Answer
a) $f^{-1}(x)=\dfrac{2x+1}{x-1}$
b) See graph
c) The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y=x$.
d) $D_f=(-\infty,2)\cup(2,\infty),R_f=(-\infty,1)\cup(1,\infty)$
$D_{f^{-1}}=(-\infty,1)\cup(1,\infty),R_{f^{-1}}=(-\infty,2)\cup(2,\infty)$
Work Step by Step
We are given the function:
$f(x)=\dfrac{x+1}{x-2}$
$y=\dfrac{x+1}{x-2}$
a) Determine the inverse $f^{-1}$. Interchange $x$ and $y$:
$x=\dfrac{y+1}{y-2}$
$x(y-2)=y+1$
$xy-2x=y+1$
$xy-y=2x+1$
$y(x-1)=2x+1$
$y=\dfrac{2x+1}{x-1}$
$f^{-1}(x)=\dfrac{2x+1}{x-1}$
b) Graph both functions.
c) The graph of the function $f^{-1}$ is the reflection of the graph of $f$ across the line $y=x$.
d) Determine the domain and range of $f$:
$D_f=(-\infty,2)\cup(2,\infty)$
$R_f=(-\infty,1)\cup(1,\infty)$
Determine the domain and range of $f^{-1}$:
$D_{f^{-1}}=(-\infty,1)\cup(1,\infty)$
$R_{f^{-1}}=(-\infty,2)\cup(2,\infty)$
