Problem 81 - Thomas Calculus in SI Units

Question

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

Plot the function, find critical points via derivative, and evaluate function at critical points and endpoints to find extrema.

Step 1: Plot the Function

Start by plotting the function \( f(x) = x^4 - 8x^2 + 4x + 2 \) over the interval \([-\frac{20}{25}, \frac{64}{25}]\). Use a computer algebra system (CAS) to generate the plot. This will help us visualize where the function might have peaks and troughs (local extrema) within the specified interval.

Step 2: Find Interior Points Where Derivative is Zero

To find where \( f'(x) = 0 \), first compute the derivative: \[ f'(x) = 4x^3 - 16x + 4. \] Use the CAS to solve the equation \( 4x^3 - 16x + 4 = 0 \) within the interval. This may yield multiple solutions, corresponding to potential local minima or maxima.

Step 3: Find Interior Points Where Derivative Does Not Exist

Check if there are any points in the interval where the derivative \( f'(x) \) does not exist. For polynomial functions like this, \( f'(x) \) exists everywhere, so there are no such points in this case.

Step 4: Evaluate the Function at Critical Points and Endpoints

Evaluate \( f(x) \) at each of the critical points found in Step 2 and at the endpoints \( x = -\frac{20}{25} \) and \( x = \frac{64}{25} \). This will help determine the function's value at these significant points.

Step 5: Determine Absolute Extrema

Compare the values from Step 4 to find the absolute maximum and minimum over the given interval. The highest value is the absolute maximum, and the lowest is the absolute minimum. Identify where these extreme values occur in terms of \(x\).

Key Concepts

Critical Points

Critical points are essential when analyzing the extrema of a function. These are the points in the domain of the function where the derivative is zero or does not exist. Identifying these points helps determine possible locations of local minima and maxima.

Critical points occur when the slope of the tangent to the function is zero, indicating that the function is no longer increasing or decreasing at that point. Mathematically, we find these points by setting the derivative of the function equal to zero:

• Calculate the derivative of the function, such as in the problem where the derivative is given by \( f'(x) = 4x^3 - 16x + 4 \).


• Solve the equation \( 4x^3 - 16x + 4 = 0 \) to find the values of \( x \) where the slope is zero.

Knowing where the derivative does not exist is also crucial, but as often in polynomial functions like this one, it exists everywhere within the specified domain.

Derivative

The derivative of a function is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point and is key to finding critical points. The process of differentiation helps us identify both potential extrema and intervals of increase or decrease.

Calculating the derivative involves applying rules and formulas to the given function. For example, given the function \( f(x) = x^4 - 8x^2 + 4x + 2 \), you compute the derivative \( f'(x) \) as follows:

• Apply the power rule: \( d/dx (x^n) = nx^{n-1} \) to each term of the polynomial.


• The derivative becomes \( f'(x) = 4x^3 - 16x + 4 \).

This derivative function helps us find points where the slope is zero (where local extrema might occur) and guide us in understanding the behavior of the original function on the interval.

Interval Evaluation

Evaluating a function over a specific interval is essential for identifying the behavior and extrema of that function over a limited domain. Once we identify critical points where the function changes direction or reaches extremum values, evaluation becomes imperative.

To evaluate a function effectively:

• Calculate the function's value at all critical points found within the interval.


• Evaluate the function at the interval’s endpoints to ensure all possible extrema are considered.

For example, in the original problem, we evaluate \( f(x) \) at each critical point within \([-\frac{20}{25}, \frac{64}{25}]\) and at the endpoints themselves. This data is crucial for determining the absolute extrema.

Absolute Maximum and Minimum

The absolute maximum and minimum refer to the highest and lowest values of a function within a given interval. Finding these values is a multi-step process that involves solving derivatives and evaluating the function at several significant points.

Steps to find absolute extrema:

• Using the derivative, identify critical points where the derivative is zero.


• Evaluate the function at these critical points and the interval endpoints.


• Compare all values: The highest value corresponds to the absolute maximum, and the lowest corresponds to the absolute minimum.

These calculations tell us more than just where the function peaks but show where it indeed reaches its most extreme values within the interval. In our problem, this process is done over \([-\frac{20}{25}, \frac{64}{25}]\) and helps capture the essence of the function's behavior in that closed interval.