Function Analysis: F(x) = X / (3x + 9) At X = -3

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Function Analysis: f(x) = x / (3x + 9) at x = -3

Let's dive into analyzing the function f(x) = x / (3x + 9), particularly its behavior at x = -3. This involves understanding what happens as x approaches -3, whether the function is defined at that point, and if not, what kind of discontinuity we might encounter. So, grab your calculators (or don't, we'll try to keep it simple), and let's get started!

Understanding the Function

Before we zoom in on x = -3, let's get a feel for the function itself. The function f(x) is a rational function, which means it's a ratio of two polynomials. In this case, the numerator is simply x, and the denominator is 3x + 9. Rational functions are generally well-behaved, except where the denominator equals zero, because division by zero is a big no-no in the math world. Identifying where the denominator is zero is crucial, as this typically indicates a vertical asymptote or a removable discontinuity.

To find where the denominator 3x + 9 is zero, we set it equal to zero and solve for x:

3x + 9 = 0 3x = -9 x = -3

Ah, ha! We've already pinpointed that x = -3 is a critical point. This confirms that the function is undefined at x = -3 because the denominator becomes zero. Now, the big question is: what kind of undefined behavior are we looking at? Is it a vertical asymptote, a hole (removable discontinuity), or something else entirely? Keep reading, guys!

Analyzing the Behavior at x = -3

Okay, so we know that f(x) is undefined at x = -3. But what kind of undefined is it? To figure this out, we need to investigate what happens to the function as x gets really, really close to -3 from both sides – from values slightly less than -3 (the left-hand limit) and from values slightly greater than -3 (the right-hand limit). This will give us clues about whether the function is heading towards infinity (a vertical asymptote) or approaching a specific value (a removable discontinuity).

Checking for a Vertical Asymptote

A vertical asymptote occurs when the function's value zooms off towards positive or negative infinity as x approaches a certain value. To determine if we have a vertical asymptote at x = -3, we need to evaluate the left-hand and right-hand limits:

Limit as x approaches -3 from the left (x < -3):

As x approaches -3 from the left, x takes on values like -3.1, -3.01, -3.001, and so on. Let's examine what happens to the numerator and denominator:

  • Numerator (x): Approaches -3.
  • Denominator (3x + 9): Approaches 0 from the negative side (e.g., 3(-3.1) + 9 = -0.3).

So, we have a negative number divided by a very small negative number. A negative divided by a negative is positive, and a number divided by a very small number tends towards infinity. Therefore, the limit as x approaches -3 from the left is positive infinity.

Limit as x approaches -3 from the right (x > -3):

As x approaches -3 from the right, x takes on values like -2.9, -2.99, -2.999, and so on. Again:

  • Numerator (x): Approaches -3.
  • Denominator (3x + 9): Approaches 0 from the positive side (e.g., 3(-2.9) + 9 = 0.3).

In this case, we have a negative number divided by a very small positive number. This results in negative infinity. Therefore, the limit as x approaches -3 from the right is negative infinity.

Since the limits from the left and right are infinite (albeit with different signs), we can confidently conclude that there is a vertical asymptote at x = -3. The function shoots off to positive infinity on one side and negative infinity on the other. Wowza!

Checking for Removable Discontinuity (a Hole)

A removable discontinuity, often called a hole, occurs when the limit of the function exists at a certain point, but the function is not defined at that point. This usually happens when we can cancel out a common factor from the numerator and denominator. In our case, let's see if we can simplify the function:

f(x) = x / (3x + 9) f(x) = x / (3(x + 3))

Unfortunately, there are no common factors between the numerator (x) and the denominator (3(x + 3)) that we can cancel out. This means that we cannot simplify the function to remove the discontinuity at x = -3. Therefore, there is no removable discontinuity or "hole" in the graph at that point.

Conclusion

Alright, guys, we've thoroughly investigated the behavior of the function f(x) = x / (3x + 9) at x = -3. Our analysis revealed the following:

  1. The function is undefined at x = -3 because the denominator becomes zero.
  2. There is a vertical asymptote at x = -3, as the function approaches positive infinity from the left and negative infinity from the right.
  3. There is no removable discontinuity (hole) at x = -3 because we cannot simplify the function to eliminate the problematic factor in the denominator.

So, in a nutshell, the function has a wild, infinite party happening at x = -3! Understanding these concepts is essential for working with rational functions and calculus in general. Keep practicing, keep exploring, and you'll become a math whiz in no time!

Additional Considerations

While we've focused on the behavior at x = -3, it's always good to consider the broader picture. Let's briefly touch upon a couple of other relevant aspects of the function.

Horizontal Asymptote

To find the horizontal asymptote (the value that f(x) approaches as x goes to positive or negative infinity), we look at the degrees of the polynomials in the numerator and denominator. In this case:

  • Degree of numerator (x): 1
  • Degree of denominator (3x + 9): 1

Since the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is 1, and the leading coefficient of the denominator is 3. Therefore, the horizontal asymptote is y = 1/3. This means that as x gets extremely large (positive or negative), the function f(x) approaches the value 1/3.

Intercepts

Let's find the x and y-intercepts to get a better sense of where the function crosses the axes.

  • x-intercept: This occurs when f(x) = 0. So, we solve x / (3x + 9) = 0. A fraction is zero only when the numerator is zero. Therefore, x = 0. The x-intercept is at the origin (0, 0).
  • y-intercept: This occurs when x = 0. So, we evaluate f(0) = 0 / (3(0) + 9) = 0 / 9 = 0. The y-intercept is also at the origin (0, 0).

In this particular case, both intercepts coincide at the origin.

Graphing the Function

Putting all this information together, we can sketch a graph of the function. The graph will have a vertical asymptote at x = -3, a horizontal asymptote at y = 1/3, and will pass through the origin (0, 0). As x approaches -3 from the left, the graph will shoot upwards towards positive infinity. As x approaches -3 from the right, the graph will plummet downwards towards negative infinity. As x goes to positive or negative infinity, the graph will approach the line y = 1/3.

This comprehensive analysis should give you a solid understanding of the function f(x) = x / (3x + 9), especially its intriguing behavior at x = -3. Remember to always consider the domain, asymptotes, intercepts, and limits when analyzing any function. Keep up the awesome work!