# Thread: Square root's inverse right but restrictions?

1. ## Square root's inverse right but restrictions?

I get the answer which is f-1(x) = (-1/9(x+2)^2) + 1 but the solution guide says:

" f-1(x) = (-1/9(x+2)^2) + 1 x >/= 2 OR f-1(x) = (-1/9(x+2)^2) + 1 [-2, infinty["

How do I get such restrictions? This seems to be false when checked with my graphing software but I could be wrong.

Any help would be greatly appreciated!

2. let f(x)=y and the equation with respect to x

$y=3\sqrt{-(x-1)}-2$

$y+2=3\sqrt{-(x-1)}$

$(y+2)^2=9(-(x-1))$

$\frac{(y+2)^2}{9}=-(x-1)$

$-\left(\frac{(y+2)^2}{9}\right)=x-1$

$1-\left(\frac{(y+2)^2}{9}\right)=x$

change the variables

$f(x)=1-\left(\frac{(x+2)^2}{9}\right)$

3. $f(x) = 3\sqrt{-(x - 1)} - 2$

The inverse, as Amer worked out, is
$f^{-1}(x) = -\frac{(x + 2)^2}{9} + 1$
(I like to put the variables first. )

But it wouldn't be right to leave it as it is without stating the restrictions. If we consider this function without knowing beforehand that it was the inverse of some other function, the domain would be all reals and the range would be (-∞, 1], but that's wrong in our case. This function $f^{-1}(x)$ is the inverse of $f(x)$, and since $f(x)$ is one-to-one, the domains and ranges switch.

The domain of $f(x)$ is (-∞, 1] and the range is [-2, ∞). Therefore the domain of $f^{-1}(x)$ is [-2, ∞) and the range is (-∞, 1]. The graph of $f^{-1}(x)$ would be the right half of a parabola, opening downward.

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