Find \(f \circ g\) and \(g \circ f\) for \(f(x)=x^2\) and \(g(x)=\cos x\)
📺 Video Explanation
📝 Question
Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as:
\[ f(x)=x^2,\qquad g(x)=\cos x \]
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 \(g(x)=\cos x\):
\[ (f\circ g)(x)=f(\cos x) \]
Since:
\[ f(x)=x^2 \]
So:
\[ (f\circ g)(x)=(\cos x)^2 \]
Therefore:
\[ \boxed{(f\circ g)(x)=\cos^2 x} \]
🔹 Find \((g\circ f)(x)\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute \(f(x)=x^2\):
\[ (g\circ f)(x)=g(x^2) \]
Since:
\[ g(x)=\cos x \]
So:
\[ (g\circ f)(x)=\cos(x^2) \]
Therefore:
\[ \boxed{(g\circ f)(x)=\cos(x^2)} \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=\cos^2 x} \]
\[ \boxed{(g\circ f)(x)=\cos(x^2)} \]
Hence, both composite functions are defined for all real numbers, and in general:
\[ \boxed{f\circ g \ne g\circ f} \]
🚀 Exam Shortcut
- For \(f\circ g\): put \(\cos x\) into \(x^2\)
- For \(g\circ f\): put \(x^2\) into \(\cos x\)
- Composition usually depends on order