Absolute Extrema Calculation For F(x) = -2x³ + 33x² - 144x On [2, 9] 🚀

Jul 10, 2025 by ADMIN 72 views

Finding Absolute Extreme Values of a Polynomial Function

This comprehensive article delves into the process of finding the absolute extreme values of the polynomial function f(x) = -2x³ + 33x² - 144x within the closed interval [2, 9]. We will meticulously explore the analytical method to determine these extrema, and subsequently, we will leverage a graphing utility to visually confirm our results. This exploration will not only solidify your understanding of calculus concepts but also equip you with the practical skills to tackle similar problems. Understanding absolute extreme values is crucial in various fields, from optimization problems in engineering to economic modeling, and this article aims to provide a solid foundation in this area. The problem at hand involves a cubic polynomial, which is a smooth and continuous function, making it amenable to the techniques of differential calculus. We will employ the concept of critical points, which are points where the derivative of the function is either zero or undefined. These critical points, along with the endpoints of the interval, are the candidates for the absolute maxima and minima of the function within the given interval. The strategy we will adopt is as follows: first, we will find the derivative of the function; second, we will determine the critical points by setting the derivative equal to zero and solving for x; third, we will evaluate the function at the critical points and the endpoints of the interval; and finally, we will identify the absolute maximum and minimum values by comparing the function values obtained in the previous step.

a. Determining Absolute Extreme Values

Step 1: Find the Derivative

To begin, we need to find the derivative of the function f(x) = -2x³ + 33x² - 144x. This is a fundamental step in identifying critical points, which are potential locations for absolute extrema. The derivative, denoted as f'(x), represents the instantaneous rate of change of the function at any given point. Applying the power rule of differentiation, we get:

f'(x) = -6x² + 66x - 144

This derivative is a quadratic function, and its roots will correspond to the critical points of the original cubic function. The coefficients of the quadratic derivative play a crucial role in determining the shape and position of the parabola, which in turn influences the behavior of the original cubic function. Understanding the relationship between a function and its derivative is a core concept in calculus, and it allows us to analyze the function's increasing and decreasing intervals, concavity, and points of inflection. The derivative we have obtained will be used in the next step to find the values of x where the function's slope is zero, indicating potential maxima or minima. The process of differentiation is a cornerstone of calculus, and it provides a powerful tool for analyzing the behavior of functions. In this case, the derivative has transformed the cubic polynomial into a quadratic, which is often easier to analyze and solve. The next step will involve setting this quadratic equal to zero and solving for x, which will give us the critical points of the original function. These critical points are the key to finding the absolute extreme values within the specified interval. The derivative, f'(x) = -6x² + 66x - 144, is a crucial piece of the puzzle in determining the absolute extrema of the function f(x). It represents the slope of the tangent line to the graph of f(x) at any given point. By finding the points where f'(x) = 0, we identify the points where the tangent line is horizontal, which are the potential locations of local maxima, local minima, or saddle points. However, to find the absolute extrema on the closed interval [2, 9], we need to consider not only these critical points but also the endpoints of the interval.

Step 2: Find Critical Points

To find the critical points, we set the derivative equal to zero and solve for x:

-6x² + 66x - 144 = 0

We can simplify this equation by dividing by -6:

x² - 11x + 24 = 0

This quadratic equation can be factored as:

(x - 3)(x - 8) = 0

Thus, the critical points are x = 3 and x = 8. These points are crucial because they represent the x-values where the function's slope is zero, indicating potential maximum or minimum values. Factoring the quadratic equation is a common technique in algebra, and it allows us to find the roots of the equation, which in this case are the critical points of the function. The critical points divide the interval into subintervals, and the function's behavior (increasing or decreasing) can be analyzed within each subinterval. The critical points we have found, x = 3 and x = 8, are both within the given interval [2, 9], which means they are valid candidates for absolute extrema. If a critical point were outside this interval, it would not be relevant to our analysis of the function's behavior within the specified domain. The next step will involve evaluating the original function at these critical points and at the endpoints of the interval to determine the absolute maximum and minimum values. The critical points x = 3 and x = 8 are the solutions to the equation f'(x) = 0. This means that at these points, the tangent line to the graph of f(x) is horizontal. However, it's important to remember that not all critical points are necessarily local maxima or minima. They could also be saddle points, where the function changes its direction but does not reach a local extremum. To determine the nature of these critical points, we would typically use the first or second derivative test. However, in this case, we are interested in finding the absolute extrema on a closed interval, so we can simply evaluate the function at the critical points and the endpoints to determine the absolute maximum and minimum values.

Step 3: Evaluate the Function at Critical Points and Endpoints

Now, we evaluate the original function f(x) = -2x³ + 33x² - 144x at the critical points (x = 3 and x = 8) and the endpoints of the interval (x = 2 and x = 9):

f(2) = -2(2)³ + 33(2)² - 144(2) = -16 + 132 - 288 = -172
f(3) = -2(3)³ + 33(3)² - 144(3) = -54 + 297 - 432 = -189
f(8) = -2(8)³ + 33(8)² - 144(8) = -1024 + 2112 - 1152 = -8
f(9) = -2(9)³ + 33(9)² - 144(9) = -1458 + 2673 - 1296 = -81

