we differentiate each term individually. For example, with the function \[ f(\theta) = 3\theta^2 - 4\theta^3 \]we differentiate term by term:

• For \(3\theta^2\), the derivative is \(6\theta\).
• For \(-4\theta^3\), the derivative is \(-12\theta^2\).

So, the first derivative of the function is \[ f'(\theta) = 6\theta - 12\theta^2 \].
This result is crucial to solving problems related to functions' behavior, as it tells us about the function's rate of change.

Critical Points

Critical points are values of the variable where the first derivative is zero or undefined. These points are important because they might be where a function changes direction, from increasing to decreasing or vice versa. For our function, we find the critical points by solving \[ 6\theta - 12\theta^2 = 0 \].
By factoring out common terms, we get:\[ 6\theta(1 - 2\theta) = 0 \].
Setting each factor equal to zero gives us the critical points:

• \(\theta = 0\)
• \(\theta = \frac{1}{2}\)

These are the potential spots where the function could have local extreme points.
They are crucial for understanding the overall graph's shape.

Increasing and Decreasing Functions

Functions are considered increasing if, as the input increases, the output also increases; they are decreasing if the output decreases as the input increases. By evaluating the sign of the derivative at test points, we can determine where the function is increasing or decreasing. For the intervals:

• \((-\infty, 0)\): Taking a test point like \(\theta = -1\), we find \(f'(-1) < 0\), indicating the function decreases here.
• \((0, \frac{1}{2})\): A test point, like \(\theta = \frac{1}{4}\), gives \(f'(\frac{1}{4})\), suggesting the function increases on this interval.
• \((\frac{1}{2}, \infty)\): Here, \(f'(1) < 0\), so the function decreases.

This analysis is crucial for sketching the function and understanding its behavior.

Local Extrema

Local extrema refer to the highest or lowest points in a region of a function's graph. They are found by evaluating the function at critical points. For our function, visiting the critical points:

• At \(\theta = 0\): The calculated value is \(f(0) = 0\). The function is decreasing before and increasing after \(\theta = 0\), indicating a local minimum.
• At \(\theta = \frac{1}{2}\): The calculated value is \(f(\frac{1}{2}) = \frac{1}{4}\). The function increases before and decreases after \(\theta = \frac{1}{2}\), indicating a local maximum.

Identifying these local extrema is important for understanding the peaks and valleys in the function's graph, which can provide insights for optimizing real-world problems.