Problem 23 - University Calculus: Early Transcendentals

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

Increasing on \((-\infty, 0)\) and \((0, \frac{1}{2})\); decreasing on \((\frac{1}{2}, \infty)\). Local max at \(\theta=\frac{1}{2}\), local min at \(\theta=0\).

Step 1: Identify Derivative

To find where the function is increasing or decreasing, we start by finding its derivative. For the function \(f(\theta) = 3\theta^2 - 4\theta^3\), the first derivative is \(f'(\theta) = 6\theta - 12\theta^2\).

Step 2: Factor and Solve Derivative

The derivative \(f'(\theta) = 6\theta - 12\theta^2\) can be factored as \(6\theta(1 - 2\theta)\). Setting \(6\theta(1 - 2\theta) = 0\) gives critical points at \(\theta = 0\) and \(\theta = \frac{1}{2}\).

Step 3: Test Intervals Around Critical Points

We test intervals around the critical points to determine if \(f(\theta)\) is increasing or decreasing. Choose test points in the intervals: \((-\infty, 0)\), \((0, \frac{1}{2})\), and \((\frac{1}{2}, +\infty)\).- For \(\theta = -1\) in \((-\infty, 0)\), \(f'(-1) = 6(-1)(1 - 2(-1)) = 18\), indicating increasing.- For \(\theta = \frac{1}{4}\) in \((0, \frac{1}{2})\), \(f'(\frac{1}{4}) = 6(\frac{1}{4})(1 - 2(\frac{1}{4})) = \frac{3}{4}\), indicating increasing.- For \(\theta = 1\) in \((\frac{1}{2}, +\infty)\), \(f'(1) = 6(1)(1 - 2) = -6\), indicating decreasing.

Step 4: Determine Increasing and Decreasing Intervals

Based on the test points:- \(f(\theta)\) is increasing on the intervals \((-\infty, 0)\) and \((0, \frac{1}{2})\).- \(f(\theta)\) is decreasing on the interval \((\frac{1}{2}, +\infty)\).

Step 5: Find Function Values at Critical Points

To identify local maxima or minima, evaluate \(f(\theta)\) at previously found critical points:- \(f(0) = 3(0)^2 - 4(0)^3 = 0\).- \(f(\frac{1}{2}) = 3(\frac{1}{2})^2 - 4(\frac{1}{2})^3 = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}\).

Step 6: Identify Local and Absolute Extrema

Compare function values to determine extrema:- Since \(f(0) = 0\) is lower than \(f(\frac{1}{2}) = \frac{1}{4}\), there is a local minimum at \(\theta = 0\).- There is a local maximum at \(\theta = \frac{1}{2}\) where \(f(\theta) = \frac{1}{4}\). - The function decreases towards \(-\infty\) beyond \(\theta = \frac{1}{2}\), so there are no absolute maxima or minima beyond these points.

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that represents the rate at which a function changes at any given point. In simple terms, it tells us how fast or slowly a function is increasing or decreasing. To find the derivative of a function, we apply differentiation rules.
For the given function \(f(\theta) = 3\theta^2 - 4\theta^3\), we use the power rule to find the derivative. The power rule states that the derivative of \(a\theta^n\) is \(na\theta^{n-1}\). Applying this rule, the derivative, \(f'(\theta)\), is \(6\theta - 12\theta^2\). This resulting expression represents the slope of the tangent to the curve at any point \(\theta\).
Finding the derivative is essential in determining where the function is increasing or decreasing, as the sign of the derivative (positive or negative) directly informs us about the behavior of the function.

Critical Points

Critical points are where the derivative of a function equals zero or is undefined. These points are significant because they can indicate locations where the function's slope changes from increasing to decreasing or vice versa, potentially highlighting local maxima or minima.
To find the critical points for our function, we set \(f'(\theta) = 0\) and solve for \(\theta\). The derivative \(6\theta - 12\theta^2\) can be factored as \(6\theta(1 - 2\theta)\). Setting each factor equal to zero gives us critical points at \(\theta = 0\) and \(\theta = \frac{1}{2}\).
These values are potential candidates for local extremes and are crucial in analyzing the function's graph. However, additional testing is required around these points to determine the function's behavior conclusively.

Extrema

Extrema refer to the maximum and minimum values of a function—locations where the function reaches its highest or lowest points, either locally or globally. Recognizing and classifying these values can provide insight into the overall shape and behavior of the function.
For the given function, we evaluate it at the critical points identified earlier: \(f(0)\) and \(f(\frac{1}{2})\). The function values are \(f(0) = 0\) and \(f(\frac{1}{2}) = \frac{1}{4}\).
After comparing these values, we determine there is a local minimum at \(\theta = 0\) and a local maximum at \(\theta = \frac{1}{2}\). Since the function tends to \(-\infty\) as \(\theta\) increases beyond \(\frac{1}{2}\), there are no absolute maximum or minimum values over the entire domain. These findings tell us that \(\theta = \frac{1}{2}\) is where the highest nearby value occurs before the function starts to decrease.

Increasing/Decreasing Intervals

To understand where a function is increasing or decreasing, we test intervals around each critical point. We utilize the first derivative test, where we check the sign of \(f'(\theta)\) in the chosen intervals.
For the polynomial \(6\theta - 12\theta^2\), intervals for testing are \((-\theta, 0)\), \((0, \frac{1}{2})\), and \((\frac{1}{2}, +\infty)\). By substituting values from these intervals into \(f'(\theta)\), we observe:

• In \((-\theta, 0)\), the derivative is positive, indicating increasing behavior.
• For \((0, \frac{1}{2})\), the derivative remains positive, hence the function continues increasing.
• In \((\frac{1}{2}, +\infty)\), the derivative is negative, which shows a decreasing trend.

These tests allow us to conclude on which parts of the domain the function rises or falls, providing a fuller understanding of its graph. Intervals of increase and decrease help to picture the dynamics of the function graphically, like hills and valleys on a curve.