Domain Of (f ⋅ G)(x): A Step-by-Step Solution

by Omar Yusuf 46 views

Hey guys! Let's dive into a fun math problem today that involves finding the domain of a composite function. We're given two functions, $f(x) = \frac{1}{x-3}$ and $g(x) = \sqrt{x+5}$, and our mission, should we choose to accept it, is to determine the domain of the product $(f "." g)(x)$. This might sound a bit intimidating at first, but trust me, we'll break it down step by step so it's super clear and easy to understand. So, grab your thinking caps, and let's get started!

Understanding the Functions

Before we jump into the nitty-gritty, let's first make sure we fully grasp what each function is doing. This will give us a solid foundation for tackling the main problem. So, what do we have? We've got $f(x)$, which is a rational function, and $g(x)$, which is a square root function. Each of these types of functions has its own quirks and limitations when it comes to their domains, and understanding these is key to finding the domain of their product. Remember, the domain of a function is essentially the set of all possible input values (x-values) that will produce a valid output (a real number). So, let's dig a little deeper into each function.

Analyzing f(x) = 1/(x-3)

Let's start by taking a closer look at our friend $f(x) = \frac{1}{x-3}$. This is a rational function, which means it's a fraction where the numerator and/or the denominator are polynomials. Now, rational functions have a specific Achilles' heel: they can't have a zero in the denominator. Why? Because division by zero is a big no-no in the math world – it's undefined! So, for $f(x)$, we need to make sure that the denominator, which is $x-3$, doesn't equal zero. If it does, we're in trouble! That means we need to find the value(s) of $x$ that would make $x-3 = 0$, and then exclude those value(s) from the domain. It's like setting up a mathematical boundary – we can go everywhere except for that one spot. This is a crucial step in determining the overall domain of the composite function, so let's make sure we've got it down pat.

To find the values that make the denominator zero, we simply solve the equation $x - 3 = 0$. Adding 3 to both sides, we get $x = 3$. Aha! So, the one value we need to exclude from the domain of $f(x)$ is $x = 3$. This means that $f(x)$ is perfectly happy and well-defined for any value of $x$ except for 3. We can think of the domain of $f(x)$ as being all real numbers with a little hole punched out at $x = 3$. Mathematically, we can express this as $(-\infty, 3) \cup (3, \infty)$. This notation means all numbers from negative infinity up to (but not including) 3, and then all numbers from 3 (not including 3) to positive infinity. It's a concise way of saying that 3 is the only number we're not allowed to use as an input for $f(x)$. So, that's one piece of the puzzle solved! Now, let's move on to $g(x)$ and see what restrictions it brings to the table.

Analyzing g(x) = √(x+5)

Alright, let's shift our focus to the second function, $g(x) = \sqrt{x+5}$. This is a square root function, and like rational functions, square root functions have their own set of rules we need to follow. The key thing to remember about square roots is that we can only take the square root of non-negative numbers. In other words, the expression inside the square root (the radicand) must be greater than or equal to zero. If we try to take the square root of a negative number, we end up with an imaginary number, which is outside the realm of real numbers that we're typically working with when finding domains. So, for $g(x)$, we need to make sure that $x+5$ is greater than or equal to zero. This is another mathematical boundary we need to establish, ensuring that our inputs don't lead to imaginary outputs.

To ensure that the expression inside the square root is non-negative, we need to solve the inequality $x + 5 \geq 0$. Subtracting 5 from both sides, we get $x \geq -5$. This tells us that the domain of $g(x)$ consists of all real numbers that are greater than or equal to -5. In interval notation, we can write this as $[-5, \infty)$. This means that $g(x)$ is perfectly happy and well-defined for any value of $x$ that is -5 or greater. If we try to plug in a value less than -5, we'll end up taking the square root of a negative number, which, as we know, is a no-go. So, the domain of $g(x)$ is like a fenced-off area that starts at -5 and stretches all the way to positive infinity. Now that we've figured out the individual domains of $f(x)$ and $g(x)$, we're ready to tackle the main question: finding the domain of their product, $(f "." g)(x)$.

Finding the Domain of (f ⋅ g)(x)

Okay, we've successfully dissected $f(x)$ and $g(x)$ and figured out their individual domains. Now comes the exciting part: combining our knowledge to find the domain of $(f \cdot g)(x)$. Remember, $(f \cdot g)(x)$ simply means $f(x) \cdot g(x)$, which in our case is $\frac{1}{x-3} \cdot \sqrt{x+5}$. So, we're essentially multiplying these two functions together. But how does this affect the domain? Well, the key idea here is that the domain of the product is the set of all $x$-values that are valid inputs for both $f(x)$ and $g(x)$. It's like a Venn diagram situation – we need to find the overlap between the two individual domains. If an $x$-value isn't in the domain of either $f(x)$ or $g(x)$, then it certainly can't be in the domain of their product.

So, to find the domain of $(f \cdot g)(x)$, we need to consider the restrictions imposed by both functions. From our earlier analysis, we know that $f(x)$ has a restriction at $x = 3$, and $g(x)$ has a restriction for all $x$ values less than $-5$. Therefore, the domain of $(f \cdot g)(x)$ will be the intersection of the domains of $f(x)$ and $g(x)$. This means we need to find the values of $x$ that satisfy both $x \neq 3$ and $x \geq -5$. Let's visualize this on a number line to make it even clearer.

Imagine a number line stretching from negative infinity to positive infinity. We know that $g(x)$ is defined for all $x$ greater than or equal to -5, so we can shade in that portion of the number line starting from -5 and going to the right. However, $f(x)$ has a hole at $x = 3$, so we need to put an open circle at 3 to indicate that it's not included in the domain. Now, the domain of $(f \cdot g)(x)$ is the portion of the number line that is shaded and doesn't have an open circle. This gives us two intervals: from -5 up to (but not including) 3, and from 3 (not including 3) to positive infinity. In interval notation, we can write this as $[-5, 3) \cup (3, \infty)$. This is the set of all $x$-values that will give us a valid output for the product of $f(x)$ and $g(x)$.

The Final Answer

Drumroll, please! After our thorough analysis, we've arrived at the final answer. The domain of $(f \cdot g)(x)$ is $[-5, 3) \cup (3, \infty)$. This means that the product of the two functions is defined for all real numbers greater than or equal to -5, except for the number 3. We've successfully navigated the restrictions imposed by both the rational function and the square root function, and we've found the precise set of inputs that will give us a valid output for their product. Give yourselves a pat on the back, guys! You've tackled a challenging problem and emerged victorious. Remember, the key to finding the domain of composite functions is to carefully consider the restrictions imposed by each individual function and then find the overlap between their domains. With a little practice, you'll be domain-finding pros in no time!

So, looking back at the options provided, the correct answer is:

D. [-5,3) ∪(3, ∞)

We nailed it! This problem might have seemed a bit daunting at first, but by breaking it down into smaller, manageable steps, we were able to conquer it. Remember, math is all about understanding the underlying principles and applying them systematically. Keep practicing, keep exploring, and keep having fun with it! You've got this!