Find \(g \circ f\) and \(f \circ g\) for \(f(x)=\sin x\) and \(g(x)=2x\)
📺 Video Explanation
📝 Question
Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as:
\[ f(x)=\sin x,\qquad g(x)=2x \]
Find:
- \((g\circ f)(x)\)
- \((f\circ g)(x)\)
Also, check whether these are equal functions.
✅ Solution
🔹 Find \((g\circ f)(x)\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute \(f(x)=\sin x\):
\[ (g\circ f)(x)=g(\sin x) \]
Since:
\[ g(x)=2x \]
So:
\[ (g\circ f)(x)=2\sin x \]
Therefore:
\[ \boxed{(g\circ f)(x)=2\sin x} \]
🔹 Find \((f\circ g)(x)\)
By definition:
\[ (f\circ g)(x)=f(g(x)) \]
Substitute \(g(x)=2x\):
\[ (f\circ g)(x)=f(2x) \]
Since:
\[ f(x)=\sin x \]
So:
\[ (f\circ g)(x)=\sin 2x \]
Therefore:
\[ \boxed{(f\circ g)(x)=\sin 2x} \]
🔹 Compare Both
We get:
\[ (g\circ f)(x)=2\sin x \]
and
\[ (f\circ g)(x)=\sin 2x \]
Using the identity:
\[ \sin 2x=2\sin x\cos x \]
So in general:
\[ 2\sin x\ne \sin 2x \]
Therefore:
\[ \boxed{g\circ f\ne f\circ g} \]
🎯 Final Answer
\[ \boxed{(g\circ f)(x)=2\sin x} \]
\[ \boxed{(f\circ g)(x)=\sin 2x} \]
Hence, these are not equal functions. :contentReference[oaicite:1]{index=1}
🚀 Exam Shortcut
- For \(g\circ f\): put \(\sin x\) into \(2x\)
- For \(f\circ g\): put \(2x\) into \(\sin x\)
- Use \(\sin 2x=2\sin x\cos x\) to compare