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 ? 0.The domain of f^-1 and the range of f are all real numbers x such that x ? 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 ? 0.The domain of f^-1 and the range of f are all real numbers x such that x ? 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 ? 0.The domain of f^-1 and the range of f are all real numbers x such that x ? 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 ? 0.The domain of f^-1 and the range of f are all real numbers x such that x ? -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 ? 0.The domain of f^-1 and the range of f are all real numbers x such that x ? 2.

Question

Quizplus · Accepted Answer

The correct answer is A because the inverse function is $f ^ { - 1 } ( x ) = \frac { \sqrt { - 6 ( x - 2 ) } } { 6 }$ and the domain of f and the range of f^-1 are all real numbers x such that x ? 0. The domain of f^-1 and the range of f are all real numbers x such that x ? 2.

Restrict the Domain of the Function F So That the Function