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| A)f^-1(x)= 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)= 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)= 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. D)f^-1(x)= 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. E)f^-1(x)= 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

To ensure the function $f(x) = |x + 5|$ is one-to-one, we restrict its domain to $x \geq -5$, where it behaves like $f(x) = x + 5$. The inverse function is $f^{-1}(x) = x - 5$, with the domain of $f$ and range of $f^{-1}$ being $x \geq -5$, and the domain of $f^{-1}$ and range of $f$ being $x \geq 0$.

Restrict the Domain of the Function F So That the Function