Problem 75 - 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. $$\begin{array}{l}{\text { a. Plot the function over the interval to see its general behavior there. }} \\ {\text { b. Find the interior points where } f^{\prime}=0 . \text { (In some exercises, }} \\ {\text { you may have to use the numerical equation solver to approximate }} \\ {\text { a solution.) You may want to plot } f^{\prime} \text { as well. }}\\\\{\text { c. Find the interior points where } f^{\prime} \text { does not exist. }} \\ {\text { d. Evaluate the function at all points found in parts (b) and (c) }} \\ {\text { and at the endpoints of the interval. }} \\ {\text { e. Find the function's absolute extreme values on the interval }} \\ {\text { and identify where they occur. }}\end{array}$$ $$ f(x)=x^{4}-8 x^{2}+4 x+2, \quad[-20 / 25,64 / 25] $$

Step-by-Step Solution

Answer

The absolute extrema occur at the endpoints and critical points within the interval. Evaluate \( f(x) \) at these points to find the maxima and minima.

Step 1: Plot the Function

Use a computer algebra system (CAS) to plot the function \( f(x) = x^4 - 8x^2 + 4x + 2 \) over the interval \([-\frac{20}{25}, \frac{64}{25}]\). This gives a visual understanding of how the function behaves within the interval. Look for any notable peaks, valleys, and intercepts.

Step 2: Find Critical Points where f' = 0

Differentiate the function to get \( f'(x) = 4x^3 - 16x + 4 \). Set \( f'(x) = 0 \) to find critical points: 1. Solve \( 4x^3 - 16x + 4 = 0 \) using a CAS.2. Get approximate roots, which might need numerical methods like Newton's method if exact solutions are hard to find.

Step 3: Identify Points where f' Does Not Exist

Check if \( f'(x) = 4x^3 - 16x + 4 \) is undefined within the interval. Since this is a polynomial, it is always defined, so there are no points where \( f' \) does not exist.

Step 4: Evaluate the Function at Critical and Endpoint Values

Evaluate \( f(x) \) at the critical points found in Step 2 and at the endpoints of the interval \([-\frac{20}{25}, \frac{64}{25}]\). This includes:- Any interior critical points \(c\) where \( f'(c) = 0 \).- Endpoint values at \( x = -\frac{20}{25} \) and \( x = \frac{64}{25} \).

Step 5: Determine Absolute Extrema

Compare the function values obtained in Step 4 to determine the absolute maximum and minimum. Identify the corresponding \(x\)-values where these extrema occur on the interval.

Key Concepts

Absolute Extrema

In calculus, understanding absolute extrema is crucial when analyzing the behavior of functions. Absolute extrema refer to the highest (maximum) and lowest (minimum) points that a function reaches over a specified interval. Finding these points is essential as they indicate the most significant peaks and valleys in the graph of a function.

To find the absolute extrema of a function on a closed interval, you typically need to evaluate the function at several points:

• Critical points within the interval, where the derivative is zero or undefined.
• The endpoints of the interval itself.

By comparing the functional values at these points, you can determine which are the absolute maximum and minimum. These steps are part of a broader strategy to understand how a function behaves and identify any major changes within the interval.

Absolute extrema are particularly useful in optimization problems where you need to find the most efficient solution or output. By mastering the identification of absolute extrema, one gains a valuable tool for analyzing real-world scenarios in fields ranging from economics to physics.

Critical Points

Critical points are fundamental in calculus as they help in identifying potential maxima, minima, or points of inflection on a function's graph. A critical point occurs where the derivative of a function is zero or does not exist.

For the function given in the exercise, we can find the derivative, which can hint at these significant points. Once you have the derivative, setting it equal to zero helps find those points:

• The mathematical form is given by finding solutions to the equation \( f'(x) = 0 \).
• The points where the derivative is undefined are also critical points. However, for polynomial functions, the derivative is always defined within real numbers.

After determining the critical points, evaluating the function at these points helps in assessing their significance concerning the graph's shape. Critical points don't always correspond to a maximum or minimum, which is why analyzing the function's behavior around these points is necessary to determine whether they indeed represent turning points or not.

Polynomial Function

Polynomial functions play an essential role in calculus due to their well-defined properties and relatively straightforward behavior. These functions are composed of terms involving non-negative integer powers of the variable, and they can be written in standard form, such as \( f(x) = ax^n + bx^{n-1} + \.\.\,. < \). With respect to our specific function \( f(x) = x^4 - 8x^2 + 4x + 2 \), understanding polynomials allows us to easily differentiate and find roots.

Here are key points about polynomial functions:

• They are continuous and smooth throughout their domain, which makes them easy to differentiate and integrate.
• The degree of the polynomial dictates its overall shape and the number of turning points it may have.
• The end behavior of the function is determined by the leading term (the term with the highest power).

Understanding these characteristics simplifies the task of analyzing these functions and finding solutions to related calculus problems, like finding critical points or evaluating limits.

Numerical Methods

Numerical methods are a broad category of algorithms used to solve mathematical problems that might not have an exact analytical solution. In calculus, these methods are invaluable when dealing with complex equations or when approximations provide quicker insights.

For instance, while solving for critical points in polynomial functions like the one in our exercise, exact analytical solutions can be challenging or impossible to derive. Here's where numerical methods, such as Newton's method or the bisection method, come into play:

• These techniques provide approximations to roots or solutions by iterative refinement, often implemented within computer algebra systems (CAS).
• They enable calculation of solutions to a desired degree of accuracy, making them very practical for complex or real-world problems.
• Despite the intensive calculations involved, these methods are efficient when implemented computationally.

Thus, understanding and applying numerical methods are crucial skills in calculus, enhancing the analytical toolkit available for tackling a wide variety of mathematical challenges.