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. 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. $$ f(x)=x^{4}-8 x^{2}+4 x+2, \quad[-20 / 25,64 / 25] $$

Step-by-Step Solution

Answer

Evaluate the function at critical points and endpoints, compare, and identify absolute extrema.

Step 1: Plot the Function

To understand the function's behavior, start by plotting the function \(f(x) = x^4 - 8x^2 + 4x + 2\) over the interval \([-\frac{20}{25}, \frac{64}{25}]\). Use a graphing software or a CAS (Computer Algebra System) to visualize it. The plot will help identify the general shape and behavior, and where the extrema might occur.

Step 2: Find the Critical Points where f' = 0

Compute the derivative of the function, \(f'(x) = 4x^3 - 16x + 4\). Use a CAS to solve \(f'(x) = 0\) within the interval. This step provides critical points where the slope of the tangent is zero. Finding exact solutions might require numerical approximation techniques via the CAS.

Step 3: Find Points where f' Does Not Exist

For polynomials like \(f(x) = x^4 - 8x^2 + 4x + 2\), check the derivative \(f'(x) = 4x^3 - 16x + 4\). Polynomial derivatives exist everywhere, so there are no points within the given interval where \(f'(x)\) does not exist.

Step 4: Evaluate f at Critical and Endpoints

Evaluate \(f(x)\) at the critical values found in step 2, and at the endpoints of the interval \([-\frac{20}{25}, \frac{64}{25}]\). This means you compute \(f(x)\) at these values and compare results: calculate \(f(-\frac{20}{25})\), \(f(\frac{64}{25})\), and at any critical points found in step 2.

Step 5: Determine Absolute Extrema

Compare all evaluations from step 4. The smallest value will be the absolute minimum, and the largest will be the absolute maximum within the interval. Note the \(x\)-values at which these occur to identify the locations of the extrema.

Key Concepts

Critical Points

Critical points in calculus are points on the graph of a function where the derivative is either zero or does not exist. These points are essential because they can indicate where a function's local maximums or minimums occur. When examining the function \( f(x) = x^4 - 8x^2 + 4x + 2 \), you first take its derivative, which is \( f'(x) = 4x^3 - 16x + 4 \). By setting this derivative equal to zero, \( f'(x) = 0 \), you solve for \( x \) to find potential critical points.

In this case, the critical points are those values of \( x \) within the specified interval where the derivative zeroes out. These points can be found using a computer algebra system (CAS) or graphing software to handle the complex calculations involved. In polynomials such as the one here, the derivative exists everywhere within the real numbers, so there are no \( f'(x) \) values that do not exist.

Absolute Extrema

Absolute extrema refer to the function's lowest and highest values over a certain interval. To find the absolute extrema of \( f(x) = x^4 - 8x^2 + 4x + 2 \) over the interval \([-\frac{20}{25}, \frac{64}{25}]\), you'll need to evaluate the function at critical points within the interval, as well as at the interval's endpoints.

This involves computing \( f(x) \) at each location identified in Step 4 of the solution process. Once you have these values, compare them to identify the smallest (absolute minimum) and largest (absolute maximum) within the interval. These results will tell you where \( f(x) \) achieves its extreme values in both directions.

Polynomial Derivatives

Polynomial derivatives are simpler than derivatives of other more complex functions because polynomials are smooth and continuous everywhere. The main task is to apply the power rule: for a term \( ax^n \), the derivative is \( nax^{n-1} \). For this function, \( f(x) = x^4 - 8x^2 + 4x + 2 \), using the power rule gives \( f'(x) = 4x^3 - 16x + 4 \).

This initial step provides the derivative assessment necessary for finding critical points. Since polynomial functions are smooth, their derivatives exist everywhere. Therefore, when determining where the derivative does not exist, there would be no such values for a polynomial. This property simplifies some steps in solving for extrema.

Interval Evaluation

Interval evaluation in calculus is the process of testing points over a specific range to ascertain function behavior. For the function \( f(x) \) given, you need to consider both endpoints of the interval \([-\frac{20}{25}, \frac{64}{25}]\) and any critical points found.

Evaluating a function requires plugging these \( x \) values into the original \( f(x) \) equation to calculate corresponding \( y \) values. This step is essential for finding absolute extrema because only by evaluating \( f(x) \) at critical and endpoint values can the least and greatest values over a defined interval be determined. Interval evaluation thus ensures you do not miss any extrema when it naturally occurs at the endpoints.