Find The Domain Of F(x)=sqrt(-x+2)

Hey guys! Welcome back to Plastik Magazine, where we break down all things awesome, including some tricky math problems. Today, we're diving deep into the world of functions and tackling a question that might have popped up in your math class: What is the domain of the function $f(x)=\sqrt{-x+2}$? Now, before we get bogged down in symbols and technicalities, let's get real about what the 'domain' of a function actually means. Think of it as the VIP list for your function's input. It's the set of all possible values that you can plug into the function (the 'x' values) without breaking it or getting some weird, undefined result. For $f(x)=\sqrt{-x+2}$, we're looking for all the 'x' values that make this square root happy. And what makes a square root happy? It needs a non-negative number inside it! You can't take the square root of a negative number and get a real number answer, right? That's like trying to find a unicorn – theoretically possible in some abstract sense, but not in the real numbers we usually deal with in these problems. So, for $f(x)=\sqrt{-x+2}$ to give us a real number output, the stuff inside the square root, which is $-x+2$, must be greater than or equal to zero. This is the fundamental rule we need to follow. We're essentially solving the inequality $-x+2 \geq 0$. This inequality is the key to unlocking the domain. We need to isolate 'x' to see what values it can take. Let's do some algebra, shall we? Subtract 2 from both sides: $-x \geq -2$. Now, here's a crucial step that trips a lot of people up: when you multiply or divide an inequality by a negative number, you have to flip the inequality sign. So, multiplying both sides by -1 (or dividing by -1) gives us $x \leq 2$. Boom! This inequality, $x \leq 2$, tells us that for the function $f(x)=\sqrt{-x+2}$ to produce a real number, 'x' has to be less than or equal to 2. This is our domain, expressed in set notation as ${x: x \leq 2}$. So, if you're looking at the options provided (A, B, C, D), you'll see that option C matches our findings precisely. Remember, understanding the domain is super important because it defines the boundaries within which your function operates. It's like knowing the speed limit on a road; you need to stay within those limits to keep things running smoothly. Keep practicing, and these concepts will become second nature. Don't forget to check out more math breakdowns right here at Plastik Magazine!

Understanding the Core Concept: What is a Function's Domain?

Alright, let's rewind a bit and make sure we're all on the same page about what a function's domain actually is. In simple terms, guys, the domain is just the set of all possible input values (usually represented by 'x') for which the function is defined and produces a real number output. Think of a function as a machine. You put something in (the input), and the machine gives you something back (the output). The domain is simply all the stuff that this particular machine is designed to accept. If you try to put something into the machine that it wasn't designed for, it might break, or it might just spit out an error message – in math terms, this means the function is undefined for that input. For our specific function, $f(x)=\sqrt{-x+2}$, the 'machine' is the square root operation. Now, not all numbers play nicely with square roots. The core rule here is that you cannot take the square root of a negative number and get a real number. So, if the expression inside the square root, $-x+2$, turns out to be negative, our function $f(x)$ is undefined in the realm of real numbers. This is why we need to find the values of 'x' that prevent $-x+2$ from being negative. We want $-x+2$ to be zero or positive. This is the same as saying $-x+2 \geq 0$. This inequality is the engine that drives our domain calculation. It's the fundamental condition that 'x' must satisfy for $f(x)$ to be a valid, real-valued output. If we were dealing with complex numbers, the domain might be different, but in standard high school and introductory college math, we stick to real numbers unless specified otherwise. So, the entire process hinges on solving this inequality correctly. Remember those algebra basics? We need to isolate 'x'. First, let's move the constant term to the other side. Subtracting 2 from both sides of $-x+2 \geq 0$ gives us $-x \geq -2$. Now, here's the common stumbling block: dealing with the negative sign in front of 'x'. To get 'x' by itself, we have to divide both sides by -1. But here's the golden rule you absolutely must remember: when you multiply or divide both sides of an inequality by a negative number, you must reverse the direction of the inequality sign. So, $-x \geq -2$ becomes $x \leq 2$ after we divide by -1 and flip the sign. This is a super important step! If you forget to flip the sign, you'll get the wrong answer. So, the domain of $f(x)=\sqrt{-x+2}$ is all real numbers 'x' such that $x \leq 2$. This means any number less than or equal to 2 – like 2, 1, 0, -5, -100, etc. – is a valid input for this function. If you tried to plug in a number greater than 2, say $x=3$, you'd get $f(3)=\sqrt{-3+2} = \sqrt{-1}$, which is not a real number. Understanding this boundary is crucial for graphing the function and for many other mathematical applications. It's all about keeping the function 'happy' and giving us real number results.

