To solve the problem step by step, we will first define the functions and then find the required inverses and compositions.
### Step 1: Define the Functions
Given:
- Set \( A = \{1, 2, 3, 4\} \)
- Set \( B = \{3, 5, 7, 9\} \)
- Set \( C = \{7, 23, 47, 79\} \)
The functions are defined as:
- \( f: A \to B \) where \( f(x) = 2x + 1 \) for \( x \in A \)
- \( g: B \to C \) where \( g(x) = x^2 - 2 \) for \( x \in B \)
### Step 2: Find the Function \( f \)
Calculate \( f(x) \) for each \( x \in A \):
- \( f(1) = 2(1) + 1 = 3 \)
- \( f(2) = 2(2) + 1 = 5 \)
- \( f(3) = 2(3) + 1 = 7 \)
- \( f(4) = 2(4) + 1 = 9 \)
Thus, the mapping of \( f \) is:
\[ f = \{(1, 3), (2, 5), (3, 7), (4, 9)\} \]
### Step 3: Find the Function \( g \)
Calculate \( g(y) \) for each \( y \in B \):
- \( g(3) = 3^2 - 2 = 9 - 2 = 7 \)
- \( g(5) = 5^2 - 2 = 25 - 2 = 23 \)
- \( g(7) = 7^2 - 2 = 49 - 2 = 47 \)
- \( g(9) = 9^2 - 2 = 81 - 2 = 79 \)
Thus, the mapping of \( g \) is:
\[ g = \{(3, 7), (5, 23), (7, 47), (9, 79)\} \]
### Step 4: Find the Composition \( g \circ f \)
Now, we find \( g(f(x)) \):
- \( g(f(1)) = g(3) = 7 \)
- \( g(f(2)) = g(5) = 23 \)
- \( g(f(3)) = g(7) = 47 \)
- \( g(f(4)) = g(9) = 79 \)
Thus, the mapping of \( g \circ f \) is:
\[ g \circ f = \{(1, 7), (2, 23), (3, 47), (4, 79)\} \]
### Step 5: Find the Inverse \( (g \circ f)^{-1} \)
To find \( (g \circ f)^{-1} \), we reverse the ordered pairs:
\[ (g \circ f)^{-1} = \{(7, 1), (23, 2), (47, 3), (79, 4)\} \]
### Step 6: Find the Inverse \( f^{-1} \)
To find \( f^{-1} \):
- From \( f(x) = 2x + 1 \), we set \( y = 2x + 1 \) and solve for \( x \):
\[ y - 1 = 2x \]
\[ x = \frac{y - 1}{2} \]
Thus, the inverse function is:
\[ f^{-1}(y) = \frac{y - 1}{2} \]
Now, we find \( f^{-1}(y) \) for \( y \in B \):
- \( f^{-1}(3) = \frac{3 - 1}{2} = 1 \)
- \( f^{-1}(5) = \frac{5 - 1}{2} = 2 \)
- \( f^{-1}(7) = \frac{7 - 1}{2} = 3 \)
- \( f^{-1}(9) = \frac{9 - 1}{2} = 4 \)
Thus, the mapping of \( f^{-1} \) is:
\[ f^{-1} = \{(3, 1), (5, 2), (7, 3), (9, 4)\} \]
### Step 7: Find the Inverse \( g^{-1} \)
To find \( g^{-1} \):
- From \( g(x) = x^2 - 2 \), we set \( y = x^2 - 2 \) and solve for \( x \):
\[ y + 2 = x^2 \]
\[ x = \sqrt{y + 2} \]
Thus, the inverse function is:
\[ g^{-1}(y) = \sqrt{y + 2} \]
Now, we find \( g^{-1}(y) \) for \( y \in C \):
- \( g^{-1}(7) = \sqrt{7 + 2} = 3 \)
- \( g^{-1}(23) = \sqrt{23 + 2} = 5 \)
- \( g^{-1}(47) = \sqrt{47 + 2} = 7 \)
- \( g^{-1}(79) = \sqrt{79 + 2} = 9 \)
Thus, the mapping of \( g^{-1} \) is:
\[ g^{-1} = \{(7, 3), (23, 5), (47, 7), (79, 9)\} \]
### Step 8: Find the Composition \( f^{-1} \circ g^{-1} \)
Now, we find \( f^{-1}(g^{-1}(y)) \):
- For \( y = 7 \): \( g^{-1}(7) = 3 \), \( f^{-1}(3) = 1 \) → \( (7, 1) \)
- For \( y = 23 \): \( g^{-1}(23) = 5 \), \( f^{-1}(5) = 2 \) → \( (23, 2) \)
- For \( y = 47 \): \( g^{-1}(47) = 7 \), \( f^{-1}(7) = 3 \) → \( (47, 3) \)
- For \( y = 79 \): \( g^{-1}(79) = 9 \), \( f^{-1}(9) = 4 \) → \( (79, 4) \)
Thus, the mapping of \( f^{-1} \circ g^{-1} \) is:
\[ f^{-1} \circ g^{-1} = \{(7, 1), (23, 2), (47, 3), (79, 4)\} \]
### Step 9: Compare \( (g \circ f)^{-1} \) and \( f^{-1} \circ g^{-1} \)
We have:
- \( (g \circ f)^{-1} = \{(7, 1), (23, 2), (47, 3), (79, 4)\} \)
- \( f^{-1} \circ g^{-1} = \{(7, 1), (23, 2), (47, 3), (79, 4)\} \)
Since both sets are equal, we conclude that:
\[ (g \circ f)^{-1} = f^{-1} \circ g^{-1} \]
### Final Answer
Yes, \( (g \circ f)^{-1} = f^{-1} \circ g^{-1} \).
