Question 1167057
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^2\ +\ x]


The domain is *[tex \Large x\ \geq\ -\frac{1}{2}] (given) and the range is *[tex \Large y\ \geq\ -\frac{1}{4}]


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


Complete the square


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ \frac{1}{4}\ =\ x^2\ +\ x\ +\ \frac{1}{4}]


Factor the RHS


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


Take the square root


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ \frac{1}{2}\ =\ \pm\sqrt{y\ +\ \frac{1}{4}]


Isolate *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\sqrt{y\ +\ \frac{1}{4}}\ -\ \frac{1}{2}]


Swap the variables and swap the domain and range:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \pm\sqrt{x\ +\ \frac{1}{4}}\ -\ \frac{1}{2}]


The domain is *[tex \Large x\ \geq\ -\frac{1}{4}] and the range is *[tex \Large y\ \geq\ -\frac{1}{2}]


Eliminating the negative side of the square root based on the range restriction:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ \ =\ \sqrt{x\ +\ \frac{1}{4}}\ -\ \frac{1}{2}]

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


I > Ø
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