Re: Inverse function Help

$\displaystyle x = 3y^2 - 5y - 1$ , $\displaystyle x \ge \frac{5}{6}$

$\displaystyle x = 3 \left(y^2 - \frac{5}{3}y \right) - 1$

$\displaystyle x = 3 \left(y^2 - \frac{5}{3}y + \frac{25}{36}\right) - 1 - \frac{25}{12}$

$\displaystyle x = 3\left(y - \frac{5}{6}\right)^2 - \frac{37}{12}$

$\displaystyle x + \frac{37}{12} = 3\left(y - \frac{5}{6}\right)^2$

$\displaystyle \frac{x}{3} + \frac{37}{36} = \left(y - \frac{5}{6}\right)^2$

$\displaystyle \sqrt{\frac{12x+37}{36}} = y - \frac{5}{6}$

$\displaystyle \frac{5}{6} + \sqrt{\frac{12x+37}{36}} = y = f^{-1}(x)$

Re: Inverse function Help

Skeeter "completed the square". You could also solve for y using the quadratic formula- which, of course, comes from "completing the square".

Re: Inverse function Help

When talking about a function, domain and range are necessary to be mentioned, a mere calculation without a domain does not qualify for a function. The original function is defined on [5/6, infinity) --> [-37/12, infinity), and the inverse is defined from [-37/12, infinity) --> [5/6, infinity). Making the function bijective like this is necessary for an inverse being defined.

Salahuddin

Maths online