Finding Relative Extreme Points And Sketching The Graph Of F(x) = X³ - 12x + 23

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Introduction

In this article, we will delve into the process of finding the relative extreme points of the function f(x) = x³ - 12x + 23. We will systematically identify any relative minima and maxima, which are crucial for understanding the behavior and shape of the function's graph. By employing techniques from differential calculus, we can accurately locate these points and use them as key anchors for sketching the graph of the function. This exploration not only demonstrates the practical application of calculus in analyzing functions but also enhances our ability to visualize and interpret mathematical concepts. We will break down each step, ensuring clarity and providing a comprehensive guide for anyone seeking to master this skill. Understanding the nuances of finding relative extrema is vital in various fields, including optimization problems in engineering, economics, and computer science.

Step-by-Step Guide to Finding Relative Extreme Points

The process of finding the relative extreme points of a function involves several key steps, each building upon the previous one to provide a complete analysis. First, we need to compute the derivative of the function, which gives us the slope of the tangent line at any point on the curve. This is a critical step because relative extrema occur where the tangent line is horizontal, meaning the derivative is zero. Next, we identify the critical points by setting the derivative equal to zero and solving for x. These critical points are potential locations for relative maxima and minima. However, not all critical points are extrema; some may be points of inflection where the concavity of the curve changes. To determine the nature of each critical point, we employ the first derivative test or the second derivative test. The first derivative test involves examining the sign of the derivative on either side of the critical point. If the derivative changes from positive to negative, the critical point is a relative maximum; if it changes from negative to positive, it is a relative minimum. Alternatively, the second derivative test uses the sign of the second derivative at the critical point. A positive second derivative indicates a relative minimum, while a negative second derivative indicates a relative maximum. Once we have identified the extreme points, we can find their corresponding y-values by plugging the x-values back into the original function. These points, along with any additional information such as the function's end behavior and intercepts, provide a solid foundation for sketching the graph. Each step in this process is essential for a thorough understanding of the function's behavior and its graphical representation.

1. Calculate the First Derivative

The first step in finding the relative extreme points of the function f(x) = x³ - 12x + 23 is to calculate its first derivative. The derivative, denoted as f'(x), gives us the slope of the tangent line to the curve at any point x. This is crucial because relative maxima and minima occur where the tangent line is horizontal, meaning the derivative is zero. To find the derivative, we apply the power rule, which states that the derivative of xⁿ is nxⁿ⁻¹. Applying this rule to each term in the function, we get:

f'(x) = d/dx (x³) - d/dx (12x) + d/dx (23)

f'(x) = 3x² - 12 + 0

f'(x) = 3x² - 12

Thus, the first derivative of the function f(x) = x³ - 12x + 23 is f'(x) = 3x² - 12. This derivative is a quadratic function, and its roots (where f'(x) = 0) will give us the x-values of the critical points, which are potential locations for relative maxima and minima. The derivative provides essential information about the function's rate of change and is the foundation for the subsequent steps in identifying extreme points. A thorough understanding of how to calculate derivatives is fundamental in calculus and is a cornerstone for analyzing functions and their graphs. This initial step sets the stage for a more detailed examination of the function's behavior and the determination of its extreme values.

2. Find Critical Points

Following the calculation of the first derivative, the next crucial step is to find the critical points of the function f(x) = x³ - 12x + 23. Critical points are the x-values at which the derivative f'(x) is either equal to zero or undefined. These points are significant because they represent potential locations for relative maxima, relative minima, or points of inflection. To find the critical points, we set the first derivative equal to zero and solve for x. From the previous step, we found that f'(x) = 3x² - 12. Setting this equal to zero gives us:

3x² - 12 = 0

To solve this equation, we first factor out the common factor of 3:

3(x² - 4) = 0

Next, we divide both sides by 3:

x² - 4 = 0

This is a difference of squares, which can be factored as:

(x - 2)(x + 2) = 0

Setting each factor equal to zero gives us the solutions:

x - 2 = 0 => x = 2

x + 2 = 0 => x = -2

Therefore, the critical points of the function are x = 2 and x = -2. These points are the potential locations for relative extreme points. It is important to note that critical points are not automatically extreme points; further analysis is required to determine whether they are relative maxima, relative minima, or neither. This analysis typically involves using either the first derivative test or the second derivative test, which we will explore in the subsequent steps. Identifying critical points is a vital step in understanding the behavior of a function and is essential for sketching its graph.

3. Determine Relative Extrema Using the First Derivative Test

After identifying the critical points, the next step is to determine whether these points correspond to relative maxima, relative minima, or neither. One effective method for making this determination is the first derivative test. The first derivative test involves analyzing the sign of the first derivative, f'(x), on intervals around each critical point. The idea is that if the derivative changes sign at a critical point, it indicates a change in the function's direction, which corresponds to a relative extremum. For the function f(x) = x³ - 12x + 23, we found the critical points to be x = -2 and x = 2. To apply the first derivative test, we need to consider intervals to the left and right of each critical point. This divides the real number line into three intervals: (-∞, -2), (-2, 2), and (2, ∞). We will choose a test value within each interval and evaluate f'(x) at that value.

  1. Interval (-∞, -2): Choose x = -3. Then f'(-3) = 3(-3)² - 12 = 3(9) - 12 = 27 - 12 = 15. Since f'(-3) > 0, the function is increasing on this interval.
  2. Interval (-2, 2): Choose x = 0. Then f'(0) = 3(0)² - 12 = -12. Since f'(0) < 0, the function is decreasing on this interval.
  3. Interval (2, ∞): Choose x = 3. Then f'(3) = 3(3)² - 12 = 3(9) - 12 = 27 - 12 = 15. Since f'(3) > 0, the function is increasing on this interval.