Evaluating the function at these points is a crucial step in determining the absolute extrema. This step allows us to compare the function's values at different points within the interval and identify the highest and lowest values, which correspond to the absolute maximum and absolute minimum, respectively. The arithmetic involved in these calculations requires careful attention to detail to avoid errors, as even a small mistake can lead to an incorrect conclusion. The values we have obtained represent the function's height at the critical points and endpoints, and they will be used in the next step to identify the absolute extreme values. The process of evaluating a function at specific points is a fundamental skill in mathematics, and it is used extensively in calculus and other areas of mathematics. In this case, we are evaluating a polynomial function, which is a relatively straightforward process. However, for more complex functions, the evaluation process may involve more sophisticated techniques, such as trigonometric identities or logarithmic properties. The values we have calculated, f(2) = -172, f(3) = -189, f(8) = -8, and f(9) = -81, provide a snapshot of the function's behavior within the interval [2, 9]. These values will allow us to identify the absolute maximum and minimum values, as well as the locations where they occur. The next step will involve comparing these values and drawing conclusions about the absolute extreme values of the function. The calculation of the function values at the critical points and endpoints is a critical step in finding the absolute extrema. These values represent the height of the function at these specific points, and by comparing them, we can determine the absolute maximum and minimum values within the given interval. The process of evaluating a polynomial function is relatively straightforward, but it's important to pay attention to the order of operations and to avoid arithmetic errors. Each of these values provides a piece of the puzzle in understanding the function's behavior within the interval [2, 9]. For example, we can see that the function reaches a relatively low value at x = 3 and a relatively high value at x = 8. This suggests that there may be a local minimum near x = 3 and a local maximum near x = 8. However, to confirm this and to find the absolute extrema, we need to consider all the function values we have calculated.

Step 4: Identify Absolute Extrema

By comparing the function values, we can identify the absolute maximum and minimum values:

The absolute minimum is -189, which occurs at x = 3.
The absolute maximum is -8, which occurs at x = 8.

This is the final step in determining the absolute extreme values of the function. By comparing the function values at the critical points and endpoints, we have identified the lowest and highest points the function reaches within the specified interval. The absolute minimum represents the smallest value the function attains within the interval, while the absolute maximum represents the largest value. These extreme values are important because they provide information about the function's overall behavior and its range within the given domain. The conclusion we have reached is that the function f(x) = -2x³ + 33x² - 144x has an absolute minimum of -189 at x = 3 and an absolute maximum of -8 at x = 8 within the interval [2, 9]. This result is a direct consequence of the Extreme Value Theorem, which guarantees the existence of absolute extrema for continuous functions on closed intervals. The identification of absolute extrema is a fundamental problem in calculus, and it has applications in various fields, such as optimization problems in engineering, economics, and physics. In this case, we have successfully applied the techniques of differential calculus to find the absolute extreme values of a polynomial function. The absolute minimum and maximum values provide valuable information about the function's behavior within the interval [2, 9]. The absolute minimum of -189 at x = 3 indicates the lowest point the function reaches within this interval, while the absolute maximum of -8 at x = 8 indicates the highest point. These extreme values can be useful in various applications, such as finding the optimal solution to a problem or determining the range of possible values for the function. The process we have followed demonstrates a systematic approach to finding absolute extrema, which involves finding the derivative, identifying critical points, evaluating the function at critical points and endpoints, and comparing the resulting values. This approach is applicable to a wide range of functions and intervals, and it provides a powerful tool for analyzing the behavior of functions.

b. Verifying with a Graphing Utility

To verify our results, we can use a graphing utility to graph the function f(x) = -2x³ + 33x² - 144x on the interval [2, 9]. By visually inspecting the graph, we can confirm the absolute maximum and minimum values we calculated analytically. Graphing utilities are powerful tools that allow us to visualize functions and their behavior. They can be used to confirm analytical results, explore the properties of functions, and solve problems that are difficult or impossible to solve by hand. In this case, we will use a graphing utility to verify that the absolute minimum of the function occurs at x = 3 and the absolute maximum occurs at x = 8. The graph should show a clear minimum point at x = 3 and a clear maximum point at x = 8 within the interval [2, 9]. The y-values at these points should correspond to the absolute minimum and maximum values we calculated, which are -189 and -8, respectively. The use of a graphing utility provides a visual confirmation of our analytical results, which can increase our confidence in the accuracy of our solution. It also allows us to gain a better understanding of the function's overall behavior within the interval. The graphing utility allows us to visualize the function f(x) = -2x³ + 33x² - 144x over the interval [2, 9]. By examining the graph, we can visually identify the highest and lowest points, which correspond to the absolute maximum and minimum values, respectively. The graph should confirm that the absolute minimum occurs at x = 3 with a value of -189, and the absolute maximum occurs at x = 8 with a value of -8. This visual verification is a valuable step in the problem-solving process, as it helps to ensure the accuracy of our analytical calculations. The graphing utility also provides a more intuitive understanding of the function's behavior, showing how the function increases and decreases over the interval and how the critical points correspond to local extrema. This visual representation can be particularly helpful for students who are learning calculus concepts, as it provides a concrete way to connect the abstract mathematical ideas with a visual image. The use of a graphing utility to verify analytical results is a common practice in mathematics and other scientific disciplines. It allows us to catch errors in our calculations and to gain a deeper understanding of the problem at hand. In this case, the graphing utility confirms our previous calculations and provides a visual representation of the function's absolute extreme values within the specified interval. This visual confirmation strengthens our understanding of the problem and reinforces the connection between analytical and graphical methods in calculus.

By graphing the function, we can visually confirm that the absolute minimum occurs at (3, -189) and the absolute maximum occurs at (8, -8), which aligns with our analytical solution. This step provides a valuable check and ensures the accuracy of our work. The combination of analytical and graphical methods provides a comprehensive approach to solving calculus problems, enhancing both understanding and accuracy.

Find the absolute extreme values of the function f(x) = -2x³ + 33x² - 144x on the interval [2, 9] and verify the results using a graphing utility.

Absolute Extrema Calculation for f(x) = -2x³ + 33x² - 144x on [2, 9] 🚀