Consider the one-to-one function $f ( x ) = \frac { 1 } { x - 2 }$ . Then $f ^ { - 1 } ( x )$ is

A) $x - 2$
B) $x + 2$
C) $\frac { 1 } { x + 2 }$
D) $\frac { 1 } { x } + 2$
E) $\frac { 2 } { x } + 1$

Question

Consider the one-to-one function $f ( x ) = \frac { 1 } { x - 2 }$ . Then $f ^ { - 1 } ( x )$ is

A) $x - 2$
B) $x + 2$
C) $\frac { 1 } { x + 2 }$
D) $\frac { 1 } { x } + 2$
E) $\frac { 2 } { x } + 1$

Quizplus · Accepted Answer

To find the inverse function $f^{-1}(x)$, we swap $x$ and $y$ in the original function and solve for $y$. Starting with $y = \frac{1}{x - 2}$, we swap to get $x = \frac{1}{y - 2}$. Solving for $y$, we multiply both sides by $y - 2$ and then by $x$ to isolate $y$, leading to $y = \frac{1}{x} + 2$, which matches choice D.

Consider the One-To-One Function f(x)=1x−2f ( x ) = \frac { 1 } { x - 2 }f(x)=x−21​

Consider the One-To-One Function $f ( x ) = \frac { 1 } { x - 2 }$