Question 328864
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Step 1: Note the domain, *[tex \Large \{x\,|\,x\,\in\,\mathbb{R},\,x\,\geq\,2\}], the *[tex \Large x\,\geq\,2] being specified as part of the function definition, and the range *[tex \Large \{q(x)\,|\,q(x)\,\in\,\mathbb{R},\,q(x)\,\geq\,0\}]


Step 2: Replace *[tex \Large q(x)] with *[tex \Large y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ (x\ -\ 2)^2]


Step 3: Solve for *[tex \Large y] in terms of *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 2\ =\ \sqrt{y}]


Only the positive square root because of the *[tex \Large x\ \geq\ 2] restriction.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \sqrt{y}\ +\ 2]


Step 4: Swap variables and replace *[tex \Large y] with *[tex \Large q^{-1}(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ q^{-1}(x)\ =\ \sqrt{x}\ +\ 2]


The domain of the inverse is the range of the original function.  The range of the inverse is the domain of the original function.


Domain: *[tex \Large \{x\,|\,x\,\in\,\mathbb{R},\,x\,\geq\,0\}]


Range: *[tex \Large \{q^{-1}(x)\,|\,q^{-1}(x)\,\in\,\mathbb{R},\,q^{-1}(x)\,\geq\,2\}]




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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