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 - 9| + 1 A)f^-1(x)= x + 8 The domain of f and the range of f^-1 are all real numbers x such that x ≥ 9.The domain of f^-1 and the range of f are all real numbers x such that x ≥ 1. B)f^-1(x)= x - 8 The domain of f and the range of f^-1 are all real numbers x such that x ≥ 9.The domain of f^-1 and the range of f are all real numbers x such that x ≥ 1. C)f^-1(x)= -x - 8 The domain of f and the range of f^-1 are all real numbers x such that x ≥ 1.The domain of f^-1 and the range of f are all real numbers x such that x ≥ -9. D)f^-1(x)= x + 8 The domain of f and the range of f^-1 are all real numbers x such that x ≥ -9.The domain of f^-1 and the range of f are all real numbers x such that x ≥ 1. E)f^-1(x)= -x + 8 The domain of f and the range of f^-1 are all real numbers x such that x ≥ 1.The domain of f^-1 and the range of f are all real numbers x such that x ≥ 9.

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Quizplus · Accepted Answer

The function f(x) = |x - 9| + 1 is not one-to-one on its entire domain because it fails the horizontal line test. However, if we restrict the domain to x ≥ 9, then the function becomes one-to-one and has an inverse function.To find the inverse function, we can switch the roles of x and y and solve for y:y = |x - 9| + 1x = |y - 9| + 1|y - 9| = x - 1y - 9 = ±(x - 1)y = 9 ± (x - 1)y = 9 + x - 1 or y = 9 - x + 1y = x + 8 or y = -x + 8Since the domain of f is restricted to x ≥ 9, the inverse function f^-1(x) must be restricted to y ≥ 9. Therefore, the inverse function is f^-1(x) = x + 8.The domain of f is all real numbers x such that x ≥ 9, and the range of f is all real numbers y such that y ≥ 1. The domain of f^-1 is all real numbers x such that x ≥ 9, and the range of f^-1 is all real numbers y such that y ≥ 1.

Restrict the Domain of the Function F So That the Function