Verify \(h\circ(g\circ f)=(h\circ g)\circ f\) for Given Functions
📺 Video Explanation
📝 Question
Let:
\[ f:\mathbb{N}\to\mathbb{N},\qquad f(x)=2x \]
\[ g:\mathbb{N}\to\mathbb{N},\qquad g(y)=3y+4 \]
\[ h:\mathbb{N}\to\mathbb{R},\qquad h(z)=\sin z \]
Show that:
\[ h\circ(g\circ f)=(h\circ g)\circ f \]
✅ Solution
🔹 Step 1: Find \(g\circ f\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute \(f(x)=2x\):
\[ (g\circ f)(x)=g(2x)=3(2x)+4 \]
\[ (g\circ f)(x)=6x+4 \]
🔹 Step 2: Find \(h\circ(g\circ f)\)
\[ (h\circ(g\circ f))(x)=h(6x+4) \]
Since:
\[ h(z)=\sin z \]
So:
\[ (h\circ(g\circ f))(x)=\sin(6x+4) \]
🔹 Step 3: Find \(h\circ g\)
By definition:
\[ (h\circ g)(y)=h(g(y)) \]
Substitute \(g(y)=3y+4\):
\[ (h\circ g)(y)=\sin(3y+4) \]
🔹 Step 4: Find \((h\circ g)\circ f\)
\[ ((h\circ g)\circ f)(x)=(h\circ g)(f(x)) \]
Substitute \(f(x)=2x\):
\[ ((h\circ g)\circ f)(x)=\sin(3(2x)+4) \]
\[ ((h\circ g)\circ f)(x)=\sin(6x+4) \]
🎯 Final Answer
\[ (h\circ(g\circ f))(x)=\sin(6x+4) \]
and
\[ ((h\circ g)\circ f)(x)=\sin(6x+4) \]
Therefore,
\[ \boxed{h\circ(g\circ f)=(h\circ g)\circ f} \]
Hence, associativity is verified. :contentReference[oaicite:1]{index=1}
🚀 Exam Shortcut
- First apply inner function, then outer function
- Associativity means regrouping does not change the result
- Both sides must give same expression for all \(x\)