Solving (f/g)(5) For F(x)=7+4x And G(x)=1/(2x)

Hey math whizzes! Today we're diving into a super common type of problem you'll see in algebra: function operations. Specifically, we're going to tackle how to find the value of a combined function, the division of two functions in this case, at a specific point. The problem asks us to find the value of (rac{f}{g})(5) given two functions: $f(x) = 7 + 4x$ and g(x) = rac{1}{2x}. This might look a little intimidating at first, but trust me, it's all about breaking it down step-by-step. We're going to dissect what (rac{f}{g})(5) actually means, figure out the individual values of $f(5)$ and $g(5)$, and then combine them to get our final answer. So grab your calculators, maybe a snack, and let's get this math party started!

Understanding Function Notation and Operations

Before we jump into solving, let's make sure we're all on the same page with function notation. You've got $f(x)$ and $g(x)$ here. The '$x$' inside the parentheses is the input, and the expression on the other side of the equals sign is how you calculate the output. So, $f(x) = 7 + 4x$ means that for any input value '$x$', you multiply it by 4 and then add 7 to get the output of the function $f$. Similarly, g(x) = rac{1}{2x} means for any input '$x$', you multiply it by 2 and then take the reciprocal (1 divided by that result) to get the output of $g$. Now, what about (rac{f}{g})(x)? This notation means you're creating a new function by dividing the function $f(x)$ by the function $g(x)$. So, (rac{f}{g})(x) = rac{f(x)}{g(x)}. When we want to find the value of this new function at a specific point, like $x=5$, we write (rac{f}{g})(5). This means we need to evaluate the expression rac{f(5)}{g(5)}. The key takeaway here is that we can't just plug 5 into some magically combined formula directly. We need to find the output of each individual function at $x=5$ first, and then perform the division. It’s like having two separate machines, one for $f$ and one for $g$, and you need to get a result from each before you can combine them. This is a fundamental concept in algebra that opens up a world of possibilities when dealing with how functions interact with each other. Understanding this foundational concept is crucial for mastering more complex mathematical concepts down the line, so let's really lock this in.

Calculating f(5)

Alright guys, let's get down to business and calculate the value of $f(5)$. Remember, our function $f(x)$ is defined as $f(x) = 7 + 4x$. The notation $f(5)$ tells us to substitute the value 5 for every instance of '$x$' in the expression for $f(x)$. So, we'll replace '$x$' with '5'. This gives us $f(5) = 7 + 4(5)$. Now, we follow the order of operations (PEMDAS/BODMAS – Parentheses/Brackets, Exponents/Orders, Multiplication and Division, Addition and Subtraction). First, we handle the multiplication: $4 imes 5 = 20$. After performing the multiplication, the expression becomes $f(5) = 7 + 20$. The final step is to perform the addition: $7 + 20 = 27$. So, the value of $f(5)$ is 27. This means that when the input to the function $f$ is 5, the output is 27. Pretty straightforward, right? It’s all about carefully substituting the input value and then meticulously following the order of operations to arrive at the correct result. This step is crucial because if we get this value wrong, our final answer for (rac{f}{g})(5) will also be incorrect. It's like building a house – the foundation has to be solid, and $f(5)=27$ is a crucial part of our mathematical foundation for this problem. We've successfully evaluated one of our functions, and the result is a clean, whole number, which is always nice. Keep this number, 27, in mind as we move on to the next part of our calculation!

Calculating g(5)

Now, let's shift our focus to the second function, $g(x)$. We are given g(x) = rac{1}{2x}. Just like we did with $f(x)$, we need to find the value of $g(5)$. This means we substitute 5 for '$x$' in the expression for $g(x)$. So, g(5) = rac{1}{2(5)}. Again, we follow the order of operations. First, perform the multiplication in the denominator: $2 imes 5 = 10$. Now, the expression for $g(5)$ becomes g(5) = rac{1}{10}. So, the value of $g(5)$ is rac{1}{10} (or 0.1 in decimal form). This tells us that when the input to the function $g$ is 5, the output is rac{1}{10}. It's important to note that $g(x)$ has a potential restriction: $x$ cannot be 0 because division by zero is undefined. Since our input is 5, which is not 0, we are good to go. Evaluating $g(5)$ yielded a fraction, which is perfectly fine. Sometimes functions produce neat integers, and other times they produce fractions or decimals. The key is to be comfortable working with both. We've now successfully calculated the values of both $f(5)$ and $g(5)$. We have $f(5) = 27$ and g(5) = rac{1}{10}. These are the two essential components we need to solve the original problem. Make sure you have these values handy, as the next step involves combining them. This step is arguably the most satisfying because you're seeing the two pieces you've worked on come together to form the final answer.

Putting it All Together: Calculating (f/g)(5)

We've reached the final stage, guys! We've calculated $f(5) = 27$ and g(5) = rac{1}{10}. Now, we need to find the value of (rac{f}{g})(5). Remember from our earlier discussion that the notation (rac{f}{g})(5) means we need to calculate rac{f(5)}{g(5)}. We simply substitute the values we found: (rac{f}{g})(5) = rac{27}{rac{1}{10}}. Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of rac{1}{10} is rac{10}{1} (or just 10). So, the expression becomes (rac{f}{g})(5) = 27 imes 10. Performing this multiplication, we get $27 imes 10 = 270$. Therefore, the value of (rac{f}{g})(5) is 270. We have successfully navigated through evaluating individual functions and then combining them through division to find the final answer. This problem beautifully illustrates how function notation works and how we can perform operations on functions. It’s a fundamental skill that underpins many more advanced mathematical concepts. So, you've taken $f(x)$ and $g(x)$, figured out what they do at the input 5, and then performed a division to get your final result. This is the power of understanding function operations – you can combine and manipulate functions in logical ways to answer specific questions. Take a moment to appreciate the process. You started with two separate algebraic expressions, and through a series of logical steps, you arrived at a single numerical answer. That's pretty awesome!

Conclusion: A Solid Mathematical Win

So there you have it, math adventurers! We've successfully tackled the problem of finding (rac{f}{g})(5) for the given functions $f(x) = 7 + 4x$ and g(x) = rac{1}{2x}. By breaking down the problem into smaller, manageable steps – understanding the notation, calculating $f(5)$, calculating $g(5)$, and finally performing the division – we arrived at our answer: 270. This process highlights the elegance and logic of mathematics. Function operations are a key concept in algebra, allowing us to combine and analyze functions in various ways. Whether it's addition, subtraction, multiplication, or division, understanding how to perform these operations is crucial for further mathematical studies. Remember, the key is always to evaluate each function at the given input first before performing the operation. Don't be afraid of fractions or complex-looking notation; with a clear understanding of the steps involved and careful calculation, you can solve these problems with confidence. Keep practicing these types of problems, and you'll find that function operations become second nature. High fives all around for conquering this mathematical challenge! Keep exploring, keep learning, and most importantly, keep having fun with math!