Function Analysis F(x) = X³ - 3x² + 4 Domain, Zeros, Extrema

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This article provides a detailed analysis of the cubic function f(x) = x³ - 3x² + 4. We will explore its key characteristics, including its domain, zeros (roots), extrema (local maxima and minima), intervals of increase and decrease, concavity, and inflection points. This comprehensive analysis will give you a thorough understanding of the function's behavior and graphical representation. Whether you're a student learning calculus or simply interested in exploring the properties of polynomial functions, this guide will provide valuable insights and step-by-step explanations. We will use a combination of algebraic techniques and calculus concepts to dissect this function and reveal its intricacies. Let's embark on this mathematical journey and uncover the hidden features of f(x) = x³ - 3x² + 4.

1. Determining the Domain

When analyzing any function, the domain is the first aspect we need to determine. The domain of a function refers to the set of all possible input values (x-values) for which the function is defined. For polynomial functions, like our f(x) = x³ - 3x² + 4, the domain is particularly straightforward. Polynomials are defined for all real numbers, meaning we can plug in any real number for x and obtain a valid output. There are no restrictions, such as division by zero or taking the square root of a negative number, that would limit the input values. Therefore, the domain of f(x) = x³ - 3x² + 4 is all real numbers, which can be expressed in interval notation as (-∞, ∞). This signifies that the function extends infinitely in both the positive and negative x-directions. Understanding the domain is crucial because it sets the stage for further analysis. It tells us the range of x-values we need to consider when finding zeros, extrema, and other important features of the function. In this case, knowing that the domain is all real numbers allows us to confidently proceed with our analysis without worrying about any domain-related restrictions. The domain provides the foundation upon which we build our understanding of the function's behavior across the entire number line. This initial step is not just a formality but a critical aspect of function analysis, ensuring we consider the function's complete scope.

2. Finding the Zeros (Roots)

The zeros, also known as roots, of a function are the x-values where the function intersects the x-axis, meaning f(x) = 0. To find the zeros of f(x) = x³ - 3x² + 4, we need to solve the equation x³ - 3x² + 4 = 0. This is a cubic equation, which can be more challenging to solve than quadratic equations. One common technique for solving cubic equations is to look for rational roots using the Rational Root Theorem. The Rational Root Theorem states that if a polynomial equation with integer coefficients has a rational root p/q (where p and q are integers with no common factors other than 1), then p must be a factor of the constant term and q must be a factor of the leading coefficient. In our case, the constant term is 4 and the leading coefficient is 1. Therefore, the possible rational roots are the factors of 4 divided by the factors of 1, which are ±1, ±2, and ±4. We can test these potential roots by substituting them into the equation f(x) = x³ - 3x² + 4. If we find a value that makes f(x) = 0, then we have found a root. By trying x = -1, we get f(-1) = (-1)³ - 3(-1)² + 4 = -1 - 3 + 4 = 0. So, x = -1 is a root of the equation. This means that (x + 1) is a factor of the polynomial x³ - 3x² + 4. Now, we can use polynomial long division or synthetic division to divide x³ - 3x² + 4 by (x + 1) to find the remaining quadratic factor. Performing the division, we get (x³ - 3x² + 4) / (x + 1) = x² - 4x + 4. Now we have factored the cubic equation into (x + 1)(x² - 4x + 4) = 0. To find the remaining roots, we need to solve the quadratic equation x² - 4x + 4 = 0. This quadratic equation can be factored as (x - 2)² = 0. Therefore, x = 2 is a repeated root (it has a multiplicity of 2). In summary, the zeros of f(x) = x³ - 3x² + 4 are x = -1 and x = 2 (with multiplicity 2). These zeros tell us where the graph of the function intersects the x-axis, providing important reference points for sketching the curve. Understanding the nature of these roots, including the multiplicity of the root at x=2, is crucial for accurately interpreting the function's behavior near these points.

