Is the inverse of f(x) = 9-2x^2
f-1(x) = -sqrt( (9/2) - (x/2) )
Thanks,
D.
The working in your post is correct except I do not see why you can't have a plus or minus infront of that square root unless there is something about the domain you're not telling me
$\displaystyle y = 9-2x^2$
$\displaystyle 2x^2 = 9-y$
$\displaystyle x^2 = \frac{9-y}{2}$
$\displaystyle x = \pm \sqrt{\frac{9-y}{2}}$
$\displaystyle f^{-1}(x) = \pm \sqrt{\frac{9-x}{2}}$
Bear in mind that's not a function since it's not one to one - the domain would have to be restricted to make it a function. t