Finding Relative Extreme Points And Graphing F(x) = X³ - 3x + 1

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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 - 3x + 1. Understanding these extreme points is crucial for sketching an accurate graph of the function and analyzing its behavior. Relative extreme points, also known as local maxima and minima, represent the points where the function changes its direction, either from increasing to decreasing (local maximum) or from decreasing to increasing (local minimum). By identifying these points, we can gain valuable insights into the function's shape and overall characteristics. To achieve this, we will utilize the concepts of derivatives, critical points, and the first and second derivative tests. This comprehensive approach will enable us to pinpoint the exact locations of the relative extrema and ultimately sketch a representative graph of the given cubic function.

Determining the First Derivative

The cornerstone of finding relative extreme points lies in the concept of derivatives. The first derivative of a function, denoted as f'(x), provides information about the function's rate of change and its increasing or decreasing nature. To find the first derivative of our function, f(x) = x³ - 3x + 1, we will apply the power rule of differentiation. This rule states that the derivative of xⁿ is nxⁿ⁻¹. Applying this rule to each term in our function, we get:

f'(x) = d/dx (x³) - d/dx (3x) + d/dx (1)

f'(x) = 3x² - 3 + 0

f'(x) = 3x² - 3

The first derivative, f'(x) = 3x² - 3, is a quadratic function that represents the slope of the tangent line to the original function f(x) at any given point x. This derivative is essential for identifying critical points, which are the potential locations of relative extrema. By analyzing the sign of f'(x), we can determine where the function is increasing (f'(x) > 0), decreasing (f'(x) < 0), or has a horizontal tangent (f'(x) = 0). These critical points and the intervals between them form the foundation for understanding the function's behavior and locating its relative extreme points.

Finding Critical Points

Critical points are the key to unlocking the locations of relative extrema. These points occur where the first derivative, f'(x), is either equal to zero or undefined. In our case, f'(x) = 3x² - 3 is a polynomial function, which means it is defined for all real numbers. Therefore, we only need to find the points where f'(x) = 0. Setting the first derivative equal to zero, we have:

3x² - 3 = 0

To solve this quadratic equation, we can first factor out a common factor of 3:

3(x² - 1) = 0

Dividing both sides by 3, we get:

x² - 1 = 0

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

(x - 1)(x + 1) = 0

Setting each factor equal to zero, we find the critical points:

x - 1 = 0 => x = 1

x + 1 = 0 => x = -1

Thus, we have identified two critical points: x = 1 and x = -1. These critical points are the potential locations of relative maxima or minima. To determine the nature of these points, we can use the first or second derivative test. These tests will help us determine whether the function changes from increasing to decreasing or vice versa at these points, ultimately revealing whether they correspond to relative maxima or minima.

Applying the First Derivative Test

The first derivative test is a powerful tool for determining the nature of critical points. It relies on analyzing the sign of the first derivative, f'(x), in the intervals around the critical points. By examining whether f'(x) changes its sign from positive to negative or from negative to positive, we can identify relative maxima and minima, respectively. To apply the first derivative test, we need to consider the intervals determined by our critical points, which are x = -1 and x = 1. This divides the real number line into three intervals: (-∞, -1), (-1, 1), and (1, ∞).

Let's analyze the sign of f'(x) = 3x² - 3 in each interval:

  1. Interval (-∞, -1): Choose a test value, say x = -2. Then, f'(-2) = 3(-2)² - 3 = 3(4) - 3 = 9 > 0. Since f'(x) > 0 in this interval, the function is increasing.
  2. Interval (-1, 1): Choose a test value, say x = 0. Then, f'(0) = 3(0)² - 3 = -3 < 0. Since f'(x) < 0 in this interval, the function is decreasing.
  3. Interval (1, ∞): Choose a test value, say x = 2. Then, f'(2) = 3(2)² - 3 = 3(4) - 3 = 9 > 0. Since f'(x) > 0 in this interval, the function is increasing.

Based on this analysis, we can conclude the following:

  • At x = -1, the function changes from increasing to decreasing, indicating a relative maximum.
  • At x = 1, the function changes from decreasing to increasing, indicating a relative minimum.

To find the y-coordinates of these extreme points, we substitute the x-values back into the original function, f(x) = x³ - 3x + 1.

Determining Relative Extrema

Now that we have identified the critical points and used the first derivative test to determine their nature, we can find the exact coordinates of the relative extrema. We found that x = -1 corresponds to a relative maximum, and x = 1 corresponds to a relative minimum. To find the y-coordinates of these points, we substitute these x-values back into the original function, f(x) = x³ - 3x + 1.

For the relative maximum at x = -1:

f(-1) = (-1)³ - 3(-1) + 1 = -1 + 3 + 1 = 3

So, the relative maximum point is (-1, 3).

For the relative minimum at x = 1:

f(1) = (1)³ - 3(1) + 1 = 1 - 3 + 1 = -1

So, the relative minimum point is (1, -1).

Therefore, we have successfully identified the relative extrema of the function f(x) = x³ - 3x + 1. The function has a relative maximum at (-1, 3) and a relative minimum at (1, -1). These points are crucial for sketching the graph of the function, as they represent the peaks and valleys of the curve. By knowing these extreme points and the intervals where the function is increasing or decreasing, we can create a more accurate and informative graph.

Sketching the Graph

To sketch the graph of the function f(x) = x³ - 3x + 1, we can use the information we have gathered about its relative extrema and increasing/decreasing intervals. We know that the function has a relative maximum at (-1, 3) and a relative minimum at (1, -1). We also know that the function is increasing on the intervals (-∞, -1) and (1, ∞), and decreasing on the interval (-1, 1). Additionally, it's helpful to find the y-intercept, which is the point where the graph intersects the y-axis. This occurs when x = 0:

f(0) = (0)³ - 3(0) + 1 = 1

So, the y-intercept is (0, 1).

With this information, we can sketch a graph that captures the essential features of the function. Here's a step-by-step approach:

  1. Plot the relative maximum point (-1, 3) and the relative minimum point (1, -1).
  2. Plot the y-intercept (0, 1).
  3. Draw a smooth curve that passes through these points.
  4. In the interval (-∞, -1), the function is increasing, so draw the curve rising as it approaches the relative maximum at (-1, 3).
  5. In the interval (-1, 1), the function is decreasing, so draw the curve falling from the relative maximum at (-1, 3) to the relative minimum at (1, -1).
  6. In the interval (1, ∞), the function is increasing, so draw the curve rising as it moves away from the relative minimum at (1, -1).

The resulting graph will be a cubic curve with a peak at the relative maximum and a valley at the relative minimum. The graph will also pass through the y-intercept at (0, 1). This sketch provides a visual representation of the function's behavior and confirms our analytical findings regarding its extreme points and intervals of increase and decrease.

Conclusion

In this article, we have successfully found the relative extreme points of the function f(x) = x³ - 3x + 1 using the concepts of derivatives and the first derivative test. We identified the critical points by setting the first derivative equal to zero and solving for x. Then, we used the first derivative test to determine that x = -1 corresponds to a relative maximum and x = 1 corresponds to a relative minimum. We calculated the y-coordinates of these extreme points by substituting the x-values back into the original function, finding the relative maximum point to be (-1, 3) and the relative minimum point to be (1, -1). Finally, we used this information, along with the y-intercept, to sketch a graph of the function that accurately reflects its behavior. Understanding the process of finding relative extreme points is essential for analyzing and graphing functions, providing valuable insights into their properties and behavior.