Critical Point Analysis Of F(x) = X³ - 3x² + 3x - 1 At X = 1

Jul 11, 2025 by ADMIN 61 views

Understanding Critical Points of the Function f(x) = x³ - 3x² + 3x - 1

In the realm of calculus, identifying and classifying critical points of a function is a fundamental skill. Critical points provide valuable insights into the behavior of a function, including where it reaches local maxima, local minima, or exhibits other interesting characteristics. In this comprehensive exploration, we will delve into the function f(x) = x³ - 3x² + 3x - 1 and meticulously analyze its critical point located at x = 1. By employing the tools of differential calculus, we aim to definitively classify this critical point and gain a deeper understanding of the function's local behavior.

Defining Critical Points

Before we embark on our analysis, it is crucial to establish a clear understanding of what constitutes a critical point. In mathematical terms, a critical point of a function f(x) is a point in the domain of f where either the derivative of f, denoted as f'(x), is equal to zero or f'(x) is undefined. These points are of particular interest because they often correspond to locations where the function changes its direction of movement, transitioning from increasing to decreasing or vice versa. Furthermore, critical points can also indicate the presence of local extrema, which are points where the function attains a local maximum or local minimum value.

To effectively identify and classify critical points, we employ two primary techniques: the first derivative test and the second derivative test. The first derivative test analyzes the sign changes of the first derivative around the critical point, while the second derivative test examines the concavity of the function at the critical point. By combining these approaches, we can gain a comprehensive understanding of the nature of critical points.

Finding the Critical Point

Our initial task is to locate the critical points of the given function, f(x) = x³ - 3x² + 3x - 1. To achieve this, we must first determine the derivative of the function, f'(x). Applying the power rule of differentiation, we obtain:

f'(x) = 3x² - 6x + 3

Now, to find the critical points, we set the derivative equal to zero and solve for x:

3x² - 6x + 3 = 0

We can simplify this equation by dividing both sides by 3:

x² - 2x + 1 = 0

This quadratic equation can be factored as:

(x - 1)² = 0

Therefore, the only solution is:

x = 1

This indicates that the function f(x) = x³ - 3x² + 3x - 1 has a single critical point located at x = 1. Our next step is to classify this critical point using the first and second derivative tests.

Applying the First Derivative Test

The first derivative test involves analyzing the sign of the derivative, f'(x), in the intervals surrounding the critical point. This analysis helps us determine whether the function is increasing or decreasing in these intervals, providing valuable information about the nature of the critical point.

To apply the first derivative test, we consider the intervals to the left and right of the critical point x = 1. Let's examine the sign of f'(x) in these intervals:

Interval x < 1: Choose a test value, say x = 0. Then, f'(0) = 3(0)² - 6(0) + 3 = 3, which is positive. This indicates that the function is increasing in the interval x < 1.
Interval x > 1: Choose a test value, say x = 2. Then, f'(2) = 3(2)² - 6(2) + 3 = 3, which is also positive. This indicates that the function is increasing in the interval x > 1.

Since the derivative f'(x) does not change its sign at x = 1, the first derivative test suggests that the critical point at x = 1 is neither a local maximum nor a local minimum. This is because the function does not transition from increasing to decreasing or vice versa at this point.

Applying the Second Derivative Test

To further confirm our classification of the critical point, we can employ the second derivative test. This test involves evaluating the second derivative of the function, f''(x), at the critical point. The sign of the second derivative provides information about the concavity of the function at that point.

First, we need to find the second derivative of f(x). Differentiating f'(x) = 3x² - 6x + 3, we obtain:

f''(x) = 6x - 6

Now, we evaluate the second derivative at the critical point x = 1:

f''(1) = 6(1) - 6 = 0

The second derivative test is inconclusive when f''(1) = 0. This means that the test does not provide definitive information about the nature of the critical point. In such cases, we rely on the first derivative test or other analytical methods to classify the critical point.

Conclusion: Classifying the Critical Point

Based on our analysis using the first derivative test, we have determined that the critical point of the function f(x) = x³ - 3x² + 3x - 1 at x = 1 is neither a local maximum nor a local minimum. The first derivative test revealed that the function is increasing both to the left and right of x = 1, indicating that the function does not change its direction of movement at this point.

Furthermore, the second derivative test was inconclusive, as f''(1) = 0. This reinforces our reliance on the first derivative test for classifying the critical point in this case.

Therefore, we can confidently conclude that the critical point at x = 1 is an inflection point, where the concavity of the function changes. In summary, through the application of differential calculus techniques, we have successfully classified the critical point of the given function and gained valuable insights into its local behavior.

Let's dive into the fascinating world of calculus and explore the nature of critical points. Critical points, those special locations where a function's derivative equals zero or is undefined, hold valuable clues about a function's behavior. They can reveal where a function reaches its peaks (local maxima), valleys (local minima), or experiences a change in concavity. Today, we're focusing on the function f(x) = x³ - 3x² + 3x - 1 and the critical point nestled at x = 1. Our mission is to determine the type of critical point we've encountered: Is it a local maximum, a local minimum, or perhaps something else entirely?

