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 }

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\}

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\}

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\}

The Function F Is One-To-One f(x)=(x+2)3−8f ( x ) = ( x + 2 ) ^ { 3 } - 8f(x)=(x+2)3−8

Multiple Choice