Question

what is the inverse function of $f(x) = \frac{x}{x - 2}$? $f^{-1}(x) = -\frac{2x}{x - 1}$ $f^{-1}(x) = \frac{x - 1}{x}$ $f^{-1}(x) = \frac{2x}{x + 1}$ $f^{-1}(x) = \frac{2x}{x - 1}$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x)=\frac{x}{x - 2} \), so we write \( y=\frac{x}{x - 2} \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap \( x \) and \( y \), getting \( x=\frac{y}{y - 2} \).

Step3: Solve for \( y \)

First, multiply both sides by \( y - 2 \): \( x(y - 2)=y \).
Expand the left side: \( xy-2x = y \).
Move all terms with \( y \) to one side: \( xy - y=2x \).
Factor out \( y \): \( y(x - 1)=2x \).
Then divide both sides by \( x - 1 \): \( y=\frac{2x}{x - 1} \).
So the inverse function \( f^{-1}(x)=\frac{2x}{x - 1} \).

Answer:

\( f^{-1}(x)=\frac{2x}{x - 1} \) (the option with this expression)