Question 324697
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Next time, one question per post.  Why should I do two problems and only get credit for one?


1.  *[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ (x\ -\ 2)^3]


Replace *[tex \Large f(x)] with *[tex \Large y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ (x\ -\ 2)^3]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^{\frac{1}{3}}\ =\ x\ -\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ y^{\frac{1}{3}}\ +\ 2]


Swap the positions of the variables and replace *[tex \Large y] with *[tex \Large f^{-1}(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ =\ x^{\frac{1}{3}}\ +\ 2]


Can also be expressed:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ =\ \sqrt[\small{3}]{x}\ +\ 2]


Done.


2.  Use the procedure above to determine that the inverse of the given function is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ =\ \frac{x\,+\,1}{2x\,-\,1}]


Zero of the denominator is *[tex \Large x\ =\ \frac{1}{2}].  When *[tex \Large x\ \rightarrow\ \frac{1}{2}] from the left, *[tex \Large f^{-1}(x)\ \rightarrow -\infty], but when *[tex \Large x\ \rightarrow\ \frac{1}{2}] from the right, *[tex \Large f^{-1}(x)\ \rightarrow \infty].


Since the numerator and denominator polynomials are of the same degree, the rational function is asymptotic to the ratio of the lead coefficients, namely *[tex \Large y\ =\ \frac{1}{2}] in this case.  Hence, *[tex \Large y\ =\ \frac{1}{2}] is excluded from the range.


Range, in interval notation:*[tex \Large\ \ \ \left(-\infty,\,\frac{1}{2}\right)\ \small{\cup}\Large\  \left(\frac{1}{2},\,\infty\right)]


Range, in set builder notation: *[tex \Large \{y\,|\,y\,\in\,\mathbb{R},\,y\ \neq\ \frac{1}{2}\}]


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