Restrict the Domain of the Function F So That the Function

Restrict the domain of the function f so that the function is one-to-one and has an inverse function.Then find the inverse function f^-1.State the domains and ranges of f and f^-1.

F(x) = -6x² + 2

A) $f ^ { - 1 } ( x ) = \frac { \sqrt { - 6 ( x - 2 ) } } { 6 }$ The domain of f and the range of f^-1 are all real numbers x such that x $\ge$ 0.The domain of f^-1 and the range of f are all real numbers x such that x $\le$ 2.
B) $f ^ { - 1 } ( x ) = \frac { \sqrt { - 2 ( x - 6 ) } } { 2 }$ The domain of f and the range of f^-1 are all real numbers x such that x $\ge$ 0.The domain of f^-1 and the range of f are all real numbers x such that x $\le$ 2.
C) $f ^ { - 1 } ( x ) = \frac { \sqrt { - 6 ( x - 2 ) } } { - 6 }$ The domain of f and the range of f^-1 are all real numbers x such that x $\ge$ 0.The domain of f^-1 and the range of f are all real numbers x such that x $\le$ 2.
D) $f ^ { - 1 } ( x ) = \frac { \sqrt { - 6 ( x - 2 ) } } { 6 }$ The domain of f and the range of f^-1 are all real numbers x such that x $\ge$ 0.The domain of f^-1 and the range of f are all real numbers x such that x $\le$ -2.
E) $f ^ { - 1 } ( x ) = \frac { \sqrt { - 6 ( x + 2 ) } } { 6 }$ The domain of f and the range of f^-1 are all real numbers x such that x $\ge$ 0.The domain of f^-1 and the range of f are all real numbers x such that x $\le$ 2.