Function Operations: Find (fg)(10) Step-by-Step

Hey guys! Let's dive into the fascinating world of function composition, where we'll explore how to combine functions and evaluate them at specific points. In this article, we're going to tackle a problem involving two functions, $f(x) = x^2 - 20$ and $g(x) = 19 - x$, and we'll be figuring out the value of $(fg)(10)$. Buckle up, because we're about to unravel this mathematical puzzle step by step!

Understanding Function Composition: The Building Blocks

Before we jump into the specifics of our problem, let's make sure we're all on the same page about what function composition actually means. Function composition is essentially a way of combining two functions by applying one function to the result of the other. Think of it like a mathematical assembly line, where the output of one function becomes the input of the next.

When we write $(f \circ g)(x)$, which is often read as "f of g of x," we're instructing ourselves to first evaluate the function $g(x)$ and then use that result as the input for the function $f(x)$. It's crucial to remember the order of operations here – we work from the inside out. So, $g(x)$ gets the first shot, and then $f(x)$ takes over using $g(x)$ 's output. This concept might seem a bit abstract at first, but with a few examples and some practice, it'll become second nature.

In mathematical notation, we can express function composition as follows:

$$(f \circ g)(x) = f(g(x))$$

This equation simply formalizes what we've already discussed: to find $(f \circ g)(x)$, we first determine $g(x)$, and then we plug that value into $f(x)$. Function composition is a fundamental concept in mathematics, and it pops up in various areas like calculus, algebra, and even computer science. Mastering this concept will open up a whole new world of mathematical possibilities for you.

Now, let's bring this understanding to our specific problem and see how we can apply it to find $(fg)(10)$. We'll break it down into manageable steps, making sure everyone can follow along.

Cracking the Code: Deciphering (fg)(10)

Now that we've got a handle on the basics of function composition, let's zoom in on our specific problem: finding the value of $(fg)(10)$. But wait a minute, you might be thinking, isn't $(fg)(10)$ a little different from the $(f \circ g)(x)$ we just talked about? Great question! You've got a keen eye for detail. The notation $(fg)(10)$ can be a little misleading at first glance. It might look like we're simply multiplying the functions $f$ and $g$ together and then plugging in $x = 10$. However, in this context, $(fg)(10)$ actually represents the product of the functions evaluated at $x = 10$.

In other words, we need to find the values of $f(10)$ and $g(10)$ separately, and then multiply those results together. This is a subtle but important distinction from function composition, where we plug the entire function $g(x)$ into $f(x)$. So, to be crystal clear, here's the breakdown:

$$(fg)(10) = f(10) \cdot g(10)$$

This equation tells us that to find $(fg)(10)$, we first need to evaluate $f(10)$, then evaluate $g(10)$, and finally multiply the two results. This is a much simpler process than function composition, as we're not dealing with nested functions. We're simply evaluating each function at a specific point and then performing a straightforward multiplication.

Now that we've clarified what $(fg)(10)$ means in this context, let's roll up our sleeves and get to the actual calculations. We'll start by finding $f(10)$, and then we'll move on to $g(10)$.

Evaluating f(10): Plugging in the Value

Alright, let's kick things off by finding the value of $f(10)$. Remember, our function $f(x)$ is defined as $f(x) = x^2 - 20$. To evaluate $f(10)$, we simply need to substitute $x = 10$ into the expression for $f(x)$. This is a fundamental skill in algebra, and it's something you'll use time and time again in your mathematical journey.

So, let's do it! We replace every instance of $x$ in the equation for $f(x)$ with the value 10:

$$f(10) = (10)^2 - 20$$

Now, we just need to simplify this expression using the order of operations (PEMDAS/BODMAS). First up, we have the exponent: $(10)^2$ means 10 multiplied by itself, which is 100. So, we can rewrite our equation as:

$$f(10) = 100 - 20$$

Next, we perform the subtraction: 100 minus 20 equals 80. Therefore, we have:

$$f(10) = 80$$

Fantastic! We've successfully evaluated $f(10)$. This means that when we plug in $x = 10$ into the function $f(x)$, the output is 80. This is a crucial piece of the puzzle, and we're one step closer to finding $(fg)(10)$. Now, let's shift our focus to the other function, $g(x)$, and determine its value at $x = 10$.

