Find \(g \circ f\) and \(f \circ g\) for \(f(x)=2x+3\) and \(g(x)=x^2+5\)
📺 Video Explanation
📝 Question
Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as:
\[ f(x)=2x+3,\qquad g(x)=x^2+5 \]
Find:
- \((g\circ f)(x)\)
- \((f\circ g)(x)\)
✅ Solution
🔹 Find \((g\circ f)(x)\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute \(f(x)=2x+3\):
\[ g(f(x))=g(2x+3) \]
Since:
\[ g(x)=x^2+5 \]
So:
\[ g(2x+3)=(2x+3)^2+5 \]
Expand:
\[ (2x+3)^2=4x^2+12x+9 \]
Thus:
\[ (g\circ f)(x)=4x^2+12x+14 \]
🔹 Find \((f\circ g)(x)\)
By definition:
\[ (f\circ g)(x)=f(g(x)) \]
Substitute \(g(x)=x^2+5\):
\[ f(x^2+5) \]
Since:
\[ f(x)=2x+3 \]
So:
\[ f(x^2+5)=2(x^2+5)+3 \]
Simplify:
\[ =2x^2+10+3 \]
\[ (f\circ g)(x)=2x^2+13 \]
🎯 Final Answer
\[ \boxed{(g\circ f)(x)=4x^2+12x+14} \]
\[ \boxed{(f\circ g)(x)=2x^2+13} \]
🚀 Exam Shortcut
- \(g\circ f\): put \(f(x)\) inside \(g\)
- \(f\circ g\): put \(g(x)\) inside \(f\)
- Always simplify carefully after substitution