Finding Critical Values And Relative Extrema For F(x) = X^2 - 2x + 3

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In calculus, identifying critical values and relative extrema is crucial for understanding the behavior of a function. This article will guide you through the process of finding these key features for the quadratic function f(x) = x² - 2x + 3. We will cover the steps involved in determining critical values using the derivative and then utilize these values to locate any relative maxima or minima. This comprehensive guide aims to provide a clear and understandable explanation, ensuring you grasp the underlying concepts and can apply them to other functions.

1. Understanding Critical Values

Critical values are the x-values at which the derivative of a function is either zero or undefined. These points are significant because they often indicate where the function's slope changes direction, potentially leading to relative maxima or minima. To find the critical values, we first need to compute the derivative of the given function.

For the function f(x) = x² - 2x + 3, the derivative, denoted as f'(x), can be found using the power rule. The power rule states that if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. Applying this rule to each term in our function, we get:

  • The derivative of x² is 2x¹ or 2x.
  • The derivative of -2x is -2.
  • The derivative of the constant 3 is 0.

Therefore, the derivative of f(x) = x² - 2x + 3 is f'(x) = 2x - 2. Now that we have the derivative, we can find the critical values by setting f'(x) equal to zero and solving for x.

Setting the derivative to zero, we have the equation:

2x - 2 = 0

To solve for x, we add 2 to both sides of the equation:

2x = 2

Then, we divide both sides by 2:

x = 1

Thus, the critical value for the function f(x) = x² - 2x + 3 is x = 1. This single critical value suggests a potential turning point in the graph of the function. In this case, since the function is a parabola opening upwards (due to the positive coefficient of the x² term), this critical value is likely to represent the location of a relative minimum. However, to confirm this, we must proceed with further analysis, which we will cover in the next section.

Understanding the concept of critical values is essential in calculus as it forms the foundation for determining the intervals where a function is increasing or decreasing, and for locating points of local extrema. The derivative, as we've seen, plays a crucial role in this process. It provides us with information about the rate of change of the function at any given point, and by finding where the derivative is zero or undefined, we pinpoint the locations where this rate of change may be transitioning, thereby identifying potential maxima and minima. This first step of finding critical values is the cornerstone of optimizing functions and solving a wide array of problems in various fields, including physics, engineering, and economics. In the next sections, we'll delve into how these critical values are used to identify relative extrema.

2. Determining Relative Extrema

Relative extrema, which include relative maxima and relative minima, are the points where a function reaches a local maximum or minimum value within a specific interval. To find the relative extrema, we use the critical values we identified in the previous section. There are two primary methods for determining the nature of the extrema: the first derivative test and the second derivative test. For this example, we will focus on the first derivative test, as it offers a straightforward way to analyze the behavior of the function around the critical value.

The first derivative test involves examining the sign of the derivative, f'(x), on intervals to the left and right of the critical value. If the derivative changes sign from negative to positive at the critical value, it indicates a relative minimum. Conversely, if the derivative changes sign from positive to negative, it indicates a relative maximum. If the derivative does not change sign, the critical value corresponds to neither a maximum nor a minimum.

In our case, the critical value is x = 1. We need to test the sign of f'(x) = 2x - 2 on intervals to the left and right of x = 1. Let's choose test points x = 0 (to the left) and x = 2 (to the right).

  • For x = 0, f'(0) = 2(0) - 2 = -2. The derivative is negative, indicating that the function is decreasing in this interval.
  • For x = 2, f'(2) = 2(2) - 2 = 2. The derivative is positive, indicating that the function is increasing in this interval.

Since the derivative changes from negative to positive at x = 1, we can conclude that there is a relative minimum at this point. To find the y-coordinate of the relative minimum, we substitute x = 1 back into the original function:

f(1) = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2

Therefore, the relative minimum occurs at the point (1, 2). There are no other critical values, so there are no other potential relative extrema.

The first derivative test is a powerful tool in calculus, enabling us to understand the behavior of a function around its critical points. By analyzing the sign changes of the derivative, we can definitively determine whether a critical point corresponds to a relative maximum, a relative minimum, or neither. This information is crucial for sketching the graph of a function, identifying its key features, and solving optimization problems. The ability to apply the first derivative test effectively provides a deeper understanding of how a function changes and behaves, adding a critical dimension to calculus studies. The interpretation of derivative sign changes is not merely a mathematical exercise; it offers a visual and intuitive grasp of the function's dynamics. This knowledge is invaluable in a range of applications where understanding functional behavior is key, from economics to engineering. In the following section, we will summarize our findings and provide a clear answer to the problem.

3. Summary and Conclusion

In summary, for the function f(x) = x² - 2x + 3, we have determined the following:

  • Critical Value: The critical value is x = 1. This was found by taking the derivative of the function, f'(x) = 2x - 2, setting it equal to zero, and solving for x.
  • Relative Extrema: There is a relative minimum at the point (1, 2). This was determined using the first derivative test, which showed that the derivative changes from negative to positive at x = 1.

Therefore, the function f(x) = x² - 2x + 3 has one critical value and one relative extremum, which is a minimum. This analysis provides a complete understanding of the function's turning points and overall behavior. Understanding the function's behavior through its critical points and extrema is crucial for various applications, including optimization problems, curve sketching, and understanding the nature of mathematical models.

The process of finding critical values and relative extrema is a fundamental aspect of calculus, allowing us to gain insights into the behavior of functions. By following the steps outlined in this article, you can effectively analyze various functions and determine their critical points and extrema. This skill is not only essential for academic success in calculus but also valuable in various fields that utilize mathematical modeling and optimization techniques. The ability to interpret and analyze functions based on their derivatives and critical points provides a solid foundation for more advanced mathematical studies and practical applications in science, engineering, economics, and beyond. The combination of critical value determination and the application of the first derivative test offers a robust approach to function analysis, enabling a comprehensive understanding of functional behavior and its implications in real-world scenarios.