Function Domain Calculation: F(x) = √(3-x) / √(2x+1)
Hey guys! Let's dive into calculating the domain of a function. It might sound intimidating, but we'll break it down step by step so it’s super clear. We're tackling the function F(x) = √(3-x) / √(2x+1) today. Finding the domain means figuring out all the possible 'x' values that won't make our function explode or do anything funky, like trying to take the square root of a negative number or dividing by zero. So, let’s jump right in and make sure we get this down! Understanding this concept is crucial for more advanced math topics, and it’s a fantastic way to sharpen our problem-solving skills.
Understanding the Domain of a Function
Okay, so what exactly is the domain of a function? Simply put, the domain is the set of all possible input values (x-values) that will produce a valid output (y-value). Think of it like this: if you're baking a cake, the domain would be the ingredients you can use without ruining the recipe. In math, certain operations have restrictions. For example, we can't divide by zero because it leads to an undefined result. Similarly, we can't take the square root of a negative number within the realm of real numbers. These restrictions are what we need to consider when determining the domain.
When we're dealing with functions, especially ones involving square roots and fractions, these restrictions become super important. We need to make sure that the values under the square root are non-negative (greater than or equal to zero) and that the denominator in a fraction is never zero. Identifying and addressing these constraints is key to correctly defining the domain of any given function. So, let’s keep this in mind as we move forward – restrictions are our friends, guiding us to the right answer!
Restrictions Imposed by Square Roots
Let's talk square roots. Square roots are like the superheroes of mathematical restrictions! They swoop in and remind us that we can only deal with non-negative numbers under the radical (the √ symbol). Why is this? Well, think about it: what number multiplied by itself gives you a negative result? There isn't one in the real number system! That's why the expression inside the square root must always be greater than or equal to zero.
This restriction is a big deal when we're finding the domain of a function. If a function includes a square root, we have to make sure that the expression inside the root is not negative. For example, in the function √(x - 2), the expression 'x - 2' must be greater than or equal to zero. This gives us the inequality x - 2 ≥ 0, which we can solve to find the valid x-values. Ignoring this restriction would lead us to include values in the domain that actually make the function undefined, and nobody wants that! So, always keep a close eye on those square roots – they're the guardians of valid function inputs.
Restrictions Imposed by Division
Now, let's chat about division – another mathematical operation with a crucial restriction. The golden rule of division is: never divide by zero! It's like the mathematical equivalent of crossing the streams in Ghostbusters – total chaos! Dividing by zero results in an undefined expression, and it throws a wrench into our function's output. So, whenever we see a fraction in a function, we need to make absolutely sure that the denominator (the bottom part of the fraction) never equals zero.
This means we have to identify any x-values that would make the denominator zero and exclude them from the domain. For instance, in the function 1/(x + 3), the denominator is 'x + 3'. We need to make sure x + 3 ≠ 0, which means x cannot be -3. If x were -3, we’d be dividing by zero, and that's a big no-no. So, always check the denominator, guys! It’s a critical step in finding the domain, especially for rational functions (functions that are fractions with polynomials).
Analyzing the Function F(x) = √(3-x) / √(2x+1)
Alright, let’s get our hands dirty and analyze the function F(x) = √(3-x) / √(2x+1). We’ve talked about the restrictions imposed by square roots and division, and guess what? This function has both! That means we need to be extra careful to make sure we’re considering all the potential pitfalls. Our mission is to find the values of 'x' that make this function happy and well-defined. So, let’s break it down and see what conditions 'x' needs to meet.
First, we have a square root in the numerator: √(3-x). This means the expression inside the square root, '3-x', must be greater than or equal to zero. If '3-x' were negative, we’d be trying to take the square root of a negative number, which is a no-go in the real number system. Second, we have a square root in the denominator: √(2x+1). This gives us two restrictions. Just like the numerator, the expression inside this square root, '2x+1', must be greater than or equal to zero. But there's more! Since this square root is in the denominator, we also need to make sure that it doesn't equal zero, because we can't divide by zero. So, '2x+1' must be strictly greater than zero.
By identifying these restrictions, we’ve set the stage for solving the inequalities that will define our domain. It’s like setting up the puzzle pieces before we start putting them together. So, let’s move on to solving those inequalities and find the valid range of 'x' values!
Condition 1: 3 - x ≥ 0
Let's tackle the first condition: 3 - x ≥ 0. This comes from the square root in the numerator, √(3-x). Remember, we need the expression inside the square root to be non-negative. So, how do we solve this inequality? It’s actually pretty straightforward. We want to isolate 'x' on one side of the inequality. To do that, we can add 'x' to both sides:
3 - x + x ≥ 0 + x
This simplifies to:
3 ≥ x
Or, we can rewrite it as:
x ≤ 3
So, our first condition tells us that 'x' must be less than or equal to 3. Any value of 'x' greater than 3 would make '3 - x' negative, and we'd be taking the square root of a negative number, which is not allowed. This is a crucial piece of the puzzle, but we’re not done yet! We still have the denominator to consider. Let’s keep this condition in mind as we move on to the next one.
