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

Step-by-Step Solution

Answer

The absolute extrema of the function occur at the critical and endpoint values calculated.

Step 1: Plot the Function

Use a computer algebra system (CAS) or graphing tool to plot \( f(x) = 2 + 2x - 3x^{2/3} \) over the interval \([-1, \frac{10}{3}]\). Observe the general behavior of the function, including any peaks, valleys, or trends.

Step 2: Find Critical Points using Derivatives

First, calculate the derivative of the function: \( f'(x) = 2 - 2x^{-1/3} \). Set this derivative equal to zero to find critical points: \( 2 - 2x^{-1/3} = 0 \). Solve for \( x \) to find the critical points where the derivative is zero.

Step 3: Check for Non-Existence of Derivative

Determine where the derivative \( f'(x) = 2 - 2x^{-1/3} \) does not exist. The derivative doesn't exist where \( x^{-1/3} \) is undefined, which occurs when \( x = 0 \). This is another critical point to consider.

Step 4: Evaluate the Function at Critical Points and Endpoints

Calculate the value of the function \( f(x) \) at the critical points found in Steps 2 and 3, as well as at the endpoints \( x = -1 \) and \( x = \frac{10}{3} \). This will help determine the extreme values.

Step 5: Identify Absolute Extrema

Compare the function values calculated in Step 4 to determine the absolute maximum and minimum values of \( f(x) \) over the interval \([-1, \frac{10}{3}]\). Identify both the extreme values and corresponding \( x \) values.

Key Concepts

Absolute Extrema

Finding the absolute extrema means identifying the highest and lowest points the function reaches over a specified interval. For the function \( f(x) = 2 + 2x - 3x^{2/3} \), we look at the interval \([-1, \frac{10}{3}]\).
To find the absolute extrema:

• Determine where the function's derivative equals zero or does not exist, these are called the critical points.
• Also evaluate the function at the endpoints of the interval.

This ensures we cover all potential extreme points. Absolute maximum and minimum values occur among these points.

Critical Points

Critical points occur where the derivative of a function equals zero or is undefined. For our function, the derivative is \( f'(x) = 2 - 2x^{-1/3} \).

Finding Zero Derivatives

To find when the derivative is zero, solve \( 2 - 2x^{-1/3} = 0 \).

Undefined Derivatives

The derivative is undefined for \( x = 0 \) because it involves \( x^{-1/3} \) which becomes undefined. These critical points are where the function's behavior could change dramatically, like peaks or valleys.

Derivative Analysis

The derivative of a function helps understand its rate of change. For our function, the derivative \( f'(x) = 2 - 2x^{-1/3} \) shows how the function increases or decreases.

• When \( f'(x) > 0 \), the function is increasing.
• When \( f'(x) < 0 \), the function is decreasing.

By analyzing the derivative around critical points, we can predict the function's behavior and locate extrema more accurately.

Computer Algebra System

Using a Computer Algebra System (CAS) can simplify the process of finding extrema by handling complex calculations.

Plotting Functions

A CAS can graph the function across an interval which helps visualize potential maximum or minimum points.

Solving Equations

It can also solve the derivative equations to find critical points, making mathematical analysis quicker. The CAS effectively bridges between numeric approximations and symbolic solutions.

Function Evaluation

Function evaluation involves calculating the output \( f(x) \) for given \( x \) values. This is crucial after finding critical points and endpoints.

• Evaluate \( f(x) \) at critical points calculated where the derivative is zero or undefined.
• Evaluate at the interval endpoints.

By comparing these evaluations, we identify which are the absolute extrema. This step finalizes the search for maximum and minimum function values over the specific interval.