Answer
a) $f^{-1}(x)=\dfrac{-5x-2}{3x-1}$
b) See graph
c) The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y=x$.
d) $D_f=\left(-\infty,-\dfrac{5}{3}\right)\cup\left(-\dfrac{5}{3},\infty\right),R_f=\left(-\infty,\dfrac{1}{3}\right)\cup\left(\dfrac{1}{3},\infty\right)$
$D_{f^{-1}}=\left(-\infty,\dfrac{1}{3}\right)\cup\left(\dfrac{1}{3},\infty\right),R_{f^{-1}}=\left(-\infty,-\dfrac{5}{3}\right)\cup\left(-\dfrac{5}{3},\infty\right)$
Work Step by Step
We are given the function:
$f(x)=\dfrac{x-2}{3x+5}$
$y=\dfrac{x-2}{3x+5}$
a) Determine the inverse $f^{-1}$. Interchange $x$ and $y$:
$x=\dfrac{y-2}{3y+5}$
$x(3y+5)=y-2$
$3xy+5x=y-2$
$3xy-y=-5x-2$
$y(3x-1)=-5x-2$
$y=\dfrac{-5x-2}{3x-1}$
$f^{-1}(x)=\dfrac{-5x-2}{3x-1}$
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 $f$:
$D_f=\left(-\infty,-\dfrac{5}{3}\right)\cup\left(-\dfrac{5}{3},\infty\right)$
$R_f=\left(-\infty,\dfrac{1}{3}\right)\cup\left(\dfrac{1}{3},\infty\right)$
Determine the domain and range of $f^{-1}$:
$D_{f^{-1}}=\left(-\infty,\dfrac{1}{3}\right)\cup\left(\dfrac{1}{3},\infty\right)$
$R_{f^{-1}}=\left(-\infty,-\dfrac{5}{3}\right)\cup\left(-\dfrac{5}{3},\infty\right)$
