Find \(f \circ g\) and \(g \circ f\) for \(f(x)=\sin^{-1}x\) and \(g(x)=x^2\)
📺 Video Explanation
📝 Question
Let functions be defined as:
\[ f(x)=\sin^{-1}x,\qquad g(x)=x^2 \]
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 \(g(x)=x^2\):
\[ (f\circ g)(x)=\sin^{-1}(x^2) \]
For \(\sin^{-1}(x^2)\) to be defined:
\[ -1\le x^2\le 1 \]
Since \(x^2\ge 0\), we need:
\[ x^2\le 1 \]
So:
\[ -1\le x\le 1 \]
Therefore:
\[ \boxed{(f\circ g)(x)=\sin^{-1}(x^2),\quad -1\le x\le 1} \]
🔹 Find \((g\circ f)(x)\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute \(f(x)=\sin^{-1}x\):
\[ (g\circ f)(x)=g(\sin^{-1}x) \]
Since:
\[ g(x)=x^2 \]
So:
\[ (g\circ f)(x)=(\sin^{-1}x)^2 \]
For \(\sin^{-1}x\) to exist:
\[ -1\le x\le 1 \]
Therefore:
\[ \boxed{(g\circ f)(x)=(\sin^{-1}x)^2,\quad -1\le x\le 1} \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=\sin^{-1}(x^2),\quad -1\le x\le 1} \]
\[ \boxed{(g\circ f)(x)=(\sin^{-1}x)^2,\quad -1\le x\le 1} \]
Hence, both compositions are defined only on:
\[ \boxed{[-1,1]} \]
🚀 Exam Shortcut
- Always check domain of inverse trig functions
- \(\sin^{-1}t\) needs \(-1\le t\le 1\)
- Order matters in composition