Finding Relative Extrema And Graphing F(x) = X³ - 27x + 60

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Introduction to Extreme Points

In the realm of calculus, understanding the behavior of functions is paramount. One crucial aspect is identifying extreme points, which include relative maxima and minima. These points represent the peaks and valleys of a function's graph, providing valuable insights into its behavior. In this article, we will embark on a journey to find the relative extreme points of the function f(x) = x³ - 27x + 60 and then sketch its graph. This process involves leveraging the power of derivatives and critical points to unravel the function's characteristics. By the end of this exploration, you'll gain a deeper understanding of how calculus tools can be applied to analyze and visualize functions effectively. Let's dive into the step-by-step process of finding these key points and bringing the function to life on a graph.

Step 1: Finding the First Derivative

The first step in locating the relative extreme points is to calculate the first derivative of the function. The derivative, denoted as f'(x), provides us with information about the slope of the function at any given point. Specifically, the critical points, where the derivative is either zero or undefined, are potential locations of relative maxima or minima. To find f'(x) for our function f(x) = x³ - 27x + 60, we apply the power rule of differentiation. The power rule states that if f(x) = axⁿ, then f'(x) = naxⁿ⁻¹. Applying this rule to each term in our function, we get:

  • The derivative of x³ is 3x².
  • The derivative of -27x is -27.
  • The derivative of the constant 60 is 0.

Combining these results, the first derivative of f(x) is: f'(x) = 3x² - 27. This derivative is a quadratic function, which is continuous and defined for all real numbers. Therefore, we only need to focus on the points where f'(x) = 0 to find our critical points. The first derivative will play a crucial role in determining where the function's slope changes, indicating the presence of relative extreme points.

Step 2: Finding Critical Points

Now that we have the first derivative, f'(x) = 3x² - 27, our next task is to find the critical points. Critical points are the x-values where the derivative is either equal to zero or undefined. These points are crucial because they represent potential locations of relative maxima or minima. To find the critical points, we set f'(x) equal to zero and solve for x:

3x² - 27 = 0

To solve this equation, we can first add 27 to both sides:

3x² = 27

Next, we divide both sides by 3:

x² = 9

Taking the square root of both sides, we get two possible solutions:

x = ±3

Thus, the critical points are x = 3 and x = -3. These are the x-values where the function's slope might change direction, indicating a potential maximum or minimum. In the next step, we will use the second derivative test to determine the nature of these critical points and whether they correspond to relative maxima or minima. These critical points are the foundation for understanding the function's local behavior and sketching an accurate graph.

Step 3: Using the Second Derivative Test

To determine whether the critical points x = 3 and x = -3 correspond to relative maxima or minima, we employ the second derivative test. This test involves finding the second derivative of the function, f''(x), and evaluating it at each critical point. The sign of the second derivative at a critical point provides valuable information about the concavity of the function at that point. If f''(x) > 0, the function is concave up, indicating a relative minimum. Conversely, if f''(x) < 0, the function is concave down, indicating a relative maximum.

First, we find the second derivative of f(x) = x³ - 27x + 60. We already know that the first derivative is f'(x) = 3x² - 27. Now, we differentiate f'(x) to find f''(x). Applying the power rule again:

  • The derivative of 3x² is 6x.
  • The derivative of -27 is 0.

Thus, the second derivative is: f''(x) = 6x.

Now, we evaluate f''(x) at each critical point:

  • At x = 3, f''(3) = 6(3) = 18. Since 18 > 0, the function is concave up at x = 3, indicating a relative minimum.
  • At x = -3, f''(-3) = 6(-3) = -18. Since -18 < 0, the function is concave down at x = -3, indicating a relative maximum.

This test confirms that we have a relative minimum at x = 3 and a relative maximum at x = -3. In the next step, we will find the corresponding y-values for these points to fully define the coordinates of the relative extrema.

Step 4: Finding the y-coordinates

Now that we have identified the x-coordinates of the relative extreme points, we need to find their corresponding y-coordinates. To do this, we plug the x-values of the critical points back into the original function, f(x) = x³ - 27x + 60. This will give us the y-values that complete the coordinates of the relative maximum and minimum.

First, let's find the y-coordinate for the relative minimum at x = 3:

f(3) = (3)³ - 27(3) + 60

f(3) = 27 - 81 + 60

f(3) = 6

So, the relative minimum point is located at (3, 6).

Next, let's find the y-coordinate for the relative maximum at x = -3:

f(-3) = (-3)³ - 27(-3) + 60

f(-3) = -27 + 81 + 60

f(-3) = 114

Therefore, the relative maximum point is located at (-3, 114).

Now we have both the x and y coordinates for the relative extrema. The relative minimum is at (3, 6), and the relative maximum is at (-3, 114). These points are crucial for sketching the graph of the function, as they indicate where the function changes direction. In the next section, we will use this information, along with other key points, to sketch the graph of f(x) = x³ - 27x + 60.

Step 5: Sketching the Graph

With the critical points and their nature determined, we can now proceed to sketch the graph of the function f(x) = x³ - 27x + 60. We know the following:

  • Relative Maximum: (-3, 114)
  • Relative Minimum: (3, 6)

To sketch the graph, we also need to consider the function's end behavior. Since f(x) is a cubic function with a positive leading coefficient, we know that as x approaches positive infinity, f(x) also approaches positive infinity. Conversely, as x approaches negative infinity, f(x) approaches negative infinity. This gives us a general idea of the graph's overall shape.

Additionally, it can be helpful to find the y-intercept by setting x = 0:

f(0) = (0)³ - 27(0) + 60 = 60

So, the y-intercept is at (0, 60). This point provides another reference for sketching the graph.

Now, we can sketch the graph. Start by plotting the relative maximum at (-3, 114), the relative minimum at (3, 6), and the y-intercept at (0, 60). Knowing the end behavior, we can sketch a curve that rises from negative infinity, reaches the relative maximum, descends to the relative minimum, and then rises again towards positive infinity. The graph should be smooth and continuous, reflecting the nature of a polynomial function.

By connecting these points and considering the end behavior, we obtain a comprehensive sketch of the function f(x) = x³ - 27x + 60. The graph visually represents the function's behavior, showing the relative extrema and how the function changes direction. This graphical representation provides a valuable complement to the analytical methods we used to find the critical points.

Conclusion

In this article, we successfully found the relative extreme points of the function f(x) = x³ - 27x + 60 and sketched its graph. We began by finding the first derivative, f'(x) = 3x² - 27, and identifying the critical points at x = 3 and x = -3. Using the second derivative test, we determined that x = -3 corresponds to a relative maximum and x = 3 corresponds to a relative minimum. We then calculated the y-coordinates of these points, finding the relative maximum at (-3, 114) and the relative minimum at (3, 6). Finally, we used this information, along with the y-intercept and the function's end behavior, to sketch the graph.

This process demonstrates the power of calculus in analyzing functions and understanding their behavior. By finding derivatives and critical points, we can gain valuable insights into a function's local extrema and overall shape. Sketching the graph provides a visual representation of these analytical findings, enhancing our understanding of the function. This approach is applicable to a wide range of functions and is a fundamental skill in calculus and related fields. The ability to find relative extreme points and sketch graphs is essential for problem-solving in mathematics, physics, engineering, and other disciplines.