# Inverse function Help

• Oct 27th 2012, 02:57 PM
VARSOFFON
Inverse function Help
For this problem, y = f(x) = 3x2 − 5x − 1
on the domain of all real numbers.

1. Restrict
y = f(x)to the domain {ax} and find the formula for the inverse function fhttp://www.webassign.net/images/thinsp.gif−1(y).

I found the vertex (5/6, -37/12)
But I am not sure what to do now for the problem. Any help is appreciated, thank you!
• Oct 27th 2012, 03:27 PM
skeeter
Re: Inverse function Help
$x = 3y^2 - 5y - 1$ , $x \ge \frac{5}{6}$

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

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

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

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

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

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

$\frac{5}{6} + \sqrt{\frac{12x+37}{36}} = y = f^{-1}(x)$
• Oct 27th 2012, 04:46 PM
HallsofIvy
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".
• Oct 30th 2012, 09:26 PM
Salahuddin559
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