The Quest for Critical Points: A Derivative's Tale

The first step in our exploration is to find the derivative of our function. The derivative, denoted as f'(x), provides us with the slope of the function at any given point. Critical points are where this slope momentarily flattens out, becoming zero, or where the slope becomes undefined (think sharp corners or vertical tangents). Using the power rule of differentiation, we find:

f'(x) = 3x² - 6x + 3

Now, to pinpoint our critical points, we set the derivative equal to zero and solve for x:

3x² - 6x + 3 = 0

We can simplify this equation by dividing both sides by 3, giving us:

x² - 2x + 1 = 0

This quadratic equation factors beautifully into:

(x - 1)² = 0

This reveals that our function has a single critical point at:

x = 1

This is the location we'll be investigating. But what kind of critical point is it? That's where our investigative tools, the first and second derivative tests, come into play.

The First Derivative Test: Charting the Function's Course

The first derivative test is like a compass, guiding us by analyzing the sign changes of the first derivative around our critical point. Remember, the derivative tells us whether the function is increasing (positive derivative) or decreasing (negative derivative). By observing how the derivative's sign changes as we move across x = 1, we can deduce the function's behavior.

Let's examine the intervals surrounding x = 1:

Interval x < 1: We'll pick a test value, say x = 0. Plugging this into our derivative, we get:

f'(0) = 3(0)² - 6(0) + 3 = 3

The result is positive! This indicates that the function is increasing as we approach x = 1 from the left.
Interval x > 1: Now, let's choose a test value greater than 1, such as x = 2. Evaluating the derivative:

f'(2) = 3(2)² - 6(2) + 3 = 3

Again, the derivative is positive! The function is also increasing as we move away from x = 1 to the right.

Here's the key observation: The derivative does not change sign at x = 1. The function is increasing on both sides of our critical point. This rules out the possibility of a local maximum (where the function would increase then decrease) or a local minimum (where it would decrease then increase). The first derivative test suggests that our critical point is something else.

The Second Derivative Test: Unveiling Concavity

To gain further insight, we turn to the second derivative test. The second derivative, f''(x), provides information about the concavity of the function. A positive second derivative indicates that the function is concave up (like a smile), while a negative second derivative means the function is concave down (like a frown).

First, we need to find the second derivative of our function. Differentiating f'(x) = 3x² - 6x + 3, we get:

f''(x) = 6x - 6

Now, we evaluate the second derivative at our critical point, x = 1:

f''(1) = 6(1) - 6 = 0

Uh oh! The second derivative is zero at x = 1. This means the second derivative test is inconclusive. It doesn't give us a definitive answer about the nature of the critical point. In such cases, we must rely on other methods, and our first derivative test result is our strongest clue.

The Verdict: Neither Max Nor Min

Putting it all together, we've learned that:

The function f(x) = x³ - 3x² + 3x - 1 has a critical point at x = 1.
The first derivative test showed that the function is increasing on both sides of x = 1.
The second derivative test was inconclusive.

Based on this evidence, we can confidently conclude that the critical point at x = 1 is neither a local maximum nor a local minimum. It's a point where the function momentarily flattens out, but it doesn't represent a peak or valley. This type of critical point is often referred to as an inflection point, where the concavity of the function changes.

In the fascinating field of calculus, understanding the behavior of functions is a central theme. One of the key aspects of this understanding lies in the identification and classification of critical points. Critical points are those special locations on a function's graph where the derivative is either zero or undefined. These points are significant because they often mark potential turning points, such as local maxima, local minima, or points of inflection. In this comprehensive analysis, we will focus on the function f(x) = x³ - 3x² + 3x - 1 and delve into the nature of its critical point at x = 1. By employing the tools of differential calculus, including the first and second derivative tests, we aim to definitively classify this critical point and gain a deeper understanding of the function's local behavior around this point.

Unveiling Critical Points: The Role of the Derivative

Before we embark on our investigation, it's essential to solidify our understanding of what constitutes a critical point. Mathematically speaking, a critical point of a function f(x) is a point in the domain of f where either the derivative of f, denoted as f'(x), is equal to zero or f'(x) is undefined. These points are of particular interest because they often correspond to locations where the function changes its direction of movement, transitioning from increasing to decreasing or vice versa. Moreover, critical points can also indicate the presence of local extrema, which are points where the function attains a local maximum or local minimum value.

To effectively identify and classify critical points, we rely on two primary techniques: the first derivative test and the second derivative test. The first derivative test analyzes the sign changes of the first derivative around the critical point, while the second derivative test examines the concavity of the function at the critical point. By combining these approaches, we can gain a comprehensive understanding of the nature of critical points.

