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

Step-by-Step Solution

Answer

The absolute maximum is approximately 6.35 at \( x = -2 \), and the absolute minimum is 0 at \( x = 0 \).

Step 1: Plot the Function

Using a graphing tool, plot the function \( f(x) = x^{2/3}(3 - x) \) over the interval \([-2, 2]\). Observe the overall shape and behavior of the graph within this interval. This will help to visually identify any peaks or valleys that may indicate extrema.

Step 2: Find Critical Points with Derivative Equal to Zero

Compute the derivative of the function: \( f'(x) = \frac{2}{3}x^{-1/3}(3-x) - x^{2/3} \). Set \( f'(x) = 0 \) and solve for \( x \). This leads to solving \( \frac{2}{3}x^{-1/3}(3-x) = x^{2/3} \). Simplifying and using a numerical solver gives critical points such as \( x = 0 \).

Step 3: Find Points Where Derivative Does Not Exist

Examine \( f'(x) = \frac{2}{3}x^{-1/3}(3-x) - x^{2/3} \). The term \( x^{-1/3} \) indicates that \( f'(x) \) does not exist at \( x = 0 \). However, since \( x = 0 \) is already found as a critical point, it does not add additional points in this case.

Step 4: Evaluate the Function at Critical and Endpoints

Calculate \( f(x) \) at \( x = 0 \), and the interval endpoints \( x = -2 \) and \( x = 2 \).- \( f(-2) = (-2)^{2/3}(3 - (-2)) = \approx 6.3496 \)- \( f(0) = 0^{2/3}(3 - 0) = 0 \)- \( f(2) = 2^{2/3}(3 - 2) = 2^{2/3} \approx 1.5874 \)

Step 5: Determine Absolute Extrema

Compare the values \( f(-2) \approx 6.3496 \), \( f(0) = 0 \), and \( f(2) \approx 1.5874 \). The largest value occurs at \( x = -2 \), making it the absolute maximum, and the smallest value is \( f(0) = 0 \), the absolute minimum on the interval.

Key Concepts

Critical Points

Critical points are locations on a graph where the behavior of a function changes. They can be places where the function reaches a maximum, minimum, or an inflection point.
The important aspect of finding critical points is to look for where the derivative of the function equals zero or does not exist. At these points, the slope of the tangent line to the graph is horizontal, indicating potential extrema or a different change in direction.

• Set the derivative to zero: This calculates points where there could be peaks or valleys.
• Check where the derivative doesn't exist: Sometimes, the derivative is undefined at a critical point, particularly if the function includes terms like \( x^{n} \) for negative \( n \).

By analyzing these instances, one can predict where a function might exhibit interesting behavior like maxima or minima.

Derivative

The derivative of a function measures the instantaneous rate of change of the function concerning its variable.
Mathematically, the derivative is represented as \( f'(x) \), and it describes how the function \( f(x) \) grows or shrinks at any point in its domain.

• Positive Derivative: Indicates an increasing function.
• Negative Derivative: Points to a decreasing function.
• Zero Derivative: Suggests that the function is flat or has horizontal tangents which might indicate critical points.

For example, to find critical points of \( f(x) = x^{2/3}(3 - x) \), we compute the derivative \( f'(x) = \frac{2}{3}x^{-1/3}(3-x) - x^{2/3} \). Setting this derivative to zero or finding where it doesn’t exist helps identify interesting points on the graph.

Numerical Solver

When dealing with complex functions, it might be challenging to solve for critical points algebraically.
This is where numerical solvers come in handy. These are tools, sometimes graphing software or calculators, that approximate the roots of equations when symbolic solutions are tough to find.
For instance, after deriving the equation \( \frac{2}{3}x^{-1/3}(3-x) = x^{2/3} \), calculating its roots analytically might be complicated. Using a numerical solver introduces a practical way to find approximate solutions for \( x \).

• Convenient: Numerical solvers save time and simplify complex equations.
• Approximations: They provide results close enough for practical purposes, offering insight into where real solutions lie.

Solvers are an indispensable part of modern calculus, especially when working with non-linear or difficult equations.

Absolute Extrema

Absolute extrema of a function are the highest or lowest values that the function attains over a specified interval.
This includes both the endpoints of the interval and any critical points found within it. To determine absolute extrema:

• Evaluate the function at the critical points: This includes points where both the derivative equals zero and where it does not exist.
• Check the endpoints: They can also be sites of extrema, particularly for functions with closed intervals.

For example, when examining \( f(x) = x^{2/3}(3-x) \) over the interval \([-2, 2]\), we evaluate it at critical points and endpoints. The outcomes are then compared to identify the absolute maximum and minimum values.
The absolute maximum is where \( f \text{(-2)} \approx 6.3496 \), and the absolute minimum is where \( f \text{(0)} = 0 \). This approach provides a complete picture of a function’s behavior across a given range.