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$\ge$ 5.The domain of f^-1 and the range of f are all real numbers x such that x $\ge$ 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 $\ge$ 0.The domain of f^-1 and the range of f are all real numbers x such that x $\ge$ -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 $\ge$ 5.The domain of f^-1 and the range of f are all real numbers x such that x $\ge$ 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 $\ge$0.The domain of f^-1 and the range of f are all real numbers x such that x$\ge$ 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 $\ge$ -5.The domain of f^-1 and the range of f are all real numbers x such that x $\ge$ 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 $x \geq 5$, because the square function is one-to-one in either the non-negative or non-positive domain. To find the inverse, we solve for $x$ in terms of $y$: $y = (x - 5)^2$. Taking the square root of both sides gives $x - 5 = \sqrt{y}$, so $x = \sqrt{y} + 5$, which means $f^{-1}(x) = \sqrt{x} + 5$. The domain of $f$ (which becomes the range of $f^{-1}$) is $x \geq 5$, and since $f(x)$ squares the input, the range of $f$ (and thus the domain of $f^{-1}$) is $x \geq 0$.

Restrict the Domain of the Function F So That the Function