Question 627712
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You forgot to do the last step in the process of finding the inverse function, namely to swap the variables.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \sqrt{\frac{y\ -\ 5}{-4}}\ +\ 3]


Just like you had it, but then swap *[tex \LARGE x] and *[tex \LARGE y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \sqrt{\frac{x\ -\ 5}{-4}}\ +\ 3]


And then replace *[tex \LARGE y] with *[tex \LARGE f^{-1}(x)]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ =\ \sqrt{\frac{x\ -\ 5}{-4}}\ +\ 3]


From there, I would pretty it up a little, mostly so that the domain of the inverse function is more obvious:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ =\ \frac{1}{2}\left(\sqrt{5\ -\ x}\right)\ +\ 3]


But it is still the same function and should make a tidy graph that is a reflection of half of your original graph right across the line *[tex \LARGE y\ =\ x].  Why only half?  Because way back when we took the square root of both sides, we only considered the positive half.  We could have just as easily done:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f_{\small(-)\LARGE}^{-1}(x)\ =\ -\frac{1}{2}\left(\sqrt{5\ -\ x}\right)\ +\ 3]


But we could NOT have done


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f_{\small(\pm)\LARGE}^{-1}(x)\ =\ \pm\frac{1}{2}\left(\sqrt{5\ -\ x}\right)\ +\ 3]


Why?  Because *[tex \LARGE f_{\small(\pm)\LARGE}^{-1}(x)] is NOT a function.  Why?  That question is an exercise for the student.


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