Find \((g \circ f)^{-1}\) and Verify \((g \circ f)^{-1}=f^{-1}\circ g^{-1}\)
📺 Video Explanation
📝 Question
Let:
\[ A=\{1,2,3,4\},\quad B=\{3,5,7,9\},\quad C=\{7,23,47,79\} \]
and:
\[ f:A\to B,\qquad f(x)=2x+1 \]
\[ g:B\to C,\qquad g(x)=x^2-2 \]
Find:
\[ (g\circ f)^{-1}\quad \text{and}\quad f^{-1}\circ g^{-1} \]
as sets of ordered pairs, and verify:
\[ (g\circ f)^{-1}=f^{-1}\circ g^{-1} \]
✅ Solution
🔹 Step 1: Find \(f\) as ordered pairs
Using:
\[ f(x)=2x+1 \]
for \(x\in A\):
- \(f(1)=3\)
- \(f(2)=5\)
- \(f(3)=7\)
- \(f(4)=9\)
So:
\[ f=\{(1,3),(2,5),(3,7),(4,9)\} \]
Hence:
\[ \boxed{f^{-1}=\{(3,1),(5,2),(7,3),(9,4)\}} \]
🔹 Step 2: Find \(g\) as ordered pairs
Using:
\[ g(x)=x^2-2 \]
for \(x\in B\):
- \(g(3)=7\)
- \(g(5)=23\)
- \(g(7)=47\)
- \(g(9)=79\)
So:
\[ g=\{(3,7),(5,23),(7,47),(9,79)\} \]
Hence:
\[ \boxed{g^{-1}=\{(7,3),(23,5),(47,7),(79,9)\}} \]
🔹 Step 3: Find \(g\circ f\)
By composition:
\[ (g\circ f)(x)=g(f(x)) \]
- \((g\circ f)(1)=g(3)=7\)
- \((g\circ f)(2)=g(5)=23\)
- \((g\circ f)(3)=g(7)=47\)
- \((g\circ f)(4)=g(9)=79\)
Thus:
\[ g\circ f=\{(1,7),(2,23),(3,47),(4,79)\} \]
Therefore:
\[ \boxed{ (g\circ f)^{-1} = \{(7,1),(23,2),(47,3),(79,4)\} } \]
🔹 Step 4: Find \(f^{-1}\circ g^{-1}\)
Apply \(g^{-1}\) first:
- \(7\to3\)
- \(23\to5\)
- \(47\to7\)
- \(79\to9\)
Then apply \(f^{-1}\):
- \(3\to1\)
- \(5\to2\)
- \(7\to3\)
- \(9\to4\)
So:
\[ f^{-1}\circ g^{-1} = \{(7,1),(23,2),(47,3),(79,4)\} \]
🎯 Final Answer
\[ \boxed{ (g\circ f)^{-1} = \{(7,1),(23,2),(47,3),(79,4)\} } \]
\[ \boxed{ f^{-1}\circ g^{-1} = \{(7,1),(23,2),(47,3),(79,4)\} } \]
Therefore:
\[ \boxed{(g\circ f)^{-1}=f^{-1}\circ g^{-1}} \]
🚀 Exam Shortcut
- First find mappings clearly
- Inverse = reverse ordered pairs
- For composition inverse: reverse order