3. Identifying Extrema (Local Maxima and Minima)

Extrema, which include local maxima and local minima, are crucial points on a function's graph where the function changes its direction. To find the extrema of f(x) = x³ - 3x² + 4, we use calculus. The first step is to find the first derivative of the function, denoted as f'(x). The first derivative represents the slope of the tangent line to the function at any given point. We have f(x) = x³ - 3x² + 4, so applying the power rule for differentiation, we get f'(x) = 3x² - 6x. Critical points are the points where the first derivative is either equal to zero or undefined. In this case, f'(x) is a polynomial, so it is defined for all x. To find the critical points, we set f'(x) = 0 and solve for x: 3x² - 6x = 0. We can factor out a 3x from the equation: 3x(x - 2) = 0. This gives us two critical points: x = 0 and x = 2. These are the potential locations of local maxima or minima. To determine whether these critical points are local maxima, local minima, or neither, we can use the first derivative test or the second derivative test. Let's use the first derivative test. The first derivative test involves examining the sign of f'(x) in the intervals determined by the critical points. Our critical points are x = 0 and x = 2, which divide the number line into three intervals: (-∞, 0), (0, 2), and (2, ∞). We choose a test value within each interval and evaluate f'(x) at that value. For the interval (-∞, 0), let's choose x = -1. f'(-1) = 3(-1)² - 6(-1) = 3 + 6 = 9, which is positive. This means that the function is increasing on the interval (-∞, 0). For the interval (0, 2), let's choose x = 1. f'(1) = 3(1)² - 6(1) = 3 - 6 = -3, which is negative. This means that the function is decreasing on the interval (0, 2). For the interval (2, ∞), let's choose x = 3. f'(3) = 3(3)² - 6(3) = 27 - 18 = 9, which is positive. This means that the function is increasing on the interval (2, ∞). Now we can analyze the sign changes of f'(x). At x = 0, f'(x) changes from positive to negative, indicating a local maximum. To find the y-coordinate of the local maximum, we evaluate f(0): f(0) = (0)³ - 3(0)² + 4 = 4. So, there is a local maximum at the point (0, 4). At x = 2, f'(x) changes from negative to positive, indicating a local minimum. To find the y-coordinate of the local minimum, we evaluate f(2): f(2) = (2)³ - 3(2)² + 4 = 8 - 12 + 4 = 0. So, there is a local minimum at the point (2, 0). In summary, the function f(x) = x³ - 3x² + 4 has a local maximum at (0, 4) and a local minimum at (2, 0). These extrema provide key points for understanding the function's shape and behavior. The identification of these turning points is essential for sketching an accurate graph of the function and for various applications where optimization is required. The use of the first derivative test allows us to determine not only the location of these extrema but also the intervals over which the function is increasing or decreasing.

4. Determining Intervals of Increase and Decrease

The intervals of increase and decrease are directly related to the first derivative, f'(x), which we calculated in the previous section. As we determined earlier, f'(x) = 3x² - 6x. The sign of f'(x) tells us whether the function is increasing or decreasing. If f'(x) > 0, the function is increasing. If f'(x) < 0, the function is decreasing. We already found the critical points by setting f'(x) = 0, which gave us x = 0 and x = 2. These critical points divide the number line into three intervals: (-∞, 0), (0, 2), and (2, ∞). To determine the intervals of increase and decrease, we analyze the sign of f'(x) in each interval. We did this in the previous section using the first derivative test. In the interval (-∞, 0), we chose a test value of x = -1 and found that f'(-1) = 9, which is positive. Therefore, the function is increasing on the interval (-∞, 0). In the interval (0, 2), we chose a test value of x = 1 and found that f'(1) = -3, which is negative. Therefore, the function is decreasing on the interval (0, 2). In the interval (2, ∞), we chose a test value of x = 3 and found that f'(3) = 9, which is positive. Therefore, the function is increasing on the interval (2, ∞). In summary, the function f(x) = x³ - 3x² + 4 is increasing on the intervals (-∞, 0) and (2, ∞), and it is decreasing on the interval (0, 2). These intervals of increase and decrease provide a clear picture of the function's behavior. It tells us where the function is going uphill and downhill, which is crucial for sketching the graph and understanding the overall trend of the function. The interplay between increasing and decreasing intervals, along with the critical points, helps define the shape of the curve. Understanding these intervals is also important in various applications, such as optimization problems, where we want to find the maximum or minimum value of a function. The analysis of these intervals is a key step in building a comprehensive understanding of the function's behavior.

