## Algebra and Trigonometry 10th Edition

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) $D_f=\left(-\infty,-\dfrac{5}{3}\right)\cup\left(-\dfrac{5}{3},\infty\right),R_f=\left(-\infty,\dfrac{1}{3}\right)\cup\left(\dfrac{1}{3},\infty\right)$ $D_{f^{-1}}=\left(-\infty,\dfrac{1}{3}\right)\cup\left(\dfrac{1}{3},\infty\right),R_{f^{-1}}=\left(-\infty,-\dfrac{5}{3}\right)\cup\left(-\dfrac{5}{3},\infty\right)$
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 $f^{-1}$: $D_{f^{-1}}=\left(-\infty,\dfrac{1}{3}\right)\cup\left(\dfrac{1}{3},\infty\right)$ $R_{f^{-1}}=\left(-\infty,-\dfrac{5}{3}\right)\cup\left(-\dfrac{5}{3},\infty\right)$