Now, we analyze the sign changes:

  • At x = -2, the derivative changes from positive to negative, indicating a relative maximum.
  • At x = 2, the derivative changes from negative to positive, indicating a relative minimum.

Thus, using the first derivative test, we have identified that x = -2 corresponds to a relative maximum and x = 2 corresponds to a relative minimum. This analysis is crucial for understanding the local behavior of the function and is a key step in sketching its graph.

4. Determine Relative Extrema Using the Second Derivative Test

Another method for determining whether critical points correspond to relative maxima or minima is the second derivative test. This test involves evaluating the second derivative, f''(x), at each critical point. The sign of the second derivative provides information about the concavity of the function at that point. A positive second derivative indicates that the function is concave up, suggesting a relative minimum, while a negative second derivative indicates that the function is concave down, suggesting a relative maximum. To apply the second derivative test to the function f(x) = x³ - 12x + 23, we first need to calculate the second derivative. Recall that the first derivative is f'(x) = 3x² - 12. The second derivative is the derivative of the first derivative:

f''(x) = d/dx (3x² - 12)

f''(x) = 6x

Now, we evaluate the second derivative at each critical point, which we previously found to be x = -2 and x = 2.

  1. At x = -2: f''(-2) = 6(-2) = -12 Since f''(-2) < 0, the function is concave down at x = -2, indicating a relative maximum.
  2. At x = 2: f''(2) = 6(2) = 12 Since f''(2) > 0, the function is concave up at x = 2, indicating a relative minimum.

The second derivative test confirms the results obtained from the first derivative test: x = -2 corresponds to a relative maximum, and x = 2 corresponds to a relative minimum. This test provides an alternative approach to identifying relative extrema and reinforces the importance of understanding the relationship between a function's derivatives and its behavior. The second derivative test is particularly useful when the second derivative is easy to compute and evaluate.

5. Find the y-coordinates of the Relative Extrema

After determining the x-coordinates of the relative extreme points, the next step is to find their corresponding y-coordinates. This will give us the actual points (x, y) on the graph of the function where these extrema occur. To find the y-coordinates, we substitute the x-values of the critical points into the original function, f(x) = x³ - 12x + 23. We identified the critical points as x = -2 and x = 2. Let's evaluate the function at these points:

  1. At x = -2: f(-2) = (-2)³ - 12(-2) + 23 f(-2) = -8 + 24 + 23 f(-2) = 39 So, the relative maximum point is (-2, 39).
  2. At x = 2: f(2) = (2)³ - 12(2) + 23 f(2) = 8 - 24 + 23 f(2) = 7 So, the relative minimum point is (2, 7).

Therefore, the relative extreme points of the function f(x) = x³ - 12x + 23 are (-2, 39), which is a relative maximum, and (2, 7), which is a relative minimum. These points are crucial for sketching the graph of the function, as they indicate the highest and lowest points in the local regions around these x-values. Finding the y-coordinates completes the process of identifying the relative extrema and provides a comprehensive understanding of the function's behavior. This information, along with the critical points and the analysis of the derivatives, allows for an accurate and informative graphical representation of the function.

Sketching the Graph of the Function

With the relative extreme points identified, we can now proceed to sketch the graph of the function f(x) = x³ - 12x + 23. Sketching the graph involves combining the information we've gathered about the function's behavior, including its critical points, relative extrema, and general shape. To create an accurate sketch, we can follow these steps:

  1. Plot the Relative Extreme Points: We found the relative maximum at (-2, 39) and the relative minimum at (2, 7). These points will serve as key anchors for our graph.
  2. Consider the End Behavior: Since f(x) is a cubic function with a positive leading coefficient (1), we know that as x approaches -∞, f(x) also approaches -∞, and as x approaches ∞, f(x) also approaches ∞. This tells us the general direction of the graph as we move away from the center.
  3. Determine the y-intercept: The y-intercept is the point where the graph crosses the y-axis, which occurs when x = 0. Plugging x = 0 into the function gives us f(0) = (0)³ - 12(0) + 23 = 23. So, the y-intercept is (0, 23).
  4. Analyze Intervals of Increase and Decrease: From the first derivative test, we know that the function is increasing on the intervals (-∞, -2) and (2, ∞), and decreasing on the interval (-2, 2). This tells us the direction of the graph between the critical points.
  5. Analyze Concavity: From the second derivative test, we know that the function is concave down at x = -2 and concave up at x = 2. This information helps us to understand the curve's shape.

Putting all this information together, we can sketch the graph. Start by plotting the relative maximum at (-2, 39) and the relative minimum at (2, 7). Then, plot the y-intercept at (0, 23). Knowing the end behavior, we can draw the graph starting from the bottom left, increasing until the relative maximum, then decreasing to the relative minimum, and finally increasing again towards the top right. The concavity helps us to shape the curve appropriately, making it concave down before x = -2 and concave up after x = 2. The resulting sketch provides a visual representation of the function's behavior and confirms the analysis of its relative extrema.

Conclusion

In conclusion, we have successfully found the relative extreme points of the function f(x) = x³ - 12x + 23 and sketched its graph. By systematically applying the principles of differential calculus, we identified the critical points, used the first and second derivative tests to determine the nature of these points, and found the corresponding y-coordinates. The relative maximum was determined to be at (-2, 39), and the relative minimum was at (2, 7). Additionally, we found the y-intercept at (0, 23) and analyzed the end behavior of the function. This comprehensive approach allowed us to create an accurate sketch of the function's graph, which visually represents its behavior and key features. The process of finding relative extrema and sketching graphs is a fundamental skill in calculus with wide-ranging applications in various fields, including optimization problems in engineering, economics, and computer science. Mastering these techniques enhances our ability to analyze and understand functions, making it a valuable asset in mathematical problem-solving.