Solving Function Division: (f/g)(5) For F(x)=7+4x, G(x)=1/(2x)

Hey guys! Today we're diving deep into the awesome world of mathematics, specifically tackling a problem involving function division. We've got two functions, $f(x)=7+4x$ and g(x)=rac{1}{2x}, and we need to find the value of $\left(\frac{f}{g}\right)(5)$. Don't let the notation scare you; it's just a fancy way of saying we need to divide function $f$ by function $g$ and then evaluate that new function at $x=5$. This is a fundamental concept in understanding how functions interact and build upon each other, which is super important as you progress through your math journey. We'll break it down step-by-step, making sure everyone can follow along, whether you're just starting with functions or looking for a refresher. So, grab your notebooks, your favorite thinking cap, and let's get this done!

Understanding Function Division $\left(\frac{f}{g}\right)(x)$

Alright, let's kick things off by understanding what $\left(\frac{f}{g}\right)(x)$ actually means. When we talk about the division of two functions, say $f(x)$ and $g(x)$, the resulting function, denoted as $\left(\frac{f}{g}\right)(x)$, is simply the result of dividing $f(x)$ by $g(x)$. Mathematically, this is expressed as: $ \left(\fracf}{g}\right)(x) = \frac{f(x)}{g(x)} $ But here's a crucial detail, guys this operation is only defined when $g(x)$ is not equal to zero. Division by zero is a big no-no in mathematics, so we always need to be mindful of the domain of our new function. In our specific problem, we have $f(x)=7+4x$ and $g(x)=rac{1{2x}$. So, to find $\left(\frac{f}{g}\right)(x)$, we'll substitute these expressions into our definition:

\left(\frac{f}{g}\right)(x) = \frac{7+4x}{\frac{1}{2x}} $ Now, dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of $\frac{1}{2x}$ is simply $2x$. So, our expression simplifies to: $ \left(\frac{f}{g}\right)(x) = (7+4x) \times (2x) $ To make it even cleaner, we can distribute the $2x$ into the parentheses: $ \left(\frac{f}{g}\right)(x) = 7(2x) + 4x(2x) $ $ \left(\frac{f}{g}\right)(x) = 14x + 8x^2

So, the function $\left(\frac{f}{g}\right)(x)$ is $8x^2 + 14x$. Before we move on to evaluating this at $x=5$, let's quickly check the condition that $g(x) \neq 0$. For $g(x) = \frac{1}{2x}$, this means $2x \neq 0$, which implies $x \neq 0$. Since we'll be evaluating at $x=5$, we're in the clear! The domain of our combined function $\left(\frac{f}{g}\right)(x)$ excludes $x=0$. This understanding of domain is super important in more complex problems, so always keep it in the back of your mind. It ensures that our mathematical operations are valid and that we don't run into any undefined situations. We've successfully combined our two functions into one, and now we're ready for the next exciting step!

Evaluating $\left(\frac{f}{g}\right)(5)$

Now that we have our combined function $\left(\frac{f}{g}\right)(x) = 8x^2 + 14x$, the next step is to find its value when $x=5$. This is often referred to as evaluating the function at a specific point. For us, this means plugging in $5$ wherever we see an $x$ in our expression. So, we'll calculate $\left(\frac{f}{g}\right)(5)$:

