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)= (x - 5)² ? A) $f ^ { - 1 } ( x ) = \sqrt { x } - 5$ ? The domain of f and the range of f^-1 are all real numbers x such that x ? 5.The domain of f^-1 and the range of f are all real numbers x such that x ? 0. B) $f ^ { - 1 } ( x ) = \sqrt { x } + 5$ ? 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 ? -5. C) $f ^ { - 1 } ( x ) = \sqrt { x } + 5$ ? The domain of f and the range of f^-1 are all real numbers x such that x ? 5.The domain of f^-1 and the range of f are all real numbers x such that x ? 0. D) $f ^ { - 1 } ( x ) = \sqrt { x } + 5$ ? 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 ? 5. E) $f ^ { - 1 } ( x ) = \sqrt { x } - 5$ ? The domain of f and the range of f^-1 are all real numbers x such that x ? -5.The domain of f^-1 and the range of f are all real numbers x such that x ? 0.

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Quizplus · Accepted Answer

Given $f(x) = (x - 5)^2$, the function is one-to-one if we restrict the domain to $x \geq 5$, ensuring it's increasing on this interval. The inverse function is found by solving $y = (x - 5)^2$ for $x$, giving $x = \sqrt{y} + 5$, thus $f^{-1}(x) = \sqrt{x} + 5$. The domain of $f$ and the range of $f^{-1}$ are $x \geq 5$, and the domain of $f^{-1}$ and the range of $f$ are $x \geq 0$, matching option C.

Restrict the Domain of the Function F So That the Function