Question 606275
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Step 1:  Replace *[tex \LARGE f(x)] with *[tex \LARGE y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^3\ -\ 12]


Step 2: Solve for *[tex \LARGE x] in terms of *[tex \LARGE y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^3\ -\ 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ -\ 12\ =\ y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ =\ y\ +\ 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \sqrt[3]{y\ +\ 12}]


Step 3:  Swap positions of the variables.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \sqrt[3]{x\ +\ 12}]


Step 4:  Replace  *[tex \LARGE y] with *[tex \LARGE f^{-1}(x) ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ =\ \sqrt[3]{x\ +\ 12}]


Both domain and range of the original function are all reals so no difficulties with this one.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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