Prove \(g \circ f=f+f\) for Any Real Function \(f\) and \(g(x)=2x\)
📺 Video Explanation
📝 Question
Let \(f\) be any real-valued function and let:
\[ g(x)=2x \]
Prove that:
\[ g\circ f=f+f \]
✅ Solution
🔹 Step 1: Write the composite function
By definition of composition:
\[ (g\circ f)(x)=g(f(x)) \]
🔹 Step 2: Substitute \(g(x)=2x\)
Since:
\[ g(t)=2t \]
for any real number \(t\), put \(t=f(x)\):
\[ (g\circ f)(x)=2f(x) \]
🔹 Step 3: Write \(f+f\)
By definition of sum of functions:
\[ (f+f)(x)=f(x)+f(x) \]
So:
\[ (f+f)(x)=2f(x) \]
🔹 Step 4: Compare both sides
We get:
\[ (g\circ f)(x)=2f(x) \]
and
\[ (f+f)(x)=2f(x) \]
Thus, for every real number \(x\):
\[ (g\circ f)(x)=(f+f)(x) \]
🎯 Final Answer
\[ (g\circ f)(x)=g(f(x))=2f(x) \]
and
\[ (f+f)(x)=f(x)+f(x)=2f(x) \]
Therefore:
\[ \boxed{g\circ f=f+f} \]
Hence proved. :contentReference[oaicite:1]{index=1}
🚀 Exam Shortcut
- Composition means substitute \(f(x)\) into \(g\)
- \(g(x)=2x\) doubles the input
- \(f+f\) also doubles the function value