Analyzing The Rational Function F(x) = X²/(x-1)

Hey math enthusiasts! Let's dive into a detailed analysis of the rational function f(x) = x²/(x-1). We're going to explore its domain, asymptotes, intervals of increase and decrease, concavity, and critical points. This comprehensive breakdown will help you understand how to dissect and interpret a rational function's behavior. So, grab your coffee, and let's get started, guys!

Domain of the Function

First things first, let's tackle the domain of the function. The domain represents all possible input values (x-values) for which the function is defined. For rational functions, we need to watch out for values that make the denominator equal to zero because division by zero is undefined. In our case, the denominator is (x - 1). So, we need to find the x-value that makes (x - 1) = 0. That's easy peasy, it's x = 1. Therefore, x = 1 is excluded from the domain.

So, the domain of f(x) = x²/(x - 1) is all real numbers except x = 1. We can express this in a few ways: using set notation as {x | x ∈ ℝ, x ≠ 1}, in interval notation as (-∞, 1) ∪ (1, ∞), or by simply stating that x can be any real number, but NOT 1. Understanding the domain is crucial because it immediately tells us where the function is defined and where it's not. This is the foundation upon which we build our understanding of the function's behavior. Think of it as the playing field; we know the boundaries from the get-go. Keep in mind, the domain impacts everything else we'll analyze, like asymptotes and intervals of increase/decrease.

Asymptotes: Vertical, Horizontal, and Oblique

Now, let's move on to asymptotes. Asymptotes are lines that the graph of a function approaches but never touches. They give us valuable insight into the function's long-term behavior. We'll be looking for vertical, horizontal, and oblique (slant) asymptotes.

Vertical Asymptotes

Vertical asymptotes occur where the function is undefined, which we've already identified in the domain section. The denominator becomes zero. In our function, f(x) = x²/(x - 1), the denominator is (x - 1). Setting this equal to zero gives us x = 1. So, we have a vertical asymptote at x = 1. This means that as x approaches 1 from the left or the right, the function's value shoots off towards positive or negative infinity. The graph will get infinitely close to the line x = 1, but it will never actually touch it.

Horizontal Asymptotes

Horizontal asymptotes describe the function's behavior as x approaches positive or negative infinity. To find them, we can examine the limit of the function as x approaches infinity. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. If the degree is less, the horizontal asymptote is y=0. If the degree is equal, you can find the horizontal asymptote by dividing the leading coefficients. In our case, the numerator is x² (degree 2), and the denominator is (x - 1) (degree 1). Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

Oblique Asymptotes

Since we don't have a horizontal asymptote (because the degree of the numerator is greater than the degree of the denominator), we should investigate if there's an oblique (or slant) asymptote. Oblique asymptotes occur when the degree of the numerator is exactly one more than the degree of the denominator. We find them through polynomial long division or synthetic division. Let's do some long division:

Divide x² by (x - 1).

x - 1 | x²

      x² - x
      ------
          x
          x - 1
          -----
            1

This gives us a quotient of x + 1 and a remainder of 1. So, we can rewrite our function as: f(x) = x + 1 + 1/(x - 1). As x approaches positive or negative infinity, the term 1/(x - 1) approaches zero. This means that the function approaches the line y = x + 1. Therefore, we have an oblique asymptote at y = x + 1. The graph of the function will get infinitely close to the line y = x + 1 as x goes towards positive or negative infinity.

Intervals of Increase and Decrease and Critical Points

Alright, let's find out where our function is increasing or decreasing. This involves using the first derivative. We'll use the quotient rule to find f'(x).

The quotient rule states: If f(x) = u(x)/v(x), then f'(x) = (u'(x)v(x) - u(x)v'(x))/v(x)². In our case, u(x) = x² and v(x) = x - 1.

So, u'(x) = 2x and v'(x) = 1.

Applying the quotient rule: f'(x) = (2x(x - 1) - x²(1))/(x - 1)²

Simplify: f'(x) = (2x² - 2x - x²)/(x - 1)² f'(x) = (x² - 2x)/(x - 1)²

Now, to find the critical points (where the function's rate of change might be zero or undefined), we set f'(x) = 0 and solve for x.

(x² - 2x)/(x - 1)² = 0

x² - 2x = 0 x(x - 2) = 0

This gives us x = 0 and x = 2. These are potential critical points. Also, remember that the derivative is undefined where the original function is undefined, i.e., at x = 1. Therefore, x = 0, x = 1, and x = 2 are our key points to analyze.

Now, we'll make a sign chart for f'(x) to determine the intervals of increase and decrease.

From this sign chart:

• The function is increasing on the interval (-∞, 0) and (2, ∞).
• The function is decreasing on the interval (0, 1) and (1, 2).

So, we have a local maximum at x = 0. To find the y-value, plug x = 0 into the original function: f(0) = 0²/(0 - 1) = 0. So, the local maximum is at the point (0, 0).

We also have a local minimum at x = 2. To find the y-value, plug x = 2 into the original function: f(2) = 2²/(2 - 1) = 4. So, the local minimum is at the point (2, 4).

Concavity and Points of Inflection

To determine the concavity of the function, we need to analyze the second derivative, f''(x). Let's find it. Remember, f'(x) = (x² - 2x)/(x - 1)². We will use the quotient rule again.

Let u(x) = x² - 2x and v(x) = (x - 1)². Then u'(x) = 2x - 2 and v'(x) = 2(x - 1).

Applying the quotient rule: f''(x) = ((2x - 2)(x - 1)² - (x² - 2x) * 2(x - 1))/((x - 1)⁴)

Simplify: f''(x) = (2(x - 1)³ - 2(x² - 2x)(x - 1))/((x - 1)⁴) f''(x) = (2(x - 1)[(x - 1)² - (x² - 2x)])/((x - 1)⁴) f''(x) = (2(x² - 2x + 1 - x² + 2x))/((x - 1)³) f''(x) = 2/((x - 1)³)

Now, let's analyze the sign of f''(x) to determine the intervals of concavity.

• f''(x) > 0, the graph is concave up.
• f''(x) < 0, the graph is concave down.

Remember that the second derivative is undefined at x = 1.

• The function is concave down on the interval (-∞, 1).
• The function is concave up on the interval (1, ∞).

Since the concavity changes at x = 1, and the function isn't defined at x = 1, there are no points of inflection. The concavity changes at the vertical asymptote, which isn't a point on the curve.

Summary and Graphing

So, to recap our analysis of f(x) = x²/(x - 1):

• Domain: All real numbers except x = 1.
• Vertical Asymptote: x = 1
• Oblique Asymptote: y = x + 1
• Intervals of Increase: (-∞, 0) and (2, ∞).
• Intervals of Decrease: (0, 1) and (1, 2).
• Local Maximum: (0, 0).
• Local Minimum: (2, 4).
• Concave Down: (-∞, 1)
• Concave Up: (1, ∞)
• No Points of Inflection

With this information, you can now accurately sketch the graph of the function. You'll see the vertical asymptote at x = 1, the oblique asymptote, the local maximum, the local minimum, and the changes in concavity. This detailed analysis gives you a complete picture of the function's behavior. It's like having a map to navigate the function's terrain. Nice job, guys!