Find \(f \circ g\) and \(g \circ f\) for \(f(x)=e^x\) and \(g(x)=\ln x\)
📺 Video Explanation
📝 Question
Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:(0,\infty)\to\mathbb{R}\) be defined as:
\[ f(x)=e^x,\qquad g(x)=\ln x \]
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:
\[ (f\circ g)(x)=f(\ln x) \]
Since:
\[ f(x)=e^x \]
So:
\[ (f\circ g)(x)=e^{\ln x} \]
Using the inverse property of exponential and logarithm:
\[ e^{\ln x}=x,\qquad x>0 \]
Therefore:
\[ \boxed{(f\circ g)(x)=x,\quad x>0} \]
🔹 Find \((g\circ f)(x)\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute:
\[ (g\circ f)(x)=g(e^x) \]
Since:
\[ g(x)=\ln x \]
So:
\[ (g\circ f)(x)=\ln(e^x) \]
Using the inverse property:
\[ \ln(e^x)=x \]
for all real \(x\), because \(e^x>0\). :contentReference[oaicite:1]{index=1}
Therefore:
\[ \boxed{(g\circ f)(x)=x} \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=x,\quad x>0} \]
\[ \boxed{(g\circ f)(x)=x} \]
Hence, both compositions give the identity function on their respective domains. :contentReference[oaicite:2]{index=2}
🚀 Exam Shortcut
- \(e^x\) and \(\ln x\) are inverse functions
- \(e^{\ln x}=x\) for \(x>0\)
- \(\ln(e^x)=x\) for all real \(x\)