I have been given that f(x) = 3x^2 - 4 and x > 0 ... and I have been asked to find the value of f^-1(x) when x > -4 ...

Let y = 3x^2 - 4, then $\displaystyle 3x^2 - 4 - y = 0$

Solving this equation:

$\displaystyle => \frac{4 \pm \sqrt{16 + 12y}}{6}$

$\displaystyle = \frac{4 \pm \sqrt{2} \sqrt{4 + 3y}}{6} $

$\displaystyle = \sqrt{4 + 3y}$ or $\displaystyle \frac {\sqrt{4+3y}}{3} $

Either of them is wrong.. The answer should be $\displaystyle \sqrt{\frac{1}{3}(x + 4)}$