Find \((f \circ g)\), \((g \circ f)\), \((f \circ f)\), and \(f^2\) for \(f(x)=2x+5\) and \(g(x)=x^2+1\)
📺 Video Explanation
📝 Question
Let functions \(f:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be defined as:
\[ f(x)=2x+5,\qquad g(x)=x^2+1 \]
Find:
- \((f\circ g)(x)\)
- \((g\circ f)(x)\)
- \((f\circ f)(x)\)
- \(f^2(x)\)
✅ Solution
🔹 (i) 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:
\[ f(x)=2x+5 \]
So:
\[ (f\circ g)(x)=2(x^2+1)+5 \]
\[ \boxed{(f\circ g)(x)=2x^2+7} \]
🔹 (ii) Find \((g\circ f)(x)\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute \(f(x)=2x+5\):
\[ (g\circ f)(x)=g(2x+5) \]
Since:
\[ g(x)=x^2+1 \]
So:
\[ (g\circ f)(x)=(2x+5)^2+1 \]
Expand:
\[ (2x+5)^2=4x^2+20x+25 \]
Therefore:
\[ \boxed{(g\circ f)(x)=4x^2+20x+26} \]
🔹 (iii) Find \((f\circ f)(x)\)
By definition:
\[ (f\circ f)(x)=f(f(x)) \]
Substitute \(f(x)=2x+5\):
\[ (f\circ f)(x)=f(2x+5) \]
So:
\[ (f\circ f)(x)=2(2x+5)+5 \]
\[ \boxed{(f\circ f)(x)=4x+15} \]
🔹 (iv) Find \(f^2(x)\)
Here \(f^2(x)\) means square of function value:
\[ f^2(x)=[f(x)]^2 \]
So:
\[ f^2(x)=(2x+5)^2 \]
Expand:
\[ \boxed{f^2(x)=4x^2+20x+25} \]
🎯 Final Answer
\[ \boxed{(f\circ g)(x)=2x^2+7} \]
\[ \boxed{(g\circ f)(x)=4x^2+20x+26} \]
\[ \boxed{(f\circ f)(x)=4x+15} \]
\[ \boxed{f^2(x)=4x^2+20x+25} \]
Also,
\[ \boxed{(f\circ f)(x)\ne f^2(x)} \]
🚀 Exam Shortcut
- \(f\circ g\): substitute \(g(x)\) into \(f(x)\)
- \(g\circ f\): substitute \(f(x)\) into \(g(x)\)
- \(f\circ f\): compose function with itself
- \(f^2(x)\): square the output of \(f(x)\)