Find \(f \circ g\) and \(g \circ f\) for \(f(x)=\sqrt{x+3}\) and \(g(x)=x^2+1\)
📺 Video Explanation
📝 Question
Let:
\[ f(x)=\sqrt{x+3},\qquad g(x)=x^2+1 \]
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 \(g(x)=x^2+1\):
\[ (f\circ g)(x)=f(x^2+1) \]
Since:
\[ f(x)=\sqrt{x+3} \]
So:
\[ (f\circ g)(x)=\sqrt{(x^2+1)+3} \]
\[ \boxed{(f\circ g)(x)=\sqrt{x^2+4}} \]
Since \(x^2+4>0\) for all real \(x\), domain is all real numbers.
🔹 Find \((g\circ f)(x)\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute \(f(x)=\sqrt{x+3}\):
\[ (g\circ f)(x)=g(\sqrt{x+3}) \]
Since:
\[ g(x)=x^2+1 \]
So:
\[ (g\circ f)(x)=(\sqrt{x+3})^2+1 \]
\[ (g\circ f)(x)=x+3+1 \]
\[ \boxed{(g\circ f)(x)=x+4} \]
For \(\sqrt{x+3}\) to exist:
\[ x+3\ge 0 \]
So:
\[ x\ge -3 \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=\sqrt{x^2+4},\quad x\in\mathbb{R}} \]
\[ \boxed{(g\circ f)(x)=x+4,\quad x\ge -3} \]
🚀 Exam Shortcut
- Substitute inner function carefully
- Simplify square root and square where possible
- Always check square root domain