Question 1166431
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(t)\ =\ t^2\ -\ 2t\ +\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ = t^2\ -\ 2t\ +\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 5\ =\ t^2\ -\ 2t]


Now complete the square


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 4\ =\ t^2\ -\ 2t\ +\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 4\ =\ (t\ -\ 1)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ -\ 1\ =\ \sqrt{y\,-\,4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \pm\sqrt{y\,-\,4}\ +\ 1]


Swap the variables


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \pm\sqrt{t\,-\,4}\ +\ 1]


But you can't say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{-1}(t)\ =\ \pm\sqrt{t\,-\,4}\ +\ 1]


Because the graph won't pass the vertical line test so it is not a function. So you have a relation that is an inverse of the given function, but you do NOT have an inverse function.


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">


I > Ø
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