Let's work through the three problems provided in the image. The goal is to find:

1. $(f \circ g)(x)$, which means $f(g(x))$,
2. $(g \circ f)(x)$, which means $g(f(x))$, and
3. $(f \circ g)(2)$, which means applying $f(g(2))$.

1. $f(x) = 2x$, $g(x) = x + 5$

• $(f \circ g)(x)$: This means $f(g(x)) = f(x + 5)$. $f(x + 5) = 2(x + 5) = 2x + 10$ So, $(f \circ g)(x) = 2x + 10$.

• $(g \circ f)(x)$: This means $g(f(x)) = g(2x)$. $g(2x) = 2x + 5$ So, $(g \circ f)(x) = 2x + 5$.

• $(f \circ g)(2)$: First, find $g(2)$, then apply $f$. $g(2) = 2 + 5 = 7$ Now, find $f(7)$: $f(7) = 2(7) = 14$ So, $(f \circ g)(2) = 14$.

2. $f(x) = 4x$, $g(x) = x - 3$

• $(f \circ g)(x)$: This means $f(g(x)) = f(x - 3)$. $f(x - 3) = 4(x - 3) = 4x - 12$ So, $(f \circ g)(x) = 4x - 12$.

• $(g \circ f)(x)$: This means $g(f(x)) = g(4x)$. $g(4x) = 4x - 3$ So, $(g \circ f)(x) = 4x - 3$.

• $(f \circ g)(2)$: First, find $g(2)$, then apply $f$. $g(2) = 2 - 3 = -1$ Now, find $f(-1)$: $f(-1) = 4(-1) = -4$ So, $(f \circ g)(2) = -4$.

3. $f(x) = 2x + 1$, $g(x) = 3x - 2$

• $(f \circ g)(x)$: This means $f(g(x)) = f(3x - 2)$. $f(3x - 2) = 2(3x - 2) + 1 = 6x - 4 + 1 = 6x - 3$ So, $(f \circ g)(x) = 6x - 3$.

• $(g \circ f)(x)$: This means $g(f(x)) = g(2x + 1)$. $g(2x + 1) = 3(2x + 1) - 2 = 6x + 3 - 2 = 6x + 1$ So, $(g \circ f)(x) = 6x + 1$.

• $(f \circ g)(2)$: First, find $g(2)$, then apply $f$. $g(2) = 3(2) - 2 = 6 - 2 = 4$ Now, find $f(4)$: $f(4) = 2(4) + 1 = 8 + 1 = 9$ So, $(f \circ g)(2) = 9$.

Let me know if you'd like more details or if you have any further questions!

Here are 5 questions that could expand your understanding:

1. How do the compositions of functions differ when $f(x)$ and $g(x)$ are nonlinear?
2. What happens to the composition if one of the functions is a constant?
3. How would you approach finding $(f \circ g)(x)$ if $f(x)$ and $g(x)$ involve fractions?
4. How can compositions of functions be applied in real-world scenarios?
5. Can you derive the inverse of a composed function $(f \circ g)^{-1}(x)$?

Tip: When composing functions, always start from the innermost function and work outward step by step to avoid mistakes.