Step-by-Step Solution: Solving the Inequality

Let's break down how we actually get to the domain, step by step, for $f(x)=\sqrt{-x+2}$. The core idea, as we've discussed, is that the expression inside the square root must be non-negative. That is, the radicand must be greater than or equal to zero. In our case, the radicand is $-x+2$. So, we set up the inequality:

$-x + 2 \geq 0$

Our goal now is to isolate 'x' to find the range of values it can take. It's like trying to get a clear answer from a chatty friend – you need to focus on the main point!

1. Isolate the term with 'x': We want to get the '-x' term by itself on one side of the inequality. To do this, we subtract 2 from both sides: $-x + 2 - 2 \geq 0 - 2$ This simplifies to: $-x \geq -2$

2. Solve for 'x': Now we have '-x' which is greater than or equal to -2. To find the value of 'x', we need to get rid of the negative sign. We do this by dividing both sides by -1. This is the critical step, guys! Remember the golden rule of inequalities: when you multiply or divide by a negative number, you must flip the inequality sign. rac{-x}{-1} \leq rac{-2}{-1} (See how the '$\\geq$' flipped to '$\\leq$'? That's the magic!) This gives us: $x \leq 2$

And there you have it! The solution to the inequality is $x \leq 2$. This means that any real number that is less than or equal to 2 will result in a non-negative value inside the square root, thus making our function $f(x)=\sqrt{-x+2}$ defined in the real number system. So, the domain of the function is all values of x such that x is less than or equal to 2. This is typically written in set-builder notation as ${x: x \leq 2}$.

Connecting the Domain to the Options Provided

Now that we've done the hard work and figured out the domain of $f(x)=\sqrt{-x+2}$ is $x \leq 2$, let's look back at the multiple-choice options given in the problem. We need to find the option that perfectly matches our result. The options are:

A. ${x: x \geq -2}$ B. ${x: -2 \leq x \leq 2}$ C. ${x: x \leq 2}$ D. ${x: 0 \leq x \leq 2}$

Let's analyze each one:

• Option A: ${x: x \geq -2}$ This says that 'x' must be greater than or equal to -2. For example, if $x=0$, which is $\geq -2$, then $f(0)=\sqrt{-0+2}=\sqrt{2}$, which is valid. However, if $x=3$, which is also $\geq -2$, then $f(3)=\sqrt{-3+2}=\sqrt{-1}$, which is not a real number. So, this option is incorrect because it includes values of x (like 3) that are outside the domain.

• Option B: ${x: -2 \leq x \leq 2}$ This option restricts 'x' to be between -2 and 2, inclusive. While it correctly includes many valid inputs, it's too restrictive. For instance, $x=-5$ is less than -2. Let's test it: $f(-5)=\sqrt{-(-5)+2} = \sqrt{5+2} = \sqrt{7}$. This is a perfectly valid real number! Since this option excludes numbers like -5, it's not the complete domain. So, Option B is incorrect.

• Option C: ${x: x \leq 2}$ This option states that 'x' must be less than or equal to 2. This is exactly what we found through our inequality solving: $x \leq 2$. This means any number, no matter how small (like -100, -1000, etc.), as long as it's 2 or less, will work. For example, if $x=2$, $f(2)=\sqrt{-2+2}=\sqrt{0}=0$, which is valid. If $x=-10$, $f(-10)=\sqrt{-(-10)+2}=\sqrt{10+2}=\sqrt{12}$, which is also valid. This matches our derived domain perfectly. This is our winner!

• Option D: ${x: 0 \leq x \leq 2}$ Similar to option B, this option restricts 'x' to a specific range, this time between 0 and 2, inclusive. It correctly identifies some valid inputs, but it unnecessarily excludes all negative numbers. As we saw when testing Option B, negative numbers like $x=-5$ are perfectly valid inputs for this function. Therefore, Option D is incorrect because it doesn't include all possible valid inputs.

So, after carefully analyzing the function and the provided choices, Option C is the correct answer. It accurately represents the set of all real numbers for which $f(x)=\sqrt{-x+2}$ is defined.