Let's solve each problem step by step by finding both $f \circ g$ and $g \circ f$, which represent the compositions of functions $f(x)$ and $g(x)$.

1. $f(x) = 3x - 2$ and $g(x) = \frac{1}{3}(x + 2)$

• Find $f \circ g$: $f(g(x)) = f\left(\frac{1}{3}(x + 2)\right)$ Substitute $g(x)$ into $f(x)$: $f\left(\frac{1}{3}(x + 2)\right) = 3\left(\frac{1}{3}(x + 2)\right) - 2$ Simplify: $f(g(x)) = (x + 2) - 2 = x$ So, $f \circ g = x$.

• Find $g \circ f$: $g(f(x)) = g(3x - 2)$ Substitute $f(x)$ into $g(x)$: $g(3x - 2) = \frac{1}{3}((3x - 2) + 2)$ Simplify: $g(f(x)) = \frac{1}{3}(3x) = x$ So, $g \circ f = x$.

2. $f(x) = 4x$ and $g(x) = x^2 - 1$

• Find $f \circ g$: $f(g(x)) = f(x^2 - 1)$ Substitute $g(x)$ into $f(x)$: $f(x^2 - 1) = 4(x^2 - 1)$ Simplify: $f(g(x)) = 4x^2 - 4$ So, $f \circ g = 4x^2 - 4$.

• Find $g \circ f$: $g(f(x)) = g(4x)$ Substitute $f(x)$ into $g(x)$: $g(4x) = (4x)^2 - 1$ Simplify: $g(f(x)) = 16x^2 - 1$ So, $g \circ f = 16x^2 - 1$.

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

• Find $f \circ g$: $f(g(x)) = f(x)$ Since $g(x) = x$, this is just $f(x)$: $f(g(x)) = -x^2 + 1$ So, $f \circ g = -x^2 + 1$.

• Find $g \circ f$: $g(f(x)) = g(-x^2 + 1)$ Since $g(x) = x$, this is simply: $g(f(x)) = -x^2 + 1$ So, $g \circ f = -x^2 + 1$.

Summary:

1. $f(x) = 3x - 2$, $g(x) = \frac{1}{3}(x + 2)$

   • $f \circ g = x$
   • $g \circ f = x$

2. $f(x) = 4x$, $g(x) = x^2 - 1$

   • $f \circ g = 4x^2 - 4$
   • $g \circ f = 16x^2 - 1$

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

   • $f \circ g = -x^2 + 1$
   • $g \circ f = -x^2 + 1$

Would you like further details or have any questions? Here are 5 related questions to consider:

1. What is the general definition of function composition?
2. How do we verify if two functions are inverses using composition?
3. Can $f \circ g$ and $g \circ f$ be equal for all functions?
4. What happens to the domain and range when composing two functions?
5. Can you find the inverse functions of $f(x) = 3x - 2$ and $g(x) = \frac{1}{3}(x + 2)$?

Tip: When composing functions, always substitute one function entirely into the other, replacing every occurrence of $x$.