Find \(f \circ g\) and \(g \circ f\) for \(f(x)=x+1\) and \(g(x)=e^x\)
📺 Video Explanation
📝 Question
Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as:
\[ f(x)=x+1,\qquad g(x)=e^x \]
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 \(g(x)=e^x\):
\[ (f\circ g)(x)=f(e^x) \]
Since:
\[ f(x)=x+1 \]
So:
\[ (f\circ g)(x)=e^x+1 \]
Therefore:
\[ \boxed{(f\circ g)(x)=e^x+1} \]
🔹 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:
\[ g(x)=e^x \]
So:
\[ (g\circ f)(x)=e^{x+1} \]
Therefore:
\[ \boxed{(g\circ f)(x)=e^{x+1}} \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=e^x+1} \]
\[ \boxed{(g\circ f)(x)=e^{x+1}} \]
Hence, both are defined for all real numbers and:
\[ \boxed{f\circ g \ne g\circ f} \]
🚀 Exam Shortcut
- For \(f\circ g\): put \(e^x\) into \(x+1\)
- For \(g\circ f\): put \(x+1\) into exponent
- Order matters in composition