Verify Associativity of Composite Functions for Given Mappings
📺 Video Explanation
📝 Question
Let:
\[ f:\mathbb{N}\to \mathbb{Z}_0,\qquad f(x)=2x \]
\[ g:\mathbb{Z}_0\to \mathbb{Q},\qquad g(x)=\frac{1}{x} \]
\[ h:\mathbb{Q}\to \mathbb{R},\qquad h(x)=e^x \]
Verify associativity:
\[ h\circ(g\circ f)=(h\circ g)\circ f \]
✅ Solution
🔹 Step 1: Check that compositions are defined
Since:
- \(f:\mathbb{N}\to \mathbb{Z}_0\)
- \(g:\mathbb{Z}_0\to \mathbb{Q}\)
- \(h:\mathbb{Q}\to \mathbb{R}\)
Both:
\[ h\circ(g\circ f)\quad \text{and}\quad (h\circ g)\circ f \]
are well-defined from \(\mathbb{N}\to\mathbb{R}\). :contentReference[oaicite:1]{index=1}
🔹 Step 2: Find \(g\circ f\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Substitute \(f(x)=2x\):
\[ (g\circ f)(x)=g(2x)=\frac{1}{2x} \]
🔹 Step 3: Find \(h\circ(g\circ f)\)
\[ (h\circ(g\circ f))(x)=h\left(\frac{1}{2x}\right) \]
Since:
\[ h(x)=e^x \]
So:
\[ (h\circ(g\circ f))(x)=e^{\frac{1}{2x}} \]
🔹 Step 4: Find \(h\circ g\)
\[ (h\circ g)(x)=h(g(x)) \]
Substitute:
\[ (h\circ g)(x)=h\left(\frac{1}{x}\right)=e^{\frac{1}{x}} \]
🔹 Step 5: Find \((h\circ g)\circ f\)
\[ ((h\circ g)\circ f)(x)=(h\circ g)(f(x)) \]
Substitute \(f(x)=2x\):
\[ ((h\circ g)\circ f)(x)=e^{\frac{1}{2x}} \]
🎯 Final Answer
\[ (h\circ(g\circ f))(x)=e^{\frac{1}{2x}} \]
and
\[ ((h\circ g)\circ f)(x)=e^{\frac{1}{2x}} \]
Therefore,
\[ \boxed{h\circ(g\circ f)=(h\circ g)\circ f} \]
Hence, associativity is verified. :contentReference[oaicite:2]{index=2}
🚀 Exam Shortcut
- First check if range of one function fits domain of next
- Compute inside composition first
- Associativity means both grouping methods give same result