Find X In The Domain Of F(x) = Sqrt(x-5)

Hey there, math whizzes and Plastik Magazine readers! Today, we're diving into a super common but sometimes tricky topic: finding the domain of a function, specifically when it involves a square root. You know, those problems that pop up in algebra and calculus, asking "Which value of $x$ is in the domain of $f(x)=\sqrt{x-5}$?" Don't sweat it, guys, we're going to break it down piece by piece so it all makes perfect sense. We'll explore the options A. $x=-5$, B. $x=0$, C. $x=5$, and D. $x=1$, and figure out which one actually works for this function. Get ready to flex those brain muscles, because understanding domains is key to unlocking a whole universe of mathematical possibilities. We'll be talking about why certain numbers work and others don't, all while keeping it fun and engaging. So grab your favorite thinking cap, maybe a snack, and let's get this math party started!

Understanding the Domain of a Square Root Function

Alright guys, let's get down to the nitty-gritty of why the domain of a square root function is such a big deal. When we're dealing with a function like $f(x)=\sqrt{x-5}$, the most important thing to remember is that we cannot take the square root of a negative number in the realm of real numbers. This is the golden rule, the non-negotiable principle that governs our entire investigation. Think about it – have you ever tried to find a real number that, when multiplied by itself, gives you a negative result? It's impossible! For instance, $2 \times 2 = 4$, and $(-2) \times (-2) = 4$ too. So, any number inside that square root symbol, also known as the radicand, must be zero or positive. This is where the concept of the domain comes into play. The domain is simply the set of all possible input values (the $x$-values) for which the function is defined and produces a real number output. For $f(x)=\sqrt{x-5}$, we need to ensure that the expression inside the square root, which is $x-5$, is greater than or equal to zero. Mathematically, we write this as $x-5 \ge 0$. This inequality is our key to unlocking the domain of this particular function. It's like a secret code that tells us which $x$-values are allowed to play in the sandbox of our function. We'll be manipulating this inequality shortly, but for now, just internalize this fundamental rule: radicand $\ge$ 0. This simple rule is the foundation upon which we build our understanding of square root functions and their domains. It's the first step in solving our problem and ensures that we're always working within the boundaries of real numbers. So, keep that in mind as we move forward, because it's the bedrock of our entire discussion.

Solving the Inequality for the Domain

Now that we've established the cardinal rule for square root functions – the expression inside the square root must be non-negative – let's apply it directly to our function, $f(x)=\sqrt{x-5}$. We need to solve the inequality we derived: $x-5 \ge 0$. This is a pretty straightforward linear inequality, and solving it is a piece of cake, guys. To isolate $x$, we simply need to add 5 to both sides of the inequality. When we do that, we get: $x - 5 + 5 \ge 0 + 5$, which simplifies to $x \ge 5$. This inequality, $x \ge 5$, tells us everything we need to know about the domain of $f(x)=\sqrt{x-5}$. It means that any value of $x$ that is greater than or equal to 5 is a valid input for this function. Conversely, any value of $x$ that is less than 5 will result in a negative number inside the square root, making the function undefined in the real number system. So, the domain of $f(x)=\sqrt{x-5}$ is the set of all real numbers greater than or equal to 5. We can express this in interval notation as $[5, \infty)$. The square bracket at 5 indicates that 5 is included in the domain, and the infinity symbol with a parenthesis indicates that we go on forever in the positive direction. This is the mathematical definition of the domain for our function. It's the set of all valid $x$'s. Now that we have this definitive condition, we can easily check the given options to see which one fits.

Evaluating the Given Options

Okay, team, we've done the heavy lifting by figuring out that for $f(x)=\sqrt{x-5}$ to be defined in the real numbers, we must have $x \ge 5$. Now comes the fun part: checking our multiple-choice options to see which one satisfies this condition. Let's take them one by one:

• A. $x = -5$: Is $-5 \ge 5$? Nope, definitely not! If we plug in $-5$ into the function, we get $f(-5) = \sqrt{-5 - 5} = \sqrt{-10}$. As we've hammered home, we can't take the square root of a negative number and get a real result. So, $x=-5$ is not in the domain.
• B. $x = 0$: Is $0 \ge 5$? Again, no way, José! Plugging $0$ in gives us $f(0) = \sqrt{0 - 5} = \sqrt{-5}$. This is also undefined in the real numbers. So, $x=0$ is not in the domain.
• C. $x = 5$: Is $5 \ge 5$? Yes, it is! $5$ is equal to $5$, so it satisfies the condition. Let's check it in the function: $f(5) = \sqrt{5 - 5} = \sqrt{0}$. And what's the square root of $0$? It's $0$! Since we get a real number output (0), $x=5$ is in the domain of the function.
• D. $x = 1$: Is $1 \ge 5$? Absolutely not. Plugging $1$ in yields $f(1) = \sqrt{1 - 5} = \sqrt{-4}$. Another negative under the radical, so $x=1$ is not in the domain.

See? By systematically checking each option against our derived condition ($x \ge 5$), we can definitively determine which value of $x$ is valid. It's all about following the rules and applying them consistently. The process is clear, and the answer becomes obvious once you've done the initial setup.

Conclusion: The Correct Value of x

So, after dissecting the problem and applying the fundamental rule of square root functions, we've arrived at the answer. The domain of $f(x)=\sqrt{x-5}$ requires that the expression under the square root, $x-5$, must be greater than or equal to zero. Solving the inequality $x-5 \ge 0$ gives us $x \ge 5$. This means any number that is 5 or larger is a valid input for our function. When we examined the given options – A. $x=-5$, B. $x=0$, C. $x=5$, and D. $x=1$ – only one satisfied the condition $x \ge 5$. That lucky number is C. $x=5$. At $x=5$, the function gives us $f(5) = \sqrt{5-5} = \sqrt{0} = 0$, which is a perfectly valid real number. For all other options, plugging them into the function would result in taking the square root of a negative number, which is a no-go in the world of real numbers. Understanding domains like this isn't just about passing tests, guys; it's about understanding the behavior of functions and the limitations they have. It's a crucial building block for more advanced math topics, so give yourselves a pat on the back for tackling this! Keep practicing, keep questioning, and you'll master these concepts in no time. Math is all about logical steps, and once you know the rules, the solutions reveal themselves. Awesome job, everyone!