Show \(f \circ g = g \circ f = I_{\mathbb{R}}\) for \(f(x)=x+1\) and \(g(x)=x-1\)
📺 Video Explanation
📝 Question
Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined by:
\[ f(x)=x+1,\qquad g(x)=x-1 \]
Show that:
\[ f\circ g=g\circ f=I_{\mathbb{R}} \]
✅ Solution
🔹 Find \((f\circ g)(x)\)
By definition:
\[ (f\circ g)(x)=f(g(x)) \]
Substitute \(g(x)=x-1\):
\[ (f\circ g)(x)=f(x-1) \]
Since:
\[ f(x)=x+1 \]
So:
\[ f(x-1)=(x-1)+1=x \]
Thus:
\[ (f\circ g)(x)=x \]
🔹 Find \((g\circ f)(x)\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute \(f(x)=x+1\):
\[ (g\circ f)(x)=g(x+1) \]
Since:
\[ g(x)=x-1 \]
So:
\[ g(x+1)=(x+1)-1=x \]
Thus:
\[ (g\circ f)(x)=x \]
🔹 Compare with Identity Function
Identity function on real numbers is:
\[ I_{\mathbb{R}}(x)=x \]
Since:
\[ (f\circ g)(x)=x=(g\circ f)(x) \]
for every \(x\in\mathbb{R}\), therefore:
\[ \boxed{f\circ g=g\circ f=I_{\mathbb{R}}} \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=x} \]
\[ \boxed{(g\circ f)(x)=x} \]
Hence,
\[ \boxed{f\circ g=g\circ f=I_{\mathbb{R}}} \]
🚀 Exam Shortcut
- \(+1\) and \(-1\) cancel each other
- First apply inner function, then outer
- If result is same input, it is identity function