\left(\frac{f}{g}\right)(5) = 8(5)^2 + 14(5) $ Remember your order of operations (PEMDAS/BODMAS)! First, we handle the exponent: $ 5^2 = 25 $ Now, substitute that back into our equation: $ \left(\frac{f}{g}\right)(5) = 8(25) + 14(5) $ Next, we perform the multiplications: $ 8 \times 25 = 200 $ $ 14 \times 5 = 70 $ Finally, we add the results: $ \left(\frac{f}{g}\right)(5) = 200 + 70 $ $ \left(\frac{f}{g}\right)(5) = 270 $ And there you have it, guys! The value of $\left(\frac{f}{g}\right)(5)$ is **270**. It's pretty straightforward once you break down the notation and follow the steps. We first defined what function division means, then we simplified the expression for $\left(\frac{f}{g}\right)(x)$, and finally, we substituted the value $x=5$ to get our answer. This process highlights the power of algebraic manipulation and function composition in **mathematics**. By understanding these building blocks, you can tackle much more complex problems. This particular problem is a great way to practice order of operations and substitution, skills that are absolutely essential in all areas of math. Keep practicing, and you'll be a function-division pro in no time! ### Alternative Method: Evaluate First, Then Divide Another super cool way to solve this problem is to evaluate $f(5)$ and $g(5)$ *separately* first, and *then* perform the division. This can sometimes be an easier route, especially if the functions are complex. Let's give it a shot! We need to find $\left(\frac{f}{g}\right)(5)$, which is equivalent to $\frac{f(5)}{g(5)}$. First, let's find the value of $f(5)$. Our function is $f(x) = 7 + 4x$. Substitute $x=5$ into this equation: $ f(5) = 7 + 4(5) $ $ f(5) = 7 + 20 $ $ f(5) = 27 $ So, $f(5)$ equals **27**. Easy peasy! Next, let's find the value of $g(5)$. Our function is $g(x) = rac{1}{2x}$. Substitute $x=5$ into this equation: $ g(5) = rac{1}{2(5)} $ $ g(5) = rac{1}{10} $ So, $g(5)$ equals **$rac{1}{10}$**. Awesome! Now, we just need to divide $f(5)$ by $g(5)$: $ \left(\frac{f}{g}\right)(5) = \frac{f(5)}{g(5)} = \frac{27}{\frac{1}{10}} $ Again, dividing by a fraction means multiplying by its reciprocal. The reciprocal of $\frac{1}{10}$ is $10$. So: $ \left(\frac{f}{g}\right)(5) = 27 \times 10 $ $ \left(\frac{f}{g}\right)(5) = 270

Voila! We get the exact same answer, 270. This alternative method confirms our previous result and shows you that there can be multiple paths to the same correct answer in mathematics. Which method do you prefer, guys? Some might find simplifying the combined function first more intuitive, while others might prefer evaluating each function individually. Both are perfectly valid and demonstrate a solid understanding of function operations. Remember, the key is to be comfortable with both approaches, as different problems might lend themselves better to one method over the other. Keep exploring these different techniques; it's what makes learning math so engaging and rewarding!

Key Takeaways and Further Exploration

So, what have we learned today, gang? We've successfully navigated the concept of function division, specifically calculating $\left(\frac{f}{g}\right)(5)$ for the given functions $f(x)=7+4x$ and g(x)=rac{1}{2x}. We saw that \left(\frac{f}{g}\right)(x) = rac{f(x)}{g(x)}, and we learned the importance of the domain, ensuring $g(x) \neq 0$. We also explored two effective methods to arrive at the answer: first, by forming the combined function $\left(\frac{f}{g}\right)(x)$ and then evaluating it at $x=5$, and second, by evaluating $f(5)$ and $g(5)$ individually before performing the division. Both methods yielded the same result, 270, which is a fantastic outcome!

This problem is a stepping stone to understanding more complex function operations like composition and inverse functions. Keep practicing these foundational skills. Try variations of this problem: What happens if $g(x)$ is a different function? What if you need to find $\left(\frac{g}{f}\right)(x)$? What about other values for $x$, especially values that might be excluded from the domain? For instance, what would happen if we tried to evaluate $\left(\frac{f}{g}\right)(0)$? As we noted, $x$ cannot be $0$ because $g(x)$ would be undefined. This concept of domain restriction is crucial and pops up everywhere in calculus and beyond. Understanding when and why functions are undefined is just as important as knowing when they are defined.

Don't be afraid to experiment and push the boundaries of your understanding. The world of mathematics is vast and full of fascinating connections. Keep asking questions, keep exploring, and most importantly, keep that mathematical curiosity alive! You're all doing great, and every problem you solve is a victory. Happy problem-solving, everyone!