Question

Given the function:

\[ f(x) = \frac{2x - 3}{3x - 2} \]

and if \( f_n(x) = (f \circ f \circ \ldots \circ f)(x) \) is applied \( n \) times, find \( f_{32}(x) \).

• A. \(\frac{2x-3}{3x-2}\)
• B. \(x\)
• C. \(\frac{3x+2}{2x+3}\)
• D. \(f_{23}(x)\)

Hint:

When a function's repeated composition with itself leads back to the input variable, it exhibits periodic behavior which in this case is every 32 iterations.

Answer 1

The Correct Option is B

To find \( f_{32}(x) \) for the function \( f(x) = \frac{2x - 3}{3x - 2} \), we need to determine the pattern or periodicity of the function when composed multiple times. Step 1: Compute \( f_2(x) = f(f(x)) \) \[ f(f(x)) = f\left( \frac{2x - 3}{3x - 2} \right) = \frac{2\left( \frac{2x - 3}{3x - 2} \right) - 3}{3\left( \frac{2x - 3}{3x - 2} \right) - 2} \] Simplify the numerator and denominator: \[ \text{Numerator} = \frac{4x - 6 - 9x + 6}{3x - 2} = \frac{-5x}{3x - 2} \] \[ \text{Denominator} = \frac{6x - 9 - 6x + 4}{3x - 2} = \frac{-5}{3x - 2} \] Thus: \[ f(f(x)) = \frac{ \frac{-5x}{3x - 2} }{ \frac{-5}{3x - 2} } = x \] Step 2: Determine the Periodicity Since \( f_2(x) = x \), the function \( f(x) \) is its own inverse, and composing it twice returns the original input. This implies that: \[ f_{2k}(x) = x \quad \text{for any integer } k \] \[ f_{2k+1}(x) = f(x) \quad \text{for any integer } k \] Step 3: Compute \( f_{32}(x) \) Since 32 is an even number, we can write it as \( 2 \times 16 \). Therefore: \[ f_{32}(x) = f_{2 \times 16}(x) = x \] Final Answer: \[ \boxed{x} \] This corresponds to option (2).