Discontinuities Of Rational Functions: A Detailed Guide

Hey guys! Ever wondered about those pesky points where a function just freaks out and becomes undefined? We're talking about discontinuities! Today, we're diving deep into identifying discontinuities, specifically for the function ${f(x) = \frac{5x}{x^3 + 5x^2 + 6x}}$. We'll check the values ${x = -3, -2, 0, 2, 3, 5}$ to see if they cause any trouble. So, buckle up, and let's get started!

Understanding Discontinuities

Before we jump into the specifics, let's make sure we're all on the same page about what a discontinuity actually is. In simple terms, a discontinuity is a point where a function is not continuous. This means that there's a break, a jump, or a hole in the graph of the function at that point. For rational functions (like the one we're dealing with), discontinuities usually occur where the denominator is equal to zero. This is because division by zero is undefined, leading to a break in the function's graph.

Why do we care about discontinuities? Well, they can tell us a lot about the behavior of a function. They can indicate asymptotes (lines that the function approaches but never touches), holes (removable discontinuities), and other interesting features. Identifying discontinuities is crucial in calculus, analysis, and various applications in physics and engineering.

Different types of discontinuities exist, but the most common ones we encounter with rational functions are:

1. Removable Discontinuities (Holes): These occur when a factor in the denominator can be canceled out by a factor in the numerator. At the ${x}$-value that makes this factor zero, there's a hole in the graph, but the function is defined everywhere else nearby.
2. Non-Removable Discontinuities (Vertical Asymptotes): These occur when a factor in the denominator cannot be canceled out. As ${x}$ approaches the ${x}$-value that makes this factor zero, the function approaches infinity (or negative infinity), creating a vertical asymptote.

Knowing these definitions will help us analyze our given function and identify its discontinuities.

Analyzing the Function

Let's take a closer look at our function: ${f(x) = \frac{5x}{x^3 + 5x^2 + 6x}}$. The first step in identifying discontinuities is to factor the denominator. Factoring helps us find the values of ${x}$ that make the denominator zero, which are the potential points of discontinuity. Let's factor that denominator:

${x^3 + 5x^2 + 6x = x(x^2 + 5x + 6) = x(x+2)(x+3)}$

So, our function can be rewritten as:

${f(x) = \frac{5x}{x(x+2)(x+3)}}$

Now we can see that the denominator is zero when ${x = 0}$, ${x = -2}$, or ${x = -3}$. These are the potential points of discontinuity. But before we declare them all as discontinuities, we need to check for removable discontinuities (holes). Notice that there's an ${x}$ in both the numerator and the denominator. This means we can cancel them out:

${f(x) = \frac{5}{ (x+2)(x+3)}}$, for (x eq 0)

Since we canceled out the ${x}$ term, there's a removable discontinuity (a hole) at ${x = 0}$. The remaining factors in the denominator, ${(x+2)}$ and ${(x+3)}$, cannot be canceled out. This means there are non-removable discontinuities (vertical asymptotes) at ${x = -2}$ and ${x = -3}$.

Alright, now we have a solid understanding of where the discontinuities are located based on the function's structure.

Checking the Given Values

Now that we've identified the potential discontinuities, let's check each of the given ${x}$ values (${x = -3, -2, 0, 2, 3, 5}$) to see if they are indeed discontinuities.

1. ${x = -3}$: As we found earlier, ${x = -3}$ makes the denominator zero and is not a removable discontinuity. Therefore, ${x = -3}$ is a non-removable discontinuity (vertical asymptote).

2. ${x = -2}$: Similarly, ${x = -2}$ makes the denominator zero and is not a removable discontinuity. Thus, ${x = -2}$ is a non-removable discontinuity (vertical asymptote).

3. ${x = 0}$: We determined that ${x = 0}$ is a removable discontinuity because the ${x}$ term canceled out. So, ${x = 0}$ is a removable discontinuity (hole).

4. ${x = 2}$: Let's plug ${x = 2}$ into the simplified function ${f(x) = \frac{5}{(x+2)(x+3)}}$: ${f(2) = \frac{5}{(2+2)(2+3)} = \frac{5}{4 \cdot 5} = \frac{1}{4}}$ Since the function is defined at ${x = 2}$, it is not a discontinuity.

5. ${x = 3}$: Let's plug ${x = 3}$ into the simplified function: ${f(3) = \frac{5}{(3+2)(3+3)} = \frac{5}{5 \cdot 6} = \frac{1}{6}}$ Since the function is defined at ${x = 3}$, it is not a discontinuity.

6. ${x = 5}$: Let's plug ${x = 5}$ into the simplified function: ${f(5) = \frac{5}{(5+2)(5+3)} = \frac{5}{7 \\\cdot 8} = \frac{5}{56}}$ Since the function is defined at ${x = 5}$, it is not a discontinuity.

Summary of Results

Okay, guys, let's recap our findings:

• ${x = -3}$: Non-Removable Discontinuity (Vertical Asymptote)
• ${x = -2}$: Non-Removable Discontinuity (Vertical Asymptote)
• ${x = 0}$: Removable Discontinuity (Hole)
• ${x = 2}$: Not a Discontinuity
• ${x = 3}$: Not a Discontinuity
• ${x = 5}$: Not a Discontinuity

So, there you have it! We've successfully identified the discontinuities of the given function at the specified ${x}$ values. Understanding these concepts is crucial for grasping the behavior of rational functions. Keep practicing, and you'll become a discontinuity-detecting pro in no time!