Show \(f\) and \(g\) are Bijective and Verify \((g \circ f)^{-1}=f^{-1}\circ g^{-1}\)
📺 Video Explanation
📝 Question
Let:
\[ f:\mathbb{Q}\to\mathbb{Q},\qquad f(x)=2x \]
and:
\[ g:\mathbb{Q}\to\mathbb{Q},\qquad g(x)=x+2 \]
Show that both are bijections and verify:
\[ (g\circ f)^{-1}=f^{-1}\circ g^{-1} \]
✅ Solution
🔹 Step 1: Show that \(f(x)=2x\) is bijective
One-one:
If:
\[ f(x_1)=f(x_2) \]
Then:
\[ 2x_1=2x_2 \Rightarrow x_1=x_2 \]
So \(f\) is one-one.
Onto:
Let:
\[ y\in\mathbb{Q} \]
Take:
\[ x=\frac{y}{2}\in\mathbb{Q} \]
Then:
\[ f(x)=2\cdot\frac{y}{2}=y \]
So \(f\) is onto.
Hence:
\[ \boxed{f^{-1}(x)=\frac{x}{2}} \]
🔹 Step 2: Show that \(g(x)=x+2\) is bijective
One-one:
If:
\[ g(x_1)=g(x_2) \]
Then:
\[ x_1+2=x_2+2 \Rightarrow x_1=x_2 \]
So \(g\) is one-one.
Onto:
Let:
\[ y\in\mathbb{Q} \]
Take:
\[ x=y-2\in\mathbb{Q} \]
Then:
\[ g(x)=y \]
Hence:
\[ \boxed{g^{-1}(x)=x-2} \]
🔹 Step 3: Find \(g\circ f\)
\[ (g\circ f)(x)=g(f(x)) \]
\[ =g(2x) \]
\[ =2x+2 \]
So:
\[ \boxed{(g\circ f)(x)=2x+2} \]
🔹 Step 4: Find \((g\circ f)^{-1}\)
Let:
\[ y=2x+2 \]
Then:
\[ 2x=y-2 \]
\[ x=\frac{y-2}{2} \]
Thus:
\[ \boxed{(g\circ f)^{-1}(x)=\frac{x-2}{2}} \]
🔹 Step 5: Verify \(f^{-1}\circ g^{-1}\)
\[ (f^{-1}\circ g^{-1})(x)=f^{-1}(g^{-1}(x)) \]
Substitute:
\[ =f^{-1}(x-2) \]
\[ =\frac{x-2}{2} \]
So:
\[ (f^{-1}\circ g^{-1})(x)=\frac{x-2}{2} \]
Therefore:
\[ \boxed{(g\circ f)^{-1}=f^{-1}\circ g^{-1}} \]
🎯 Final Answer
\[ \boxed{f^{-1}(x)=\frac{x}{2},\qquad g^{-1}(x)=x-2} \]
\[ \boxed{(g\circ f)^{-1}(x)=\frac{x-2}{2}} \]
and:
\[ \boxed{(g\circ f)^{-1}=f^{-1}\circ g^{-1}} \]
🚀 Exam Shortcut
- Linear functions with non-zero slope are bijections
- Find inverse by solving for \(x\)
- Composition inverse reverses order