Domain Of (cd)(x) When C(x)=5/(x-2) And D(x)=x+3

Hey guys, let's dive into a super common math problem: finding the domain of the product of two functions. We've got two functions here, $c(x)=\frac{5}{x-2}$ and $d(x)=x+3$. Our mission, should we choose to accept it, is to figure out the domain of their product, $(cd)(x)$. Now, what does the domain of a function even mean, you ask? Simply put, it's all the possible input values (the 'x' values) that the function can accept without breaking or giving us a mathematical headache. Think of it as the set of 'x's that are allowed to go into the function machine. For functions involving fractions, we gotta be extra careful about denominators. You can't have a zero down there, right? Division by zero is a big no-no in math. So, for $c(x)=\frac{5}{x-2}$, the denominator is $x-2$. To avoid division by zero, we must ensure that $x-2 \neq 0$. This means $x \neq 2$. So, the domain of $c(x)$ alone is all real numbers except for 2. Easy peasy!

Now let's look at $d(x)=x+3$. This one is a polynomial, and polynomials are pretty chill. They can handle any real number as input. There are no denominators to worry about, no square roots of negative numbers, nothing that's going to cause trouble. So, the domain of $d(x)$ is all real numbers. This is awesome because it means $d(x)$ won't impose any restrictions on our combined function's domain.

When we talk about the product of two functions, $(cd)(x)$, we're essentially multiplying $c(x)$ by $d(x)$. So, $(cd)(x) = c(x) \times d(x) = \frac{5}{x-2} \times (x+3)$. To find the domain of this new function, $(cd)(x)$, we need to consider the restrictions from both original functions. Remember, for a value of 'x' to be in the domain of $(cd)(x)$, it must be a valid input for both $c(x)$ and $d(x)$ individually. Since $d(x)$ has no restrictions (its domain is all real numbers), the only restrictions we need to worry about come from $c(x)$. And we already figured out that $c(x)$ is undefined when $x=2$. Therefore, the product function $(cd)(x)$ will also be undefined when $x=2$. So, the domain of $(cd)(x)$ is all real values of $x$ except for $x=2$. This matches option B, guys! Always remember to check each function for its own domain restrictions before combining them.

Let's get a bit more technical for a sec, shall we? The definition of the product of two functions, $(f ext{g})(x)$, is given by $(f ext{g})(x) = f(x) ext{g}(x)$. The domain of $(f ext{g})(x)$ is the intersection of the domain of $f(x)$ and the domain of $g(x)$. In our case, $f(x)$ is $c(x)=\frac{5}{x-2}$ and $g(x)$ is $d(x)=x+3$. The domain of $c(x)$, let's call it $D_c$, is the set of all real numbers $x$ such that $x-2 \neq 0$, which means $D_c = \{x \in \mathbb{R} \mid x \neq 2\}$. The domain of $d(x)$, let's call it $D_d$, is the set of all real numbers $x$ since $d(x)$ is a polynomial, so $D_d = \mathbb{R}$. The domain of $(cd)(x)$, denoted as $D_{cd}$, is $D_c \cap D_d$. The intersection of the set of all real numbers except 2 and the set of all real numbers is simply the set of all real numbers except 2. Thus, $D_{cd} = \{x \in \mathbb{R} \mid x \neq 2\}$. This confirms that option B is indeed the correct answer. It's crucial to understand that while $d(x)$ itself doesn't introduce any restrictions, the combination with $c(x)$ is still constrained by $c(x)$'s limitations. So, even if you have a function that's perfectly fine on its own, when you multiply it by another function that has a restriction, that restriction carries over to the product function. It’s like trying to build a chain – the whole chain is only as strong as its weakest link, and in the case of function domains, those restrictions are the weak links.

Let's think about why the other options are incorrect, just to solidify our understanding, guys. Option A suggests that the domain is all real values of $x$. This would be true if both $c(x)$ and $d(x)$ had domains of all real numbers. However, we've clearly established that $c(x)$ has a restriction at $x=2$. If we were to plug $x=2$ into $(cd)(x)$, we'd get $\frac{5}{2-2} \times (2+3) = \frac{5}{0} \times 5$. The $\frac{5}{0}$ part is undefined, making the entire expression undefined. So, option A is out.

Option C claims the domain is all real values of $x$ except $x=-3$. This restriction would arise if $d(x)$ had a denominator of $x+3$ or something similar that becomes zero at $x=-3$. But $d(x)=x+3$ is a simple linear function with no such issues. Plugging $x=-3$ into $(cd)(x)$ gives us $\frac{5}{-3-2} \times (-3+3) = \frac{5}{-5} \times 0 = -1 \times 0 = 0$. This is a perfectly valid real number, so $x=-3$ is definitely in the domain of $(cd)(x)$. Therefore, option C is incorrect.

Option D states that the domain is all real values of $x$ except $x=2$ and $x=-3$. This would be the correct answer if both $c(x)$ and $d(x)$ had restrictions at $x=2$ and $x=-3$ respectively, or if their combined restrictions resulted in both exclusions. However, as we've repeatedly stressed, $d(x)$ has no restrictions. The only restriction comes from $c(x)$ at $x=2$. The false inclusion of $x=-3$ as a restriction makes option D incorrect. It's super important not to add restrictions that don't exist, guys. Always stick to what the functions actually demand.

In summary, when finding the domain of a product of functions, you must find the intersection of the domains of the individual functions. Any value of $x$ that makes any of the denominators zero (or causes other undefined operations like square roots of negatives) in the original functions must be excluded from the domain of the product. In this specific problem, $c(x)$ has a restriction at $x=2$, and $d(x)$ has no restrictions. Therefore, the domain of $(cd)(x)$ is all real numbers except $x=2$. This is a fundamental concept in function composition and operations, and understanding it well will help you tackle many more advanced problems in calculus and beyond. Keep practicing, and you'll master it in no time!

Final Answer: The final answer is $\boxed{B}$