Problem 78 - Thomas Calculus

Question

In Exercises \(75-80,\) you will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. $$\begin{array}{l}{\text { a. Plot the function over the interval to see its general behavior there. }} \\ {\text { b. Find the interior points where } f^{\prime}=0 . \text { (In some exercises, }} \\ {\text { you may have to use the numerical equation solver to approximate }} \\ {\text { a solution.) You may want to plot } f^{\prime} \text { as well. }}\\\\{\text { c. Find the interior points where } f^{\prime} \text { does not exist. }} \\ {\text { d. Evaluate the function at all points found in parts (b) and (c) }} \\ {\text { and at the endpoints of the interval. }} \\ {\text { e. Find the function's absolute extreme values on the interval }} \\ {\text { and identify where they occur. }}\end{array}$$ $$ f(x)=2+2 x-3 x^{2 / 3}, \quad[-1,10 / 3] $$

Step-by-Step Solution

Answer

Minimum is \( -3 \) at \( x = -1 \); maximum is near the right endpoint.

Step 1: Plot the Function

Plot the function \( f(x) = 2 + 2x - 3x^{2/3} \) over the interval \([-1, \, 10/3]\) using a computer algebra system (CAS). This visualization helps to understand the general behavior and shape of the function over the specified interval.

Step 2: Find Critical Points

To find critical points, compute the derivative \( f'(x) \) and solve \( f'(x) = 0 \). The derivative is found using the power rule: \( f'(x) = 2 - 2x^{-1/3} \). Set this equal to zero and solve for \(x\): \[ 2 = 2x^{-1/3} \]\[ 1 = x^{-1/3} \]\[ x^{1/3} = 1 \]\[ x = 1 \].Plot \( f'(x) \) if necessary to identify further points.

Step 3: Identify Where Derivative Does Not Exist

Determine where \( f'(x) = 2 - 2x^{-1/3} \) does not exist. The derivative is undefined when \( x^{-1/3} \) is undefined, which happens at \( x = 0 \).

Step 4: Evaluate Function at Critical Points and Endpoints

Evaluate \( f(x) \) at the critical points and endpoints of the interval:\1. Endpoint \( x = -1 \): \( f(-1) = 2 + 2(-1) - 3(-1)^{2/3} = -3 \)\2. Critical Point \( x = 1 \): \( f(1) = 2 + 2(1) - 3(1)^{2/3} = 1 \)\3. Derivative Undefined \( x = 0 \): \( f(0) = 2 + 2(0) - 3(0)^{2/3} = 2 \)\4. Endpoint \( x = \frac{10}{3} \): Simplify \( \left(\frac{10}{3}\right)^{2/3} \) and evaluate \[ f\left(\frac{10}{3}\right) = 2 + 2\left(\frac{10}{3}\right) - 3\left(\frac{10}{3}\right)^{2/3}\].\Use a CAS to approximate this value.

Step 5: Determine Absolute Extrema

Compare the function values from Step 4: \(-3, 1, 2,\) and the value at \( x=\frac{10}{3} \) approximated using a CAS. The minimum is \(-3\) at \( x = -1 \) and the maximum is the value found at \( x = \frac{10}{3} \) using a CAS.

Key Concepts

Understanding Absolute Extrema

Absolute extrema refer to the highest and lowest values a function can reach over a given interval. These points, known as the absolute maximum and minimum, help understand the boundary behavior of a function.
When determining absolute extrema, it’s important to evaluate not only at the endpoints of the interval but also at any point within where the derivative is zero or undefined. This ensures that all possible extreme values are considered.
To find these points:

• Evaluate the function at the endpoints of the interval.
• Compute and analyze critical points and points where the derivative may not exist.
• Compare function values at all these locations to determine the extrema.

By comparing the function's values at these points, we can determine which are the highest and lowest within the specified domain.

Exploring Critical Points

Critical points are where the first derivative of a function equals zero or is undefined. These points are crucial because they can indicate where a function changes direction, potentially leading to local maxima or minima.
To find critical points:

• First, compute the derivative of the function using rules like the power rule.
• Solve the equation where the derivative equals zero.
• Check where the derivative might not exist - often where there is division by zero or a zero exponent.

For the function in question, we found that the derivative, when set to zero, solved to a critical point at \( x = 1 \). Additionally, we checked where it doesn’t exist, pinpointing \( x = 0 \). Identifying these points allows more analysis of the function’s behavior.

The Importance of Derivatives

Derivatives are foundational in calculus. They provide the rate of change of a function at any point and are essential for finding critical points. Their calculation can reveal a lot about the behavior of a function, such as rates of change, slopes, and tendencies to increase or decrease.
To compute the derivative:

• Use rules like the power rule to differentiate terms.
• For example, the function \( f(x) = 2 + 2x - 3x^{2/3} \) differentiates to \( f'(x) = 2 - 2x^{-1/3} \).
• The derivative equation helps analyze where the function y-values potentially reach their highest or lowest.

Learning to find and interpret derivatives is critical for deeper insights into mathematical and real-world problems.

Explaining Closed Intervals

A closed interval in mathematics is designated by brackets \([a, b]\) and includes its endpoints. This means when analyzing a function, you consider the behavior at both endpoints in addition to any critical points within.
For practical applications:

• Evaluate the function at both endpoints.
• Assess any interior points such as critical points or where the derivative doesn’t exist.

In the given problem, the interval is \([-1, \frac{10}{3}]\), so we assess the function at \( x = -1 \) and \( x = \frac{10}{3} \), alongside the interior critical points \( x = 0 \) and \( x = 1 \). Applying this method ensures a comprehensive evaluation of the function over its range.

Utilizing Computer Algebra Systems

A Computer Algebra System (CAS) is a software tool designed to perform symbolic mathematics. It can handle complex calculations, derivatives, and solve equations precisely, assisting in tasks like plotting functions or finding extrema.
CAS tools offer several advantages:

• Plot functions to get a visual understanding of their behavior over specific intervals.
• Compute derivatives easily and solve them symbolically or numerically.
• Numerically approximate solutions that can’t be solved exactly.

Using a CAS empowers users to engage more deeply with calculus problems by providing computational accuracy and visual insights that can enhance understanding.