Function Analysis: Y = 1/(x-5)

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the world of mathematics, specifically tackling a super common question that pops up in algebra: determining if a relation defines a function and finding its domain and range. We're going to break down the equation $y = \frac{1}{x-5}$ step-by-step, so by the end of this, you'll be a pro at this stuff. Let's get started!

Does This Relation Define a Function?

So, the first big question is: does the relation $y = \frac{1}{x-5}$ give us a function? In simple terms, a function is like a rule that says for every input ($x$-value), there's exactly one output ($y$-value). Think of it like a vending machine – you press one button (input), and you get one specific snack (output). You don't want to press the same button and get two different snacks, right? That would be chaos! To figure this out for our equation, we need to see if any single $x$-value can produce more than one $y$-value. Let's examine $y = \frac{1}{x-5}$. For any given number we plug in for $x$ (as long as it's allowed, which we'll get to with the domain), we will always get a single, unique value for $y$. For example, if $x=6$, then $y = \frac{1}{6-5} = \frac{1}{1} = 1$. If $x=4$, then $y = \frac{1}{4-5} = \frac{1}{-1} = -1$. There's no $x$ value that will spit out two different $y$ values. The only thing that could potentially cause an issue is if the denominator becomes zero, leading to an undefined result, but even in that case, it just means there's no $y$ value for that specific $x$, not that there are multiple $y$ values. Therefore, yes, the relation $y = \frac{1}{x-5}$ does define $y$ as a function of $x$. This is a crucial first step, and it's all about that one-to-one correspondence between inputs and outputs.

What is the Domain? (Interval Notation)

Alright, now that we've established it's a function, let's talk about its domain. The domain is basically the set of all possible $x$-values that you can plug into the function without breaking it. Think of it as the allowed inputs. For rational functions (those with a fraction where the variable is in the denominator), the biggest no-no is dividing by zero. So, we need to find the $x$-value that makes the denominator of $y = \frac{1}{x-5}$ equal to zero and then exclude it from our possible $x$-values. The denominator is $x-5$. We set it to zero: $x - 5 = 0$. Solving for $x$, we get $x = 5$. This means that $x=5$ is the only value that is not allowed in our domain because it would make the expression undefined. All other real numbers are perfectly fine. So, our domain includes all real numbers except for 5. How do we write this in interval notation, which is what they love in math class? We represent all numbers less than 5 using the interval $(-\infty, 5)$ and all numbers greater than 5 using the interval $(5, \infty)$. Since we want to include all numbers except 5, we combine these two intervals using the union symbol. So, the domain in interval notation is $(-\infty, 5) \cup (5, \infty)$. This tells us we can use any number from negative infinity up to (but not including) 5, and any number from (but not including) 5 up to positive infinity. Pretty neat, huh?

What is the Range? (Interval Notation)

Last but not least, let's figure out the range of this function. The range is the set of all possible output $y$-values that the function can produce. It's what comes out of the machine, so to speak. For $y = \frac{1}{x-5}$, we need to think about what values $y$ can actually take. We know that $y$ can never be zero. Why? Because for $y$ to be zero, the numerator of the fraction would have to be zero (rac{0}{\text{something}} = 0). However, our numerator is a constant, 1. It's never going to be zero! So, $y$ can never equal 0. Can $y$ be any other number? Let's consider it. If we want $y$ to be some specific non-zero value, say $k$, can we find an $x$ that makes it happen? We set $k = \frac{1}{x-5}$. If $k \neq 0$, we can multiply both sides by $(x-5)$ to get $k(x-5) = 1$. Then, we can divide by $k$ (since $k \neq 0$) to get $x-5 = \frac{1}{k}$. Finally, we can solve for $x$: $x = 5 + \frac{1}{k}$. Since $k$ can be any non-zero real number, $\frac{1}{k}$ can also be any non-zero real number. This means that $x$ can be found for any non-zero $y$. Therefore, the function can produce any $y$-value except for 0. Just like with the domain, we need to express this in interval notation. We want all numbers less than 0, which is $(-\infty, 0)$, and all numbers greater than 0, which is $(0, \infty)$. Combining these with the union symbol gives us the range: $(-\infty, 0) \cup (0, \infty)$. So, the function can output any real number except zero. It's awesome how we can determine these boundaries just by looking at the equation!

Putting It All Together

So, to wrap it up, guys: The relation $y = \frac{1}{x-5}$ is indeed a function because each $x$-value (that's allowed) gives only one $y$-value. Its domain, the set of all allowed $x$-values, is all real numbers except 5, written as $(-\infty, 5) \cup (5, \infty)$. And its range, the set of all possible $y$-values, is all real numbers except 0, written as $(-\infty, 0) \cup (0, \infty)$. Understanding these concepts of functions, domains, and ranges is super fundamental in mathematics, and once you get the hang of it, you'll see them popping up everywhere. Keep practicing, and don't be afraid to experiment with different equations. That's all for this one, Plastik Magazine readers! Catch you in the next article!