https://file.aiquickdraw.com/mathbot/file/1725725549039kxso2weu.jpg

Let's go through the exercise step by step.

The exercise provided is:

### Exercise:
Find (f∘g)(x) and (g∘f)(x) where:
- \( f(x) = 3x + 1 \)
- \( g(x) = 2x - 4 \)

We will solve both compositions of functions:

### 1. \( (f∘g)(x) \)
This means \( f(g(x)) \), where we substitute \( g(x) \) into \( f(x) \).

Given:
- \( f(x) = 3x + 1 \)
- \( g(x) = 2x - 4 \)

Let's substitute \( g(x) \) into \( f(x) \):
\[
f(g(x)) = f(2x - 4) = 3(2x - 4) + 1
\]
Now simplify:
\[
f(g(x)) = 3(2x - 4) + 1 = 6x - 12 + 1 = 6x - 11
\]
Thus, \( (f∘g)(x) = 6x - 11 \).

### 2. \( (g∘f)(x) \)
This means \( g(f(x)) \), where we substitute \( f(x) \) into \( g(x) \).

Given:
- \( f(x) = 3x + 1 \)
- \( g(x) = 2x - 4 \)

Let's substitute \( f(x) \) into \( g(x) \):
\[
g(f(x)) = g(3x + 1) = 2(3x + 1) - 4
\]
Now simplify:
\[
g(f(x)) = 2(3x + 1) - 4 = 6x + 2 - 4 = 6x - 2
\]
Thus, \( (g∘f)(x) = 6x - 2 \).

### Final Answers:
- \( (f∘g)(x) = 6x - 11 \)
- \( (g∘f)(x) = 6x - 2 \)

Let me know if you'd like more details or explanations about these calculations!

Here are 5 related questions for you to explore:
1. What is the general process for composing two functions?
2. How does the order of function composition affect the outcome?
3. Can the functions \( f(x) \) and \( g(x) \) be any kind of function, such as non-linear functions?
4. What happens if the inner function in the composition is undefined for certain values of \( x \)?
5. How would the results change if the functions were defined differently, for example, \( f(x) = x^2 \) and \( g(x) = 3x + 1 \)?

**Tip**: When composing functions, always be mindful of the domains of both functions to ensure that the composite function is well-defined.

Find (f∘g)(x) and (g∘f)(x) given f(x) = 3x + 1 and g(x) = 2x - 4.

Learn how to solve function composition problems by finding (f∘g)(x) and (g∘f)(x) given f(x) = 3x + 1 and g(x) = 2x - 4. This solution involves substituting one function into another, showcasing key algebraic concepts suitable for Grades 9-10.

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