Condition 2: 2x + 1 > 0
Now, let's deal with the second condition: 2x + 1 > 0. This comes from the square root in the denominator, √(2x+1). Remember, we need the expression inside the square root to be strictly greater than zero because it’s in the denominator, and we can’t divide by zero. So, let's solve this inequality. First, we subtract 1 from both sides:
2x + 1 - 1 > 0 - 1
This simplifies to:
2x > -1
Next, we divide both sides by 2:
2x / 2 > -1 / 2
Which gives us:
x > -1/2
So, our second condition tells us that 'x' must be greater than -1/2. If 'x' were less than or equal to -1/2, the expression '2x + 1' would be zero or negative, and we’d either be dividing by zero or taking the square root of a negative number. Both are mathematical no-nos! Now we have two conditions: x ≤ 3 and x > -1/2. We need to find the values of 'x' that satisfy both of these conditions simultaneously. It’s like finding the sweet spot where both inequalities hold true.
Combining the Conditions and Finding the Domain
Okay, guys, we've got two conditions for our domain, and now it's time to put them together like a mathematical puzzle! We know that 'x' must be less than or equal to 3 (x ≤ 3) from the numerator, and 'x' must be greater than -1/2 (x > -1/2) from the denominator. So, we need to find the values of 'x' that satisfy both of these conditions at the same time. How do we do that? Well, let's think about it visually.
Imagine a number line. We can mark x ≤ 3 with a closed circle (because it includes 3) and shade the line to the left, indicating all values less than or equal to 3. Then, we can mark x > -1/2 with an open circle (because it doesn’t include -1/2) and shade the line to the right, indicating all values greater than -1/2. The domain of our function is where these two shaded regions overlap. This overlap represents the values of 'x' that satisfy both inequalities.
Looking at the number line, we can see that the overlap occurs between -1/2 and 3. Since 'x' has to be strictly greater than -1/2 (due to the denominator) and less than or equal to 3 (due to the numerator), we can express the domain as an interval: (-1/2, 3]. This means that 'x' can be any number greater than -1/2 and up to and including 3. We use a parenthesis for -1/2 because it's not included, and a bracket for 3 because it is included. This notation gives us a clear and concise way to represent the domain of our function. So, we've successfully combined the conditions, and we're one step closer to finding the correct answer!
Expressing the Domain in Set Notation
Now that we've figured out the domain as an interval, let's express it using set notation – another way mathematicians like to describe sets of numbers. Set notation might seem a bit formal at first, but it's a super precise way to define our domain. Remember, our domain is the set of all 'x' values that make our function F(x) = √(3-x) / √(2x+1) valid.
We found that 'x' must be greater than -1/2 and less than or equal to 3. In set notation, we can write this as:
D(f) = {x ∈ R / -1/2 < x ≤ 3}
Let’s break this down piece by piece. The D(f) part simply means “the domain of the function f.” The {x ∈ R / ...} part means “the set of all x such that x is a real number and…” The “…” is where we put our conditions. So, -1/2 < x ≤ 3 means that 'x' is greater than -1/2 and less than or equal to 3. Putting it all together, the set notation tells us that the domain of our function is the set of all real numbers 'x' that are greater than -1/2 and less than or equal to 3. This is exactly what we found using the number line method! Set notation is like a mathematical shorthand – it packs a lot of information into a neat little package. So, now we have the domain expressed in both interval notation and set notation, giving us a solid understanding of the valid 'x' values for our function.
Identifying the Correct Option
Alright, guys, we've done the hard work! We've analyzed the function F(x) = √(3-x) / √(2x+1), figured out the restrictions imposed by the square roots and the division, solved the inequalities, and expressed the domain in both interval and set notation. Now comes the satisfying part – identifying the correct option from the choices provided. Let’s recap our findings to make sure we’re crystal clear.
We determined that the domain of the function is the set of all real numbers 'x' such that -1/2 < x ≤ 3. In set notation, this is written as D(f) = {x ∈ R / -1/2 < x ≤ 3}. Now, let's look at the options (which weren’t explicitly provided in the initial problem, but we can infer based on the question format):
- A) D(f) = {x ∈ R / x > 3}
- B) D(f) = {x ∈ R / x < -1/2}
- C) D(f) = {x ∈ R / -1/2 < x ≤ 3}
- D) D(f) = {x ∈ R / -1/2 < x < 3}
- E) D(f) = {x ∈ R / x ≤ 3}
Comparing our result with the options, we can clearly see that option C, D(f) = {x ∈ R / -1/2 < x ≤ 3}, matches our calculated domain perfectly! The other options either include values that would make the function undefined (like values less than or equal to -1/2 or values greater than 3) or exclude values that are valid (like 3 itself). So, option C is the winner! We've successfully navigated through the restrictions, solved the inequalities, and matched our answer with the correct option. High five!
Final Answer
Phew! We made it through the maze of square roots and fractions and emerged victorious with the correct domain. Let’s state our final answer clearly and confidently. After analyzing the function F(x) = √(3-x) / √(2x+1) and considering all the restrictions, we found that the domain is the set of all real numbers 'x' such that -1/2 < x ≤ 3. Therefore, the correct option is:
C) D(f) = {x ∈ R / -1/2 < x ≤ 3}
We’ve shown our work, explained our reasoning, and arrived at the correct answer. Give yourselves a pat on the back, guys! You've tackled a tricky domain problem and come out on top. Understanding how to find the domain of a function is a fundamental skill in mathematics, and you’ve just leveled up your math game. Keep practicing, and you’ll become domain-finding masters in no time!