We are given two functions $f$ and $g$ as sets of ordered pairs:

$f = \{(3, 1), (9, 3), (12, 4)\}$ $g = \{(1, 3), (3, 3), (4, 9), (5, 9)\}$

We are tasked with showing that the compositions $g \circ f$ and $f \circ g$ are defined, and then finding their actual compositions.

1. Checking if $g \circ f$ is defined

The composition $g \circ f$ means we apply $f$ first and then $g$ to the output of $f$. For this to be valid, for each output of $f(x)$, the output must be in the domain of $g$.

Steps:

1. The domain of $f$ consists of the values $3, 9,$ and $12$.
2. To form the composition $g \circ f$, we look at the outputs of $f$ for these values:
   • $f(3) = 1$
   • $f(9) = 3$
   • $f(12) = 4$

We need to check if these outputs ($1, 3, 4$) are in the domain of $g$.

• The domain of $g$ is $\{1, 3, 4, 5\}$, and all outputs from $f$ (i.e., $1, 3, 4$) are indeed in the domain of $g$.

Thus, $g \circ f$ is defined.

2. Checking if $f \circ g$ is defined

The composition $f \circ g$ means we apply $g$ first and then $f$ to the output of $g$. For this to be valid, for each output of $g(x)$, the output must be in the domain of $f$.

Steps:

1. The domain of $g$ consists of the values $1, 3, 4, 5$.
2. To form the composition $f \circ g$, we look at the outputs of $g$ for these values:
   • $g(1) = 3$
   • $g(3) = 3$
   • $g(4) = 9$
   • $g(5) = 9$

We need to check if these outputs ($3, 9$) are in the domain of $f$.

• The domain of $f$ is $\{3, 9, 12\}$, and both outputs from $g$ (i.e., $3, 9$) are indeed in the domain of $f$.

Thus, $f \circ g$ is defined.

3. Finding $g \circ f$

Now that we know both compositions are defined, we can compute them.

Steps for $g \circ f$:

• We apply $f$ first and then apply $g$ to the result:
  • $f(3) = 1$, then $g(1) = 3$ → $(3, 3)$
  • $f(9) = 3$, then $g(3) = 3$ → $(9, 3)$
  • $f(12) = 4$, then $g(4) = 9$ → $(12, 9)$

So, the composition $g \circ f$ is: $g \circ f = \{(3, 3), (9, 3), (12, 9)\}$

4. Finding $f \circ g$

Steps for $f \circ g$:

• We apply $g$ first and then apply $f$ to the result:
  • $g(1) = 3$, then $f(3) = 1$ → $(1, 1)$
  • $g(3) = 3$, then $f(3) = 1$ → $(3, 1)$
  • $g(4) = 9$, then $f(9) = 3$ → $(4, 3)$
  • $g(5) = 9$, then $f(9) = 3$ → $(5, 3)$

So, the composition $f \circ g$ is: $f \circ g = \{(1, 1), (3, 1), (4, 3), (5, 3)\}$

Final Answer:

• The composition $g \circ f = \{(3, 3), (9, 3), (12, 9)\}$
• The composition $f \circ g = \{(1, 1), (3, 1), (4, 3), (5, 3)\}$

Would you like more details on any step, or do you have any questions?

Further Exploration:

1. What is the composition of $f \circ g$ when the domain of $f$ is extended?
2. How would you handle compositions with more complex functions (e.g., polynomial functions)?
3. What happens if there is an output in $f$ or $g$ that does not exist in the other function’s domain?
4. How can you visualize the compositions of functions graphically?
5. What are other types of function operations besides composition (e.g., addition, subtraction, multiplication)?

Tip: When composing functions, always check that the output of one function is a valid input for the other function!