Question

For \(f(x)=\frac{x-1}{x+1}\), which expression equals \((f\circ f)(x)\)?

Options

• \(\frac{x-1}{x-1}\)

• \(\frac{1}{x}\)

• \(-\frac{2}{x+1}\)

• \(-\frac{1}{x}\)

Answer 1

Substituting \(f(x)\) into itself gives \(\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}\). Simplifying the numerator and denominator results in \(\frac{-2}{2x}=-\frac1x\), on \(\mathbb{R}-\{-1,0\}\).