Function Operations And Composition: Math Problems Solved

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Let's dive into some math problems focusing on function operations and composition. We'll break down each problem step-by-step, so you can easily follow along and understand the concepts. Functions might seem intimidating at first, but with a little practice, you'll be solving these like a pro!

1. Finding (f+g)(x)(f+g)(x) when f(x)=2x+20f(x) = 2x + 20 and g(x)=7xβˆ’16g(x) = 7x - 16

This problem asks us to find the sum of two functions, f(x)f(x) and g(x)g(x). Remember, function addition is pretty straightforward. We simply add the corresponding expressions for each function. To really nail this, let's break it down:

First, identify the functions we're working with. We have f(x)=2x+20f(x) = 2x + 20 and g(x)=7xβˆ’16g(x) = 7x - 16. Now, when we're finding (f+g)(x)(f+g)(x), what we're essentially doing is adding f(x)f(x) and g(x)g(x) together. So, we write it out like this: (f+g)(x)=f(x)+g(x)(f+g)(x) = f(x) + g(x). Next, substitute the actual expressions for f(x)f(x) and g(x)g(x). This gives us (f+g)(x)=(2x+20)+(7xβˆ’16)(f+g)(x) = (2x + 20) + (7x - 16). Now it’s all about combining like terms. We've got 2x2x and 7x7x, which add up to 9x9x. Then we have +20+20 and βˆ’16-16, which combine to give us +4+4. Putting it all together, (f+g)(x)=9x+4(f+g)(x) = 9x + 4. There you go! We've successfully found the sum of the two functions. It’s like merging two streams into one river; you're just combining what you already have. Don't be shy about going back and re-reading this part if you need to. The key is practice and breaking it down into digestible chunks. Think of each function as a separate ingredient, and adding them is like following a recipe. Once you've done it a couple of times, it becomes second nature. The main thing to remember is to focus on combining those like terms – that’s where the magic happens!

2. Determining (fβˆ’g)(x)(f-g)(x) given f(x)=12xβˆ’19f(x) = 12x - 19 and g(x)=10xβˆ’5g(x) = 10x - 5

Okay, so this time we're looking at function subtraction. It's super similar to addition, but with a tiny twist – we're subtracting the second function from the first. Keep your eyes peeled for those negative signs because they can be sneaky! So, we're given f(x)=12xβˆ’19f(x) = 12x - 19 and g(x)=10xβˆ’5g(x) = 10x - 5. Our mission is to find (fβˆ’g)(x)(f-g)(x), which means we need to subtract g(x)g(x) from f(x)f(x). Writing it out, we get (fβˆ’g)(x)=f(x)βˆ’g(x)(f-g)(x) = f(x) - g(x).

Now, let’s substitute the expressions for our functions: (fβˆ’g)(x)=(12xβˆ’19)βˆ’(10xβˆ’5)(f-g)(x) = (12x - 19) - (10x - 5). Here's the crucial part: we're subtracting the entire function g(x)g(x), so we need to distribute that negative sign. It’s like giving everyone in the second function a little identity makeover! Distributing the negative sign, we get (fβˆ’g)(x)=12xβˆ’19βˆ’10x+5(f-g)(x) = 12x - 19 - 10x + 5. See how that βˆ’5-5 became a +5+5? That’s the key. Now it’s smooth sailing. We combine like terms again. We have 12x12x and βˆ’10x-10x, which give us 2x2x. Then we have βˆ’19-19 and +5+5, which combine to βˆ’14-14. So, our final answer is (fβˆ’g)(x)=2xβˆ’14(f-g)(x) = 2x - 14. You nailed it! Subtraction is just like addition's slightly more cautious cousin. The big takeaway here is remembering to distribute that negative sign. Think of it like defusing a tiny bomb – handle it carefully, and you'll be just fine. And again, practice makes perfect. Do a few more of these, and you'll be subtracting functions in your sleep!

3. Evaluating (fextextdegreeg)(x)(f ext{ extdegree } g)(x) where f(x)=7xβˆ’8f(x) = 7x - 8 and g(x)=xβˆ’12g(x) = x - 12

Now, let's tackle function composition. This might look a little scarier with that little circle symbol (f∘g)(f \circ g), but trust me, it's just a new way of combining functions. This symbol means "f of g of x," which tells us we're plugging the entire function g(x)g(x) into the function f(x)f(x). It's like a function inside a function – a mathematical matryoshka doll! So, we have f(x)=7xβˆ’8f(x) = 7x - 8 and g(x)=xβˆ’12g(x) = x - 12, and we need to find (f∘g)(x)(f \circ g)(x). This means we want to find f(g(x))f(g(x)).

First, let's think about what this means. We're going to take the entire expression for g(x)g(x), which is xβˆ’12x - 12, and substitute it everywhere we see an xx in f(x)f(x). So, instead of 7xβˆ’87x - 8, we'll have 7(xβˆ’12)βˆ’87(x - 12) - 8. See how the whole g(x)g(x) expression took the place of xx in f(x)f(x)? Now, we just need to simplify. We start by distributing the 77 across the parentheses: 7(xβˆ’12)7(x - 12) becomes 7xβˆ’847x - 84. Don't forget that we still have the βˆ’8-8 hanging out at the end, so we have 7xβˆ’84βˆ’87x - 84 - 8. Finally, combine those constants: βˆ’84βˆ’8-84 - 8 equals βˆ’92-92. So, our final answer is (f∘g)(x)=7xβˆ’92(f \circ g)(x) = 7x - 92. And there we have it! Composition is like creating a chain reaction. You're taking the output of one function and feeding it as the input into another. The key is to be methodical and take it step by step. Don't rush, make sure you're substituting correctly, and always remember to simplify. You've got this! Keep practicing, and you'll be composing functions like a maestro!

Mathematical Discussions

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