Question

If \( f(x) = \frac{3x - 4}{2x - 3} \), then \( f(f(f(x))) \) will be:

• A. \( x \)
• B. \( 2x \)
• C. \( \frac{2x - 3}{3x - 4} \)
• D. \( \frac{3x - 4}{2x - 3} \)

Hint:

When applying composite functions, check if the result loops back to the original function. In this case, we observed that \( f(f(x)) = x \), and applying \( f \) again brought us back to \( f(x) \).

Answer 1

The Correct Option is D

We are given \( f(x) = \frac{3x - 4}{2x - 3} \), and we need to find \( f(f(f(x))) \).
Step 1: Compute \( f(f(x)) \).
To find \( f(f(x)) \), we first substitute \( f(x) = \frac{3x - 4}{2x - 3} \) into itself. Thus, we compute: \[ f(f(x)) = f\left( \frac{3x - 4}{2x - 3} \right) = \frac{3\left( \frac{3x - 4}{2x - 3} \right) - 4}{2\left( \frac{3x - 4}{2x - 3} \right) - 3}. \] Simplifying both the numerator and denominator: \[ \text{Numerator: } 3\left( \frac{3x - 4}{2x - 3} \right) - 4 = \frac{9x - 12}{2x - 3} - 4 = \frac{9x - 12 - 4(2x - 3)}{2x - 3} = \frac{9x - 12 - 8x + 12}{2x - 3} = \frac{x}{2x - 3}. \] \[ \text{Denominator: } 2\left( \frac{3x - 4}{2x - 3} \right) - 3 = \frac{6x - 8}{2x - 3} - 3 = \frac{6x - 8 - 3(2x - 3)}{2x - 3} = \frac{6x - 8 - 6x + 9}{2x - 3} = \frac{1}{2x - 3}. \] So, we have: \[ f(f(x)) = \frac{\frac{x}{2x - 3}}{\frac{1}{2x - 3}} = x. \]
Step 2: Compute \( f(f(f(x))) \).
Now that we know \( f(f(x)) = x \), we can apply \( f \) again: \[ f(f(f(x))) = f(x) = \frac{3x - 4}{2x - 3}. \] Thus, the value of \( f(f(f(x))) \) is \( \frac{3x - 4}{2x - 3} \), which is the same as \( f(x) \).
Step 3: Conclusion.
The answer is \( \frac{3x - 4}{2x - 3} \), which matches option (D).