Find \(f \circ g\) and \(g \circ f\) for \(f(x)=x+1\) and \(g(x)=2x+3\)
📺 Video Explanation
📝 Question
Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as:
\[ f(x)=x+1,\qquad g(x)=2x+3 \]
Find:
- \((f\circ g)(x)\)
- \((g\circ f)(x)\)
✅ Solution
🔹 Find \((f\circ g)(x)\)
By definition:
\[ (f\circ g)(x)=f(g(x)) \]
Substitute \(g(x)=2x+3\):
\[ (f\circ g)(x)=f(2x+3) \]
Since:
\[ f(x)=x+1 \]
So:
\[ (f\circ g)(x)=(2x+3)+1 \]
\[ (f\circ g)(x)=2x+4 \]
Therefore:
\[ \boxed{(f\circ g)(x)=2x+4} \]
🔹 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:
\[ g(x)=2x+3 \]
So:
\[ (g\circ f)(x)=2(x+1)+3 \]
\[ (g\circ f)(x)=2x+5 \]
Therefore:
\[ \boxed{(g\circ f)(x)=2x+5} \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=2x+4} \]
\[ \boxed{(g\circ f)(x)=2x+5} \]
Hence,
\[ \boxed{f\circ g \ne g\circ f} \]
🚀 Exam Shortcut
- For \(f\circ g\): put \(g(x)\) into \(f(x)\)
- For \(g\circ f\): put \(f(x)\) into \(g(x)\)
- Combine like terms carefully