Question

The functions \(f(x)\) and \(g(x)\) are related as \(f(g(x)) = xg(f(f(x)))\), where \(f(x) = \dfrac{x}{x-1}\). What could be the functional form \(g(x)\)?

• A. \(\dfrac{1}{x}\)
• B. \(\dfrac{x}{x+1}\)
• C. \(\dfrac{x+1}{x}\)
• D. \(\dfrac{x}{x-1}\)

Hint:

First simplify \(f(f(x))\) using the given \(f(x)\); it collapses to a very simple expression.

Answer 1

The Correct Option is C

Step 1: Simplify \(f(f(x))\).
Given \(f(x)=\dfrac{x}{x-1}\), substitute \(f(x)\) into \(f\) again.
\(f(f(x)) = \dfrac{f(x)}{f(x)-1} = \dfrac{\frac{x}{x-1}}{\frac{x}{x-1}-1} = \dfrac{\frac{x}{x-1}}{\frac{1}{x-1}} = x\).
So \(f(f(x))=x\) for every valid \(x\), meaning \(f\) undoes itself.

Step 2: Rewrite the given relation.
The relation \(f(g(x)) = xg(f(f(x)))\) becomes \(f(g(x)) = xg(x)\) once \(f(f(x))\) is replaced by \(x\).

Step 3: Solve for \(g(x)\).
Let \(t=g(x)\). Then \(f(t) = \dfrac{t}{t-1}\), so the relation reads \(\dfrac{t}{t-1} = xt\).
Dividing both sides by \(t\), which is not zero, gives \(\dfrac{1}{t-1} = x\), so \(t-1 = \dfrac{1}{x}\).
Therefore \(t = 1 + \dfrac{1}{x} = \dfrac{x+1}{x}\).

Step 4: Confirm the option.
This matches \(g(x) = \dfrac{x+1}{x}\), which is option (c) among the four kept choices.

Final Answer:
The functional form is \(g(x) = \dfrac{x+1}{x}\). \[ \boxed{g(x)=\dfrac{x+1}{x}} \]