Problem 44 - University Calculus: Early Transcendentals

Question

Find the function's absolute maximum and minimum values and say where they are assumed. $$h(\theta)=3 \theta^{2 / 3}, \quad-27 \leq \theta \leq 8$$

Step-by-Step Solution

Answer

Absolute maximum is 27 at \(\theta = -27\); absolute minimum is 0 at \(\theta = 0\).

Step 1: Define the Interval and Function

We need to find the absolute maximum and minimum values of the function \( h(\theta) = 3\theta^{2/3} \) over the interval \(-27 \leq \theta \leq 8\). The function is continuous in the given interval.

Step 2: Find the Derivative of the Function

Calculate the derivative of the function to find the critical points where the slope of the tangent is zero or undefined. The function is \( h(\theta) = 3\theta^{2/3} \), so its derivative is \( h'(\theta) = 3 \cdot \frac{2}{3}\theta^{-1/3} = 2\theta^{-1/3} \).

Step 3: Critical Points Calculation

The critical points occur where \( h'(\theta) = 0 \) or where \( h'(\theta) \) is undefined. Since \( 2\theta^{-1/3} = 0 \) has no real solution, but \( h'(\theta) \) is undefined at \( \theta = 0 \). So, \( \theta = 0 \) is a critical point.

Step 4: Evaluate the Function at Critical Points and Endpoints

Evaluate the function at the critical point and the endpoints of the interval:- \( h(-27) = 3(-27)^{2/3} = 3(9) = 27 \)- \( h(0) = 3(0)^{2/3} = 0 \)- \( h(8) = 3(8)^{2/3} = 3(4) = 12 \)

Step 5: Determine Absolute Extrema

Compare the values obtained:- At \( \theta = -27 \), \( h(-27) = 27 \)- At \( \theta = 0 \), \( h(0) = 0 \)- At \( \theta = 8 \), \( h(8) = 12 \)The absolute maximum value is \( 27 \) at \( \theta = -27 \), and the absolute minimum value is \( 0 \) at \( \theta = 0 \).

Key Concepts

Absolute Maximum

In calculus, the absolute maximum of a function refers to the highest point over a certain interval. This point gives the function's greatest output value. To find it, we need to assess both the endpoints of the interval and any critical points within it.

In the given exercise, the function is continuous and is evaluated at its endpoints and critical points. For instance, we checked the values at the boundaries

• At \( \theta = -27 \), the function's value is \( 27 \)
• At \( \theta = 8 \), the function gives \( 12 \)

Thus, the absolute maximum value in the interval \( [-27, 8] \) is \( 27 \) at \( \theta = -27 \). Finding the absolute maximum is crucial as it indicates the peak performance or capacity of a function within the specific range.

Absolute Minimum

The absolute minimum is the lowest point that a function reaches over a given interval. It's the smallest value that the function achieves within the specified range. To determine this, evaluate the function at any relevant points—critical points and the endpoints of the interval.

For the function \( h(\theta) = 3\theta^{2/3} \), we calculate:

• At \( \theta = 0 \), the output is \( 0 \)

This evaluation shows that the absolute minimum is \( 0 \), occurring at \( \theta = 0 \). The absolute minimum helps identify the lowest potential value or minimum level the function can reach within the defined interval, crucial for mathematical analysis and applications.

Critical Points

Critical points in a function are where its derivative is zero or undefined. These points are key in identifying potential maximum and minimum values. For a function like \( h(\theta) = 3\theta^{2/3} \), finding critical points involves making its derivative \( h'(\theta) \) equal to zero or identifying where it lacks definition.

The derivative \( h'(\theta) = 2\theta^{-1/3} \) is undefined at \( \theta = 0 \), marking it as a critical point. Determining critical points is a vital step in analyzing the behavior and key features of a function. These points often correspond to important changes in the slope and overall trend of a graph.

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It gives us the function's rate of change at any given point. Calculating the derivative is essential for finding critical points.

For the exercise, the derivative of \( h(\theta) = 3\theta^{2/3} \) is \( h'(\theta) = 2\theta^{-1/3} \). This expression indicates how the function's output changes with respect to \( \theta \).

The role of the derivative is critical in optimizing and understanding functions—highlighting peaks, troughs, and inflection points. By analyzing the derivative, we gain deeper insights into the dynamic nature of functions.