Originally Posted by
e^(i*pi) You're right up to here for sure.
What I was taught after putting f(x) = y is make x the subject, written in terms of y:
$\displaystyle y = \sqrt{2x^2+12}$
$\displaystyle 2x^2+12 = y^2$
$\displaystyle
2x^2 = y^2-12$
$\displaystyle
x^2 = \frac{1}{2}(y^2-12)$
$\displaystyle x = \pm \sqrt{\frac{1}{2}(y^2-12)} = \pm \frac{\sqrt2}{2}{\sqrt{y^2-12}}$
Finally replace x with $\displaystyle f^{-1}(x) $ and y with x:
$\displaystyle f^{-1}(x) = \pm \frac{\sqrt2}{2}{\sqrt{x^2-12}}$
Note that the plus/minus sign means it's not one to one and so not a function unless the domain is restricted