Let's solve the problems step by step.

Given Functions:

• $f(x) = x^2 + 3$
• $g(x) = x^2 - 4$

Now, we will address each part:

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

This represents the composition of $f$ and $g$, meaning $f(g(x))$.

1. First, find $g(x)$: $g(x) = x^2 - 4$

2. Now substitute $g(x)$ into $f(x)$: $f(g(x)) = f(x^2 - 4) = (x^2 - 4)^2 + 3$

3. Expand $(x^2 - 4)^2$: $(x^2 - 4)^2 = x^4 - 8x^2 + 16$

4. Now, add the constant from $f(x)$: $f(g(x)) = x^4 - 8x^2 + 16 + 3 = x^4 - 8x^2 + 19$

Thus, $(f \circ g)(x) = x^4 - 8x^2 + 19$.

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

This represents the composition of $g$ and $f$, meaning $g(f(x))$.

1. First, find $f(x)$: $f(x) = x^2 + 3$

2. Now substitute $f(x)$ into $g(x)$: $g(f(x)) = g(x^2 + 3) = (x^2 + 3)^2 - 4$

3. Expand $(x^2 + 3)^2$: $(x^2 + 3)^2 = x^4 + 6x^2 + 9$

4. Now subtract the constant from $g(x)$: $g(f(x)) = x^4 + 6x^2 + 9 - 4 = x^4 + 6x^2 + 5$

Thus, $(g \circ f)(x) = x^4 + 6x^2 + 5$.

(c) $(f \circ g)(2)$

This represents $f(g(2))$.

1. First, find $g(2)$: $g(2) = 2^2 - 4 = 4 - 4 = 0$

2. Now find $f(0)$: $f(0) = 0^2 + 3 = 3$

Thus, $(f \circ g)(2) = 3$.

(d) $(g \circ f)(2)$

This represents $g(f(2))$.

1. First, find $f(2)$: $f(2) = 2^2 + 3 = 4 + 3 = 7$

2. Now find $g(7)$: $g(7) = 7^2 - 4 = 49 - 4 = 45$

Thus, $(g \circ f)(2) = 45$.

Summary of Answers:

• (a) $(f \circ g)(x) = x^4 - 8x^2 + 19$
• (b) $(g \circ f)(x) = x^4 + 6x^2 + 5$
• (c) $(f \circ g)(2) = 3$
• (d) $(g \circ f)(2) = 45$

Would you like more details or explanations on any part of this? Here are 5 follow-up questions to expand your understanding:

1. How does the domain of $f \circ g$ differ from $g \circ f$?
2. What is the difference between composing functions and multiplying functions?
3. Can we reverse the composition of functions in all cases?
4. How do we handle compositions of more than two functions, such as $(h \circ f \circ g)(x)$?
5. How would these compositions change if the functions were non-polynomial, like trigonometric functions?

Tip: When working with compositions, always follow the inside-out approach, starting from the innermost function and working your way out!