Question

What is the inverse function of \( f(x)=\frac{x}{x-2} \) ? \( f^{-1}(x)=\frac{2 k}{x+1} \) \[ f^{-1}(x)=\frac{2 x}{x-1} \] \( f^{-1}(x)=\frac{z-2}{x} \) \[ f^{-1}(x)=-\frac{2 x}{x-1} \]

          What is the inverse function of \( f(x)=\frac{x}{x-2} \) ?
\( f^{-1}(x)=\frac{2 k}{x+1} \)
\[
f^{-1}(x)=\frac{2 x}{x-1}
\]
\( f^{-1}(x)=\frac{z-2}{x} \)
\[
f^{-1}(x)=-\frac{2 x}{x-1}
\]

What is the inverse function of f(x)=(x)/(x-2) ?
f^-1(x)=(2 k)/(x+1)

f^-1(x)=(2 x)/(x-1)

f^-1(x)=(z-2)/(x)

f^-1(x)=-(2 x)/(x-1)

Added by Hunter R.

Algebra and Trigonometry

James Stewart, Lothar Redlin, Saleem… 3rd Edition

Chapter 3

Instant Answer

Solved by Expert David Mccaslin

Step 1

Step 1: Set y = x/(x - 2) to represent f(x). Show more…

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What is the inverse function of \( f(x)=\frac{x}{x-2} \) ? \( f^{-1}(x)=\frac{2 k}{x+1} \) \[ f^{-1}(x)=\frac{2 x}{x-1} \] \( f^{-1}(x)=\frac{z-2}{x} \) \[ f^{-1}(x)=-\frac{2 x}{x-1} \]