Find \(f \circ g\) and \(g \circ f\) for \(f(x)=x^2\) and \(g(x)=\sqrt{x}\) on \(\mathbb{R}_+\)
📺 Video Explanation
📝 Question
Let \(\mathbb{R}_+\) be the set of all non-negative real numbers.
If functions \(f:\mathbb{R}_+\to\mathbb{R}_+\) and \(g:\mathbb{R}_+\to\mathbb{R}_+\) are defined by:
\[ f(x)=x^2,\qquad g(x)=\sqrt{x} \]
Find:
- \((f\circ g)(x)\)
- \((g\circ f)(x)\)
Also, check whether they are equal functions.
✅ Solution
🔹 Find \((f\circ g)(x)\)
By definition:
\[ (f\circ g)(x)=f(g(x)) \]
Substitute:
\[ (f\circ g)(x)=f(\sqrt{x}) \]
Since:
\[ f(x)=x^2 \]
So:
\[ f(\sqrt{x})=(\sqrt{x})^2=x \]
Therefore:
\[ \boxed{(f\circ g)(x)=x} \]
🔹 Find \((g\circ f)(x)\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute:
\[ (g\circ f)(x)=g(x^2) \]
Since:
\[ g(x)=\sqrt{x} \]
So:
\[ g(x^2)=\sqrt{x^2} \]
Because \(x\in\mathbb{R}_+\), we have \(x\ge 0\), hence:
\[ \sqrt{x^2}=x \]
Therefore:
\[ \boxed{(g\circ f)(x)=x} \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=x} \]
\[ \boxed{(g\circ f)(x)=x} \]
Hence, both are equal functions and each is the identity function on \(\mathbb{R}_+\). :contentReference[oaicite:1]{index=1}
🚀 Exam Shortcut
- Square and square root cancel each other on non-negative numbers
- \((\sqrt{x})^2=x\)
- \(\sqrt{x^2}=x\) only because \(x\ge 0\)