Locating the Critical Point: Differentiation and Equation Solving

Our initial task is to locate the critical points of the given function, f(x) = x³ - 3x² + 3x - 1. To accomplish this, we must first determine the derivative of the function, f'(x). Applying the power rule of differentiation, a fundamental tool in calculus, we obtain:

f'(x) = 3x² - 6x + 3

Now, to find the critical points, we set the derivative equal to zero and solve for x:

3x² - 6x + 3 = 0

We can simplify this equation by dividing both sides by 3:

x² - 2x + 1 = 0

This quadratic equation can be factored as:

(x - 1)² = 0

Therefore, the only solution is:

x = 1

This result indicates that the function f(x) = x³ - 3x² + 3x - 1 has a single critical point located at x = 1. Our next crucial step is to classify this critical point, determining whether it represents a local maximum, a local minimum, or some other type of critical point. To achieve this classification, we will employ the first and second derivative tests, powerful analytical tools that provide insights into the function's behavior around the critical point.

The First Derivative Test: Analyzing Sign Changes

The first derivative test is a cornerstone technique for classifying critical points. This test involves analyzing the sign of the first derivative, f'(x), in the intervals surrounding the critical point. By examining the sign changes of the derivative, we can determine whether the function is increasing or decreasing in these intervals, thereby gaining valuable information about the nature of the critical point. If the derivative changes from positive to negative at the critical point, it suggests a local maximum. Conversely, if the derivative changes from negative to positive, it indicates a local minimum. If the derivative does not change sign, the critical point may be neither a local maximum nor a local minimum.

To apply the first derivative test to our critical point at x = 1, we consider the intervals to the left and right of x = 1. Let's carefully examine the sign of f'(x) in these intervals:

Interval x < 1: We choose a test value less than 1, such as x = 0. Then, we evaluate the derivative at this test value:

f'(0) = 3(0)² - 6(0) + 3 = 3

The result is positive, indicating that the function is increasing in the interval x < 1. This means that as we approach x = 1 from the left, the function's values are getting larger.
Interval x > 1: Next, we choose a test value greater than 1, such as x = 2. We evaluate the derivative at this test value:

f'(2) = 3(2)² - 6(2) + 3 = 3

Again, the result is positive, indicating that the function is also increasing in the interval x > 1. This means that as we move away from x = 1 to the right, the function's values are also getting larger.

Here's the crucial observation: The derivative f'(x) does not change its sign at x = 1. The function is increasing both to the left and to the right of our critical point. This observation is significant because it allows us to rule out the possibility of a local maximum or a local minimum. For a local maximum, we would expect the function to increase to the left of the critical point and decrease to the right. For a local minimum, the reverse would be true. Since the function is increasing on both sides of x = 1, the first derivative test suggests that our critical point is something else entirely.

The Second Derivative Test: Assessing Concavity

To further refine our classification of the critical point, we can employ the second derivative test. This test provides valuable information about the concavity of the function at the critical point. The concavity of a function describes whether its graph is curving upwards (concave up) or downwards (concave down). The second derivative, denoted as f''(x), is the key to assessing concavity. A positive second derivative indicates that the function is concave up, while a negative second derivative indicates that the function is concave down. If the second derivative is zero at the critical point, the test is inconclusive, and we must rely on other methods, such as the first derivative test, to classify the critical point.

To apply the second derivative test, we first need to find the second derivative of our function f(x). We differentiate the first derivative, f'(x) = 3x² - 6x + 3, to obtain:

f''(x) = 6x - 6

Now, we evaluate the second derivative at our critical point, x = 1:

f''(1) = 6(1) - 6 = 0

The result of the second derivative test is zero. This is a crucial observation because it means that the second derivative test is inconclusive in this case. When the second derivative is zero at a critical point, it doesn't provide us with definitive information about whether the critical point is a local maximum, a local minimum, or neither. In such situations, we must rely on other methods, such as the first derivative test, to classify the critical point. In our case, the first derivative test provided a clear indication that the critical point is neither a local maximum nor a local minimum.

Conclusion: Classifying the Critical Point as an Inflection Point

Drawing upon the insights gained from both the first and second derivative tests, we can now confidently classify the critical point of the function f(x) = x³ - 3x² + 3x - 1 at x = 1. The first derivative test revealed that the function is increasing both to the left and right of x = 1, indicating that the function does not change its direction of movement at this point. This eliminates the possibility of a local maximum or a local minimum. The second derivative test, on the other hand, was inconclusive, as f''(1) = 0. This reinforces our reliance on the first derivative test for classifying the critical point in this specific case.

Therefore, based on our comprehensive analysis, we conclude that the critical point at x = 1 is neither a local maximum nor a local minimum. Instead, it is a point where the function's concavity changes. Such points are known as inflection points. At an inflection point, the graph of the function transitions from curving upwards to curving downwards, or vice versa. In summary, through the rigorous application of differential calculus techniques, we have successfully classified the critical point of the given function and gained valuable insights into its local behavior. The critical point at x = 1 is an inflection point, a location where the function's concavity undergoes a transformation.