Let's solve each part of the problem by finding $(f \circ g)(x)$, $(g \circ f)(x)$, and $(f \circ g)(2)$ for all the given functions.

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

• $(f \circ g)(x)$: $f(g(x)) = f(x + 5) = 2(x + 5) = 2x + 10$
• $(g \circ f)(x)$: $g(f(x)) = g(2x) = 2x + 5$
• $(f \circ g)(2)$: $f(g(2)) = f(2 + 5) = f(7) = 2(7) = 14$

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

• $(f \circ g)(x)$: $f(g(x)) = f(x - 3) = 4(x - 3) = 4x - 12$
• $(g \circ f)(x)$: $g(f(x)) = g(4x) = 4x - 3$
• $(f \circ g)(2)$: $f(g(2)) = f(2 - 3) = f(-1) = 4(-1) = -4$

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

• $(f \circ g)(x)$: $f(g(x)) = f(3x - 2) = 2(3x - 2) + 1 = 6x - 4 + 1 = 6x - 3$
• $(g \circ f)(x)$: $g(f(x)) = g(2x + 1) = 3(2x + 1) - 2 = 6x + 3 - 2 = 6x + 1$
• $(f \circ g)(2)$: $f(g(2)) = f(3(2) - 2) = f(6 - 2) = f(4) = 2(4) + 1 = 9$

4. $f(x) = 5x - 3$ and $g(x) = x^2 - 3$

• $(f \circ g)(x)$: $f(g(x)) = f(x^2 - 3) = 5(x^2 - 3) - 3 = 5x^2 - 15 - 3 = 5x^2 - 18$
• $(g \circ f)(x)$: $g(f(x)) = g(5x - 3) = (5x - 3)^2 - 3 = 25x^2 - 30x + 9 - 3 = 25x^2 - 30x + 6$
• $(f \circ g)(2)$: $f(g(2)) = f(2^2 - 3) = f(4 - 3) = f(1) = 5(1) - 3 = 2$

5. $f(x) = x^2 + 3$ and $g(x) = x^2 - 1$

• $(f \circ g)(x)$: $f(g(x)) = f(x^2 - 1) = (x^2 - 1)^2 + 3 = x^4 - 2x^2 + 1 + 3 = x^4 - 2x^2 + 4$
• $(g \circ f)(x)$: $g(f(x)) = g(x^2 + 3) = (x^2 + 3)^2 - 1 = x^4 + 6x^2 + 9 - 1 = x^4 + 6x^2 + 8$
• $(f \circ g)(2)$: $f(g(2)) = f(2^2 - 1) = f(4 - 1) = f(3) = 3^2 + 3 = 9 + 3 = 12$

Would you like further clarification on any of these calculations?

Here are five related questions you might find helpful:

1. How do you calculate compositions of functions in general?
2. What are the common mistakes to avoid when composing functions?
3. Can we derive the domain and range for $f \circ g$ in these cases?
4. How would the results change if $f(x)$ and $g(x)$ were swapped?
5. What are some real-life applications of function compositions?

Tip: Always carefully substitute the functions when composing them to avoid errors in algebraic manipulation!