Problem 77 - 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. $$ f(x)=x^{2 / 3}(3-x), \quad[-2,2] $$

Step-by-Step Solution

Answer

The absolute extremum values occur at the evaluated points from Steps 2-4.

Step 1: Plot the Function

Plot the function \( f(x) = x^{2/3}(3-x) \) over the interval \([-2, 2]\) to understand its behavior. This can be done using a graphing tool or software like a Computer Algebra System (CAS). Observe the general shape and direction of the graph within the provided interval.

Step 2: Find Points Where Derivative is Zero

Calculate the derivative of the function: \( f'(x) = \frac{d}{dx}(x^{2/3}(3-x)) \). Use the product rule and simplify it. Set the derivative equal to zero and solve for \( x \) to find the interior points where \( f'(x) = 0 \). These are potential points for extrema.

Step 3: Find Points Where Derivative Does Not Exist

Identify points where the derivative \( f'(x) \) does not exist. Since the derivative involves \( x^{2/3} \), check for points where the base \( x \) might cause non-existence, specifically where it becomes undefined or results in division by zero. Note these points.

Step 4: Evaluate the Function at Critical Points and Endpoints

Calculate the value of \( f(x) \) at all critical points found in Steps 2 and 3, and also evaluate \( f(x) \) at the endpoints \( x = -2 \) and \( x = 2 \). These evaluations will help determine the extreme values.

Step 5: Determine Absolute Extrema

Compare all the values calculated in Step 4 to identify the largest and smallest function values. These values are the absolute maxima and minima of the function on the interval \([-2, 2]\). Determine where these extrema occur by identifying the corresponding \( x \)-values.

Key Concepts

Closed Interval

A closed interval is a mathematical concept used to define a certain range of values for a variable, often denoted \([a, b]\). In this context, the endpoints are included in the interval; meaning that the value can be equal to either \(a\) or \(b\). For instance, in the exercise above, the closed interval is \([-2, 2]\).
If you were to plot this interval on a number line, both \(x = -2\) and \(x = 2\) would be part of the interval.
This is significant when finding extrema because we examine not only the behavior inside this range but also the values at the boundaries.Understanding closed intervals helps us determine where a function starts and stops, ensuring we find all possible extrema within those limits.

• Endpoints are part of the interval.
• Closed intervals provide a finite range for analysis.
• They help in defining the scope of absolute extrema exploration.

Derivative

The derivative of a function helps us understand how the function behaves—specifically, it shows the rate at which the function value is changing at each point. This is crucial for finding points where the function might have a maximum or minimum, known as the extrema.

In the given exercise, to find the derivative of \(f(x) = x^{2/3}(3-x)\), we apply the product rule, which is used when differentiating products of two functions.

• Calculate derivative using the rules of differentiation.
• Derivative indicates slope or rate of change.
• Helps locate critical points where extrema may occur.

By setting the derivative to zero, \( f'(x) = 0 \), we can find the critical points of the function, which are candidates for local maxima or minima. Understanding how to calculate and interpret derivatives allows us to effectively analyze function behavior.

Critical Points

Critical points are the \(x\)-values in a function where the derivative is either zero or undefined. Finding these points helps identify where potential maxima or minima could occur.
For the exercise function, these points occur at values that potentially alter the function's rate of change. To locate where \(f'(x) = 0\), solve the equation derived from setting the derivative to zero. Additionally, checking where the derivative does not exist is equally important, as these points can also indicate abrupt changes in the function's behavior.

• Critical points arise from \(f'(x) = 0\) or where \(f'(x)\) is undefined.
• They represent potential locations for extrema.
• Understanding how to find these points is essential in extremum analysis.

Numerical Equation Solver

A Numerical Equation Solver is a computational tool used to approximate solutions to equations that might be difficult to solve analytically. This tool is particularly useful when dealing with complex functions or derivatives without straightforward algebraic solutions.
In the exercise, if solving \(f'(x) = 0\) leads to complex or non-standard equations, a numerical equation solver can provide accurate approximations of critical points. Using this method:

• Ensures solutions where traditional methods are inefficient.
• Provides precise values when direct calculation is cumbersome.
• Introduces advanced computational techniques in analysis context.

In a CAS (Computer Algebra System), utilizing numerical solvers enhances your toolkit for tackling hard-to-solve equations and finding critical points where traditional algebra falls short.