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.
-$$f ( x ) = \frac { 3 } { x + 1 }$$
A) $f ( x ) : D = \{ x \mid x \neq - 1 \} , R = \{ y \neq 0 \}$;
$$f ^ { - 1 } ( x ) : D = \{ x \mid x \neq 0 \} , R = \{ y \mid y \neq - 1 \}$$
B) $f ( x )$ : D is all real numbers, $R = \{ y | y \neq 3 \rangle$;
$f ^ { - 1 } ( x ) : D = \{ x \mid x \neq 3 \} , R$ is all real numbers

C) $f ( x )$ : $D$ is all real numbers, $R$ is all real numbers;
$\mathrm { f } ^ { - 1 } ( \mathrm { x } )$ : $\mathrm { D }$ is all real numbers, $\mathrm { R }$ is all real numbers

D) $f ( x ) : D = \{ x \mid x \neq 3 \} , R = \{ y \mid y \neq - 1 \}$;
$$\mathrm { f } ^ { - 1 } ( \mathrm { x } ) : \mathrm { D } = \{ \mathrm { x } \mid \mathrm { x } z - 1 \} , \mathrm { R } = \{ \mathrm { y } \mid \mathrm { y } \neq 3 \}$$

Question

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.
-$$f ( x ) = \frac { 3 } { x + 1 }$$
A) $f ( x ) : D = \{ x \mid x \neq - 1 \} , R = \{ y \neq 0 \}$;
$$f ^ { - 1 } ( x ) : D = \{ x \mid x \neq 0 \} , R = \{ y \mid y \neq - 1 \}$$
B) $f ( x )$ : D is all real numbers, $R = \{ y | y \neq 3 \rangle$;
 $f ^ { - 1 } ( x ) : D = \{ x \mid x \neq 3 \} , R$ is all real numbers

C) $f ( x )$ : $D$ is all real numbers, $R$ is all real numbers;
 $\mathrm { f } ^ { - 1 } ( \mathrm { x } )$ : $\mathrm { D }$ is all real numbers, $\mathrm { R }$ is all real numbers

D) $f ( x ) : D = \{ x \mid x \neq 3 \} , R = \{ y \mid y \neq - 1 \}$;
$$\mathrm { f } ^ { - 1 } ( \mathrm { x } ) : \mathrm { D } = \{ \mathrm { x } \mid \mathrm { x } z - 1 \} , \mathrm { R } = \{ \mathrm { y } \mid \mathrm { y } \neq 3 \}$$

Quizplus · Accepted Answer

The function $f(x) = \frac{3}{x+1}$ is undefined at $x = -1$, so its domain is all real numbers except $-1$. Its range is all real numbers except $0$ because the function will never equal $0$ (there's no $x$ value that will make the numerator $0$). For the inverse function $f^{-1}(x)$, the domain and range swap, so the domain is all real numbers except $0$ (since $f(x)$ never equals $0$), and the range is all real numbers except $-1$, matching the original function's domain.

The Function F Is One-To-One f(x)=3x+1f ( x ) = \frac { 3 } { x + 1 }f(x)=x+13​

The Function F Is One-To-One $f ( x ) = \frac { 3 } { x + 1 }$