Question 700165
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Complete the square to put the function into vertex form:


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2\left(x^2\ +\ \frac{3}{2}x\ +\ \frac{9}{16}\ -\ \frac{9}{16}\ +\ \frac{1}{2}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2\left(x\ +\ \frac{3}{4}\right)^2\ +\ 2\left(-\frac{9}{16}\ +\ \frac{1}{2}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2\left(x\ +\ \frac{3}{4}\right)^2\ -\ \frac{1}{8}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ + \frac{1}{8}\ =\ 2\left(x\ +\ \frac{3}{4}\right)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{8y\ +\ 1}{8}\ =\ 2\left(x\ +\ \frac{3}{4}\right)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{8y\ +\ 1}{16}\ =\ \left(x\ +\ \frac{3}{4}\right)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pm\sqrt{\frac{8y\ +\ 1}{16}}\ =\ x\ +\ \frac{3}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{3}{4}\ \pm\sqrt{\frac{8y\ +\ 1}{16}}]


Swap positions of the variables, but you have to choose between the positive or negative root, otherwise the inverse is not a function  (you can't have more than one output for a given input of a function)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{3}{4}\ +\ \sqrt{\frac{8x\ +\ 1}{16}}\ =\ \frac{-3\ +\ \sqrt{8x\ +\ 1}}{4} ]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{3}{4}\ -\ \sqrt{\frac{8x\ +\ 1}{16}}\ =\ \frac{-3\ -\ \sqrt{8x\ +\ 1}}{4}]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ =\ \frac{-3\ +\ \sqrt{8x\ +\ 1}}{4}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ =\ \frac{-3\ -\ \sqrt{8x\ +\ 1}}{4}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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