Find \((f \circ f)\), \((f \circ f \circ f)\), \((f \circ f \circ f)(38)\), and \(f^2\) for \(f(x)=\sqrt{x-2}\)
📺 Video Explanation
📝 Question
Let:
\[ f(x)=\sqrt{x-2} \]
Find:
- \((f\circ f)(x)\)
- \((f\circ f\circ f)(x)\)
- \((f\circ f\circ f)(38)\)
- \(f^2(x)\)
Also, show that:
\[ f\circ f\ne f^2 \]
✅ Solution
🔹 (i) Find \((f\circ f)(x)\)
By definition:
\[ (f\circ f)(x)=f(f(x)) \]
Substitute:
\[ (f\circ f)(x)=f\left(\sqrt{x-2}\right) \]
Since:
\[ f(t)=\sqrt{t-2} \]
So:
\[ (f\circ f)(x)=\sqrt{\sqrt{x-2}-2} \]
Therefore:
\[ \boxed{(f\circ f)(x)=\sqrt{\sqrt{x-2}-2}} \]
🔹 (ii) Find \((f\circ f\circ f)(x)\)
By definition:
\[ (f\circ f\circ f)(x)=f((f\circ f)(x)) \]
Substitute:
\[ f\left(\sqrt{\sqrt{x-2}-2}\right) \]
So:
\[ (f\circ f\circ f)(x)=\sqrt{\sqrt{\sqrt{x-2}-2}-2} \]
Therefore:
\[ \boxed{(f\circ f\circ f)(x)=\sqrt{\sqrt{\sqrt{x-2}-2}-2}} \]
🔹 (iii) Find \((f\circ f\circ f)(38)\)
First:
\[ f(38)=\sqrt{38-2}=\sqrt{36}=6 \]
Then:
\[ f(6)=\sqrt{6-2}=\sqrt{4}=2 \]
Now:
\[ f(2)=\sqrt{2-2}=0 \]
Therefore:
\[ \boxed{(f\circ f\circ f)(38)=0} \]
🔹 (iv) Find \(f^2(x)\)
Here:
\[ f^2(x)=[f(x)]^2 \]
So:
\[ f^2(x)=\left(\sqrt{x-2}\right)^2 \]
Therefore:
\[ \boxed{f^2(x)=x-2} \]
🔹 Show that \((f\circ f)\ne f^2\)
We found:
\[ (f\circ f)(x)=\sqrt{\sqrt{x-2}-2} \]
and
\[ f^2(x)=x-2 \]
These are clearly different expressions.
Hence:
\[ \boxed{f\circ f\ne f^2} \]
🎯 Final Answer
\[ \boxed{(f\circ f)(x)=\sqrt{\sqrt{x-2}-2}} \]
\[ \boxed{(f\circ f\circ f)(x)=\sqrt{\sqrt{\sqrt{x-2}-2}-2}} \]
\[ \boxed{(f\circ f\circ f)(38)=0} \]
\[ \boxed{f^2(x)=x-2} \]
and
\[ \boxed{f\circ f\ne f^2} \]
🚀 Exam Shortcut
- \(f\circ f\): put \(f(x)\) into \(f\) again
- \(f^2(x)\): square only the output
- Composition and squaring are different operations