The Function F Is One-To-One $$f ( x ) = \sqrt { 5 x - 1 }$$

Multiple Choice

The function f is one-to-one. State the domain and the range of f and f^-1. Write the domain and range in set-builder
notation.
- $$f ( x ) = \sqrt { 5 x - 1 }$$

A) $f ( x ) : D = \{ x \mid x \geq 0 \} , R = \{ y \mid y \geq 0 \}$ ;
B) $f ( x ) : D = \left\{ x \mid x \geq \frac { 1 } { 5 } \right\} , R = \{ y \mid y \geq 0 \}$ ;

$\mathrm { f } ^ { - 1 } ( \mathrm { x } ) : \mathrm { D }$ is all real numbers, $R = \left\{ \mathrm { y } \mid \mathrm { y } \geq \frac { 1 } { 5 } \right\}$

C) $f ( x ) : D = \left\{ x \mid x \geq \frac { 1 } { 5 } \right\} , R = \{ y | y \geq 0 \rangle$ ;
$\mathrm { f } ^ { - 1 } ( \mathrm { x } ) : \mathrm { D } = \{ \mathrm { x } \mid \mathrm { x } \geq 0 \} , \mathrm { R } = \left\{ \mathrm { y } \mid \mathrm { y } \geq \frac { 1 } { 5 } \right\}$

D) $f ( x ) : D = \left\{ x \mid x \geq \frac { 1 } { 5 } \right\} , R$ is all real numbers; $f ^ { - 1 } ( x ) : D = \left\{ x | x \geq 0 \rangle , R = \left\{ y \mid y \geq \frac { 1 } { 5 } \right\} \right.$
$f ^ { - 1 } ( x ) : D$ is all real numbers, $R = \left\{ y \mid y \geq \frac { 1 } { 5 } \right\}$