5. Analyzing Concavity and Inflection Points

Concavity describes the curvature of a function's graph. A function is concave up if its graph is curved upwards, and it is concave down if its graph is curved downwards. Inflection points are the points where the concavity of the function changes. To analyze concavity, we need to find the second derivative of the function, denoted as f''(x). We already found the first derivative, f'(x) = 3x² - 6x. Now, we differentiate f'(x) to find the second derivative: f''(x) = 6x - 6. To find the inflection points, we set f''(x) = 0 and solve for x: 6x - 6 = 0. Dividing both sides by 6, we get x - 1 = 0, which gives us x = 1. So, x = 1 is a potential inflection point. To determine the intervals of concavity, we analyze the sign of f''(x) in the intervals determined by the potential inflection point. The point x = 1 divides the number line into two intervals: (-∞, 1) and (1, ∞). In the interval (-∞, 1), let's choose a test value of x = 0. f''(0) = 6(0) - 6 = -6, which is negative. This means that the function is concave down on the interval (-∞, 1). In the interval (1, ∞), let's choose a test value of x = 2. f''(2) = 6(2) - 6 = 12 - 6 = 6, which is positive. This means that the function is concave up on the interval (1, ∞). Since the concavity changes at x = 1, it is indeed an inflection point. To find the y-coordinate of the inflection point, we evaluate f(1): f(1) = (1)³ - 3(1)² + 4 = 1 - 3 + 4 = 2. So, the inflection point is at (1, 2). In summary, the function f(x) = x³ - 3x² + 4 is concave down on the interval (-∞, 1) and concave up on the interval (1, ∞). It has an inflection point at (1, 2). The analysis of concavity and inflection points provides additional insights into the shape of the function's graph. It tells us how the curve is bending, whether it's curving upwards or downwards. The inflection point marks the transition between these two behaviors. This information, combined with the information about extrema and intervals of increase and decrease, allows us to create a very accurate sketch of the function's graph. Understanding concavity is also important in various applications, such as determining the stability of structures or the efficiency of processes. The second derivative provides valuable information about the rate of change of the slope of the function, offering a deeper understanding of its behavior.

6. Sketching the Graph

By synthesizing all the information we've gathered, we can now sketch the graph of f(x) = x³ - 3x² + 4. We know the following:

  • Domain: (-∞, ∞)
  • Zeros: x = -1 and x = 2 (with multiplicity 2)
  • Local Maximum: (0, 4)
  • Local Minimum: (2, 0)
  • Increasing Intervals: (-∞, 0) and (2, ∞)
  • Decreasing Interval: (0, 2)
  • Concave Down Interval: (-∞, 1)
  • Concave Up Interval: (1, ∞)
  • Inflection Point: (1, 2)

With this information, we can start by plotting the zeros, extrema, and inflection point on a coordinate plane. The zeros are x = -1 and x = 2, so we mark those points on the x-axis. The local maximum is at (0, 4), so we plot that point. The local minimum is at (2, 0), which coincides with one of the zeros, so we already have that marked. The inflection point is at (1, 2), so we plot that point as well. Next, we consider the intervals of increase and decrease. The function is increasing from (-∞, 0), reaches a local maximum at (0, 4), and then decreases from (0, 2). This tells us that the graph rises to the left of x = 0, turns around at (0, 4), and then falls until x = 2. The function is increasing again from (2, ∞), which means that the graph rises to the right of x = 2. Since x = 2 is a zero with multiplicity 2, the graph touches the x-axis at x = 2 but does not cross it. This means that the graph