Problem 23 - Thomas Calculus in SI Units

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 \((-\infty, \frac{1}{2})\) and decreasing on \((\frac{1}{2}, \infty)\); local maximum of \(\frac{11}{16}\) occurs at \(\theta = \frac{1}{2}\).

Step 1: Find the First Derivative

To determine the intervals of increase or decrease, first find the derivative of the function. Given \( f(\theta) = 3\theta^2 - 4\theta^3 \), the first derivative is \( f'(\theta) = 6\theta - 12\theta^2 \).

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. Set \( f'(\theta) = 6\theta - 12\theta^2 = 0 \). Factoring gives \( 6\theta(1 - 2\theta) = 0 \), so \( \theta = 0 \) or \( \theta = \frac{1}{2} \).

Step 3: Test Intervals Around Critical Points

Use test points in the intervals \((-\infty, 0)\), \((0, \frac{1}{2})\), and \((\frac{1}{2}, \infty)\) to determine sign changes in \( f'(\theta) \).- For \( \theta = -1 \), \( f'(-1) = 18 \) (positive).- For \( \theta = \frac{1}{4} \), \( f'(\frac{1}{4}) = 3 \) (positive).- For \( \theta = 1 \), \( f'(1) = -6 \) (negative).The function is increasing on \((-\infty, \frac{1}{2})\) and decreasing on \((\frac{1}{2}, \infty)\).

Step 4: Determine Local Extreme Values

Local extrema occur at critical points where the derivative changes sign.- At \( \theta = 0 \), the derivative changes from positive to positive with no sign change (no local extremum).- At \( \theta = \frac{1}{2} \), the derivative changes from positive to negative, so there is a local maximum.

Step 5: Evaluate Function at Critical Points

Calculate \( f(\theta) \) at the critical points to find any local extreme values. - \( f(0) = 0 \).- \( f\left(\frac{1}{2}\right) = \frac{3}{4}\theta^2 - \frac{1}{2}\theta^3 = \frac{3}{4} - \frac{1}{2}\left(\frac{1}{2}\right)^3 = \frac{3}{4} - \frac{1}{16} = \frac{11}{16} \).The local maximum value is \( \frac{11}{16} \) at \( \theta = \frac{1}{2} \).

Step 6: Determine Absolute Extreme Values

Since the function tends to \(-\infty\) as \(\theta\) goes to \(\pm \infty\), there are no absolute maximas or minimas except the local behavior. So, evaluate the boundary behavior if needed.

Key Concepts

First Derivative Test

The first derivative test is a vital tool in calculus used to determine where a function is increasing or decreasing. This process involves finding the first derivative of the function, which symbolizes the rate of change. For the function \(f(\theta) = 3\theta^2 - 4\theta^3\), its first derivative is \(f'(\theta) = 6\theta - 12\theta^2\). Once we have the first derivative, we look for points where it equals zero or does not exist. These are potential critical points and can yield information about the function's behavior.

• Identify intervals by setting the derivative equal to zero, yielding critical points.
• Use the sign of the derivative to determine whether the function is increasing or decreasing.
• Look at sign changes between intervals to understand how the function behaves.

Understanding the first derivative aids in finding critical points and assists in applying the first derivative test.

Critical Points

Critical points are essential in analyzing the behavior of a function. They occur where the derivative is zero or undefined. For \(f(\theta) = 3\theta^2 - 4\theta^3\), finding its critical points involves solving the equation \(6\theta - 12\theta^2 = 0\). Factoring this gives us \(\theta = 0\) and \(\theta = \frac{1}{2}\). These values are the critical points where the function changes its direction of increase or decrease.

• Critical points indicate potential local maxima or minima.
• They help segment the function into intervals for further testing.
• Checking the derivative sign change at these points reveals possible local extrema.

Determining these points is crucial for examining the function's nature and locating any local extreme values.

Increasing and Decreasing Functions

Functions can either increase or decrease within a given interval. To identify these intervals, calculate the first derivative and analyze its sign. For our function \(f(\theta) = 3\theta^2 - 4\theta^3\), the derivative is \(f'(\theta) = 6\theta - 12\theta^2\). Testing points in the intervals determined by the critical points helps decide which intervals are increasing or decreasing.

• In the interval \((-\infty, 0)\), using a test value shows the function is increasing.
• From \((0, \frac{1}{2})\), the function is still increasing.
• In \((\frac{1}{2}, \infty)\), the function starts decreasing.

By pinpointing these intervals, students can graph the function's behavior accurately, showing where transitions occur between increasing and decreasing trends.

Local Extrema

Local extrema refer to the local maximum and minimum points of a function on an interval. We explore these using the critical points tabulated earlier. In our function, \(\theta = 0\) did not produce a change in sign, which indicates no local extremum at that point. However, at \(\theta = \frac{1}{2}\), the derivative shifts from positive to negative, denoting a local maximum.

• Local maxima and minima are found at critical points with derivative sign changes.
• For \(\theta = \frac{1}{2}\), \(f(\theta)\) reaches a local maximum of \(\frac{11}{16}\).
• No local minima occur because the corresponding critical points lacked an appropriate sign change.

Identifying local extrema is pivotal for understanding a function's peak and valley points, providing insights into the function's periodic or fluctuating behavior on different intervals.