Question

If \( f(x) = \frac{x+1}{x-1} \), then the value of \( f(f(x)) \) is equal to:

• A. \( x \)
• B. \( 0 \)
• C. \( -x \)
• D. \( 1 \)
• E. \( 2 \)

Hint:

Functions like \( f(x) = 1/x \) or \( f(x) = (x+1)/(x-1) \) are special because they are self-inverses. If you recognize this property, you can immediately identify that applying the function twice returns the original input \( x \).

Answer 1

The Correct Option is A

Concept: A composite function \( f(f(x)) \) is found by replacing every instance of the variable \( x \) in the original function with the entire expression of the function itself. If \( f(f(x)) = x \), the function is its own inverse (involution).

Step 1: Substitute \( f(x) \) into the function expression.
Given \( f(x) = \frac{x+1}{x-1} \), we find \( f(f(x)) \): \[ f(f(x)) = \frac{f(x) + 1}{f(x) - 1} \] \[ = \frac{\frac{x+1}{x-1} + 1}{\frac{x+1}{x-1} - 1} \]

Step 2: Simplify the complex rational expression.
To clear the fractions, multiply both the numerator and the denominator by \( (x-1) \): \[ = \frac{(x+1) + 1(x-1)}{(x+1) - 1(x-1)} \] \[ = \frac{x + 1 + x - 1}{x + 1 - x + 1} \] \[ = \frac{2x}{2} = x \]