Evaluating g(10): A Straightforward Calculation

Now that we've conquered $f(10)$, let's turn our attention to $g(10)$. Our function $g(x)$ is defined as $g(x) = 19 - x$. Just like with $f(10)$, we'll find $g(10)$ by substituting $x = 10$ into the expression for $g(x)$. This is a similar process to what we did before, but with a slightly simpler equation.

Let's plug in $x = 10$ into $g(x)$:

$$g(10) = 19 - 10$$

This is a straightforward subtraction problem. 19 minus 10 equals 9. So, we have:

$$g(10) = 9$$

Excellent! We've found that $g(10)$ is equal to 9. This means that when we input $x = 10$ into the function $g(x)$, the output is 9. We're making great progress! We now have the values of both $f(10)$ and $g(10)$, which are the two ingredients we need to calculate $(fg)(10)$.

With these pieces in hand, we're ready to put it all together and find the final answer. Let's move on to the final step, where we'll multiply the values we've calculated to get $(fg)(10)$.

Putting It All Together: The Grand Finale

We've reached the final stage of our mathematical journey! We've successfully navigated the concepts of function composition and function evaluation, and we've calculated the values of $f(10)$ and $g(10)$. Now, it's time to combine these results and find the value of $(fg)(10)$.

Remember, we established earlier that $(fg)(10)$ is equal to the product of $f(10)$ and $g(10)$. We can write this as:

$$(fg)(10) = f(10) \cdot g(10)$$

We've already found that $f(10) = 80$ and $g(10) = 9$. So, we can substitute these values into the equation:

$$(fg)(10) = 80 \cdot 9$$

Now, we just need to perform the multiplication. 80 multiplied by 9 equals 720. Therefore, we have:

$$(fg)(10) = 720$$

And there you have it! We've successfully determined that $(fg)(10)$ is equal to 720. Give yourselves a pat on the back, guys – you've tackled a challenging problem and come out on top!

Key Takeaways: Mastering Function Operations

Wow, what a journey! We've explored the fascinating world of function operations, specifically focusing on the notation $(fg)(10)$. Let's recap the key takeaways from our exploration:

• Understanding Function Composition: We started by defining function composition, where the output of one function becomes the input of another. We learned that $(f \circ g)(x) = f(g(x))$, emphasizing the order of operations (inside out).
• Deciphering (fg)(10): We clarified that $(fg)(10)$ in this context means the product of the functions evaluated at $x = 10$, i.e., $f(10) \cdot g(10)$. This is different from function composition.
• Evaluating f(10) and g(10): We practiced the fundamental skill of function evaluation by substituting $x = 10$ into the expressions for $f(x)$ and $g(x)$, finding $f(10) = 80$ and $g(10) = 9$.
• The Grand Finale: We combined our results to calculate $(fg)(10) = 80 \cdot 9 = 720$.

By understanding these key concepts and practicing these skills, you'll be well-equipped to tackle a wide range of problems involving function operations. Remember, mathematics is like building with LEGOs – each concept builds upon the previous one, creating a strong foundation for more advanced topics. So, keep practicing, keep exploring, and keep having fun with math!

This problem has shown us the importance of paying close attention to notation and understanding the subtle differences between various mathematical operations. Function composition and the product of functions are distinct concepts, and it's crucial to recognize which one is being applied in a given situation. With practice and careful attention to detail, you'll become more confident in navigating these nuances.

So, what's next? Well, you can explore more complex function compositions, work with different types of functions (like trigonometric or exponential functions), or even delve into the world of inverse functions. The possibilities are endless! The key is to keep challenging yourself and to never stop learning.

Until next time, happy problem-solving!