Problem 76 - 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. $$\begin{array}{l}{\text { a. Plot the function over the interval to see its general behavior there. }} \\ {\text { b. Find the interior points where } f^{\prime}=0 . \text { (In some exercises, }} \\ {\text { you may have to use the numerical equation solver to approximate }} \\ {\text { a solution.) You may want to plot } f^{\prime} \text { as well. }}\\\\{\text { c. Find the interior points where } f^{\prime} \text { does not exist. }} \\ {\text { d. Evaluate the function at all points found in parts (b) and (c) }} \\ {\text { and at the endpoints of the interval. }} \\ {\text { e. Find the function's absolute extreme values on the interval }} \\ {\text { and identify where they occur. }}\end{array}$$ $$ f(x)=-x^{4}+4 x^{3}-4 x+1, \quad[-3 / 4,3] $$

Step-by-Step Solution

Answer

Find critical points and evaluate the function at those points and endpoints to determine extrema.

Step 1: Plotting the Function

Use a graphing tool to plot the function \( f(x) = -x^4 + 4x^3 - 4x + 1 \) over the interval \([-\frac{3}{4}, 3]\). Notice the general shape of the curve, specifically where it appears to reach peaks (maxima) or dips (minima).

Step 2: Finding Critical Points Where f' = 0

First, find the derivative \( f'(x) = -4x^3 + 12x^2 - 4 \). Next, set \( f'(x) = 0 \) and solve for \( x \) to find critical points. This might require use of computational tools to approximate solutions because solving \( -4x^3 + 12x^2 - 4 = 0 \) analytically could be complex.

Step 3: Checking Where f' Does Not Exist

Examine the derivative \( f'(x) = -4x^3 + 12x^2 - 4 \). In this case, the derivative is defined for all real numbers, indicating there are no points in the interval where the derivative does not exist.

Step 4: Evaluating the Function at Critical Points and Endpoints

Once you have the critical points (from Step 2), evaluate the function \( f(x) \) at these points. Also evaluate \( f(x) \) at the endpoints of the interval, \( x = -\frac{3}{4} \) and \( x = 3 \) to find the function values.

Step 5: Determining Absolute Extrema

Compare the function values obtained from Step 4 to determine which is the largest (absolute maximum) and which is the smallest (absolute minimum) within the specified interval.

Key Concepts

Absolute Extrema

Absolute extrema, also known as global extrema, are the highest or lowest points that a function achieves over a given interval. These points are crucial in understanding the overall behavior of a function. For instance, if you are looking at a business profit function over a period of time, the absolute maximum gives you the peak profit, whereas the absolute minimum might indicate the lowest point of your profits in that interval.

To find absolute extrema for a function on a closed interval, follow these steps:

• Find the critical points of the function in the interval.
• Evaluate the function at these critical points as well as at the endpoints of the interval.
• Compare these values to determine the highest and lowest points, which will be your absolute maximum and minimum respectively.

These steps ensure that you capture all important variations in the function, giving a comprehensive view of its extreme values.

Derivatives

Derivatives are fundamental in calculus as they provide the rate at which a function is changing at any given point. By determining the derivative of a function, you can find its slope and understand its increasing or decreasing behavior.

In this exercise, the derivative of the function, denoted as \( f'(x) \), is essential for finding critical points. The derivative of the given function \( f(x) = -x^4 + 4x^3 - 4x + 1 \) is calculated to be \( f'(x) = -4x^3 + 12x^2 - 4 \). This expression tells us about the instantaneous rate of change of the function at any point \( x \).

Once you have the derivative, set it equal to zero to find the critical points. This is because critical points are where the derivative itself is either zero or undefined, indicating potential maxima or minima in the function.

Critical Points

Critical points of a function are specific points where the derivative \( f'(x) \) is zero or does not exist. These points are where a function might change its increasing or decreasing behavior, making them potential locations for maximum or minimum values. Finding critical points is a crucial step in locating a function's extrema.

For the function \( f(x) = -x^4 + 4x^3 - 4x + 1 \), we find the derivative \( f'(x) = -4x^3 + 12x^2 - 4 \) and set it equal to zero. Solving the equation \( -4x^3 + 12x^2 - 4 = 0 \) will give us the critical points. In this exercise, computational tools can help find the solutions since the equation can be complicated to solve analytically.

Remember, not all critical points will be extrema. After finding them, evaluate the original function at these points, as well as at the endpoints of the interval, to determine their nature.

Numerical Equation Solver

A numerical equation solver is a powerful tool for finding approximate solutions to equations that are difficult or impossible to solve analytically. When it comes to complex polynomial equations, like \( -4x^3 + 12x^2 - 4 = 0 \), numerical solvers are especially useful.

These tools apply algorithms that iteratively search for roots, or solutions, of the equation. They provide results with a high degree of accuracy, although initially working with approximations. Typically, equation solvers are built into computer algebra systems (CAS) and can handle a variety of functions and equations.

Such numerical methods are indispensable when dealing with real-world applications of calculus, as they allow you to derive insights from functions even when analytical solutions are difficult to attain.