Problem 19 - 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. $$ H(t)=\frac{3}{2} t^{4}-t^{6} $$

Step-by-Step Solution

Answer

Increasing: \((-\infty, -1)\) and \((1, \infty)\); Decreasing: \((-1, 0)\) and \((0, 1)\). Local max at \(t = -1\), local min at \(t = 1\). Both are absolute.

Step 1: Differentiate the Function

To determine where the function \( H(t) = \frac{3}{2} t^{4} - t^{6} \) is increasing or decreasing, we first find its derivative. This helps us identify critical points and intervals.The derivative is: \[ H'(t) = \frac{d}{dt}\left(\frac{3}{2} t^{4} - t^{6}\right) = 6t^3 - 6t^5 \] Simplified further, this becomes: \[ H'(t) = 6t^3(1 - t^2) \]

Step 2: Find Critical Points

Critical points occur when the derivative \( H'(t) \) is zero or undefined. Set the derivative equal to zero to find critical points: \[ 6t^3(1 - t^2) = 0 \] Solving gives: \[ t^3 = 0 \quad \Rightarrow \quad t = 0 \] \[ 1 - t^2 = 0 \quad \Rightarrow \quad t^2 = 1 \quad \Rightarrow \quad t = \pm 1 \] Therefore, the critical points are \( t = -1, 0, 1 \).

Step 3: Determine Increasing/Decreasing Intervals

Using the critical points, we analyze the sign of \( H'(t) = 6t^3(1-t^2) \) in intervals divided by \( t = -1, 0, 1 \):- **\((-\infty, -1)\):** \( H'(t) > 0 \), function increasing.- **\((-1, 0)\):** \( H'(t) < 0 \), function decreasing.- **\((0, 1)\):** \( H'(t) < 0 \), function decreasing.- **\((1, \infty)\):** \( H'(t) > 0 \), function increasing.

Step 4: Identify Local Extreme Values

Based on the intervals where \( H'(t) \) changes sign:- At \( t = -1 \), \( H'(t) \) changes from positive to negative, indicating a local maximum.- At \( t = 0 \), \( H'(t) \) changes sign from negative to negative, indicating no local extreme.- At \( t = 1 \), \( H'(t) \) changes from negative to positive, indicating a local minimum.

Step 5: Determine Absolute Extreme Values

Calculate \( H(t) \) at critical points:- \( H(-1) = \frac{3}{2}(-1)^4 - (-1)^6 = \frac{3}{2} - 1 = \frac{1}{2} \)- \( H(0) = \frac{3}{2}(0)^4 - (0)^6 = 0 \)- \( H(1) = \frac{3}{2}(1)^4 - (1)^6 = \frac{3}{2} - 1 = \frac{1}{2} \)Since as \( t \rightarrow \pm \infty\), \( H(t) \rightarrow -\infty \), the local minimum at \( t = 1 \) and local maximum at \( t = -1 \) are also absolute extrema.

Step 6: Graph the Function for Verification

Using a graphing calculator or computer software, plot the function \( H(t) = \frac{3}{2} t^{4} - t^{6} \).Visually confirm the intervals where the function is increasing or decreasing and the location of extrema:- Local maximum at \( t = -1 \) with value \( \frac{1}{2} \)- Local and absolute minimum at \( t = 1 \) with value \( \frac{1}{2} \)- The graph supports the analytical findings.

Key Concepts

Derivatives

Derivatives represent the rate of change of a function with respect to its variable. In calculus, differentiating a function helps us understand its behavior, such as when it rises and falls.

In this exercise, the function given is \( H(t) = \frac{3}{2} t^{4} - t^{6} \). To differentiate this function with respect to \( t \), we apply the power rule for each term. The power rule states that if you have \( t^n \), its derivative is \( nt^{n-1} \).

The derivative of our function is calculated as follows:

• For the term \( \frac{3}{2} t^4 \), the derivative is \( 4 \cdot \frac{3}{2} t^{3} = 6t^3 \).
• For the term \( t^6 \), the derivative is \( 6t^5 \).

Subtracting these results gives \( H'(t) = 6t^3 - 6t^5 \), which simplifies to \( 6t^3(1 - t^2) \). This derivative expression is crucial for finding critical points and understanding the function's behavior.

Critical Points

Critical points are specific values of \( t \) where the derivative of a function either equals zero or is undefined. Finding these points is essential because they often indicate where the function changes direction.

In this exercise, we set the derivative \( H'(t) = 6t^3(1 - t^2) \) equal to zero to find the critical points. We factor the derivative and solve for \( t \):

• \( 6t^3 = 0 \) gives \( t = 0 \).
• \( 1 - t^2 = 0 \) gives \( t = \pm 1 \).

This results in critical points at \( t = -1, 0, 1 \). These are the values where the function might have local maxima or minima.

Increasing and Decreasing Intervals

By using critical points, we can identify where a function is increasing or decreasing. We do this by examining the sign of the derivative \( H'(t) \) in the intervals formed between the critical points.

The critical points divide the number line into intervals: \((- fty, -1), (-1, 0), (0, 1), and (1, \infty)\). Let's explore each interval:

• In \((-fty, -1)\): The derivative \( H'(t) > 0 \), so the function is increasing.
• In \((-1, 0)\): The derivative \( H'(t) < 0 \), making the function decreasing.
• In \((0, 1)\): Again, \( H'(t) < 0 \), and the function decreases.
• In \((1, \infty)\): The derivative \( H'(t) > 0 \), so the function is increasing.

By understanding these increasing and decreasing behaviors, you can identify trends in the function which aid in the discovery of local or absolute extrema.

Local and Absolute Extrema

Local extrema are points where the function reaches a maximum or minimum relative to its immediate vicinity. Absolute extrema, on the other hand, are the highest or lowest points across the entire domain of the function.

In this problem, we determine these extrema by observing where the sign of the derivative changes at critical points:

• At \( t = -1 \), \( H'(t) \) transitions from positive to negative, indicating a local maximum.
• At \( t = 1 \), \( H'(t) \) switches from negative to positive, resulting in a local minimum.
• At \( t = 0 \), \( H'(t) \) remains negative, thus indicating no local extrema.

Thus, there is a local maximum at \( t = -1 \) and a local minimum at \( t = 1 \), both valued at \( \frac{1}{2} \). Given the nature of the function as it approaches infinity, the local minimum at \( t = 1 \) also qualifies as an absolute minimum across the function's entire range.