Algebra 2 - Find Domain and Range of Function and Its Inverse $f(x)=-x^2+1$
For some reason, the inverse $f^{-1}$ gives me a domain equal to 1 or less than with a range of all real #'s. But the domain of the original function f(x) can only be negative. As squaring the x will only give positive numbers coupled with a negative sign on the outside making them negative.
Is there a discrepancy with the book here?
 A: The function
$$f(x) = -x^2+1$$ describes a parabola, opening downwards, with a vertex at $x=0$ and $y=1$.
If we were to consider $f$ on the whole real line, we would not be able to invert it, as it is not one-to-one, that is, it does not pass the horizontal line test.

So we have a couple of options here:


*

*We could define  $f$ on $(-\infty,0]$

*or instead define it on $[0,\infty)$.


If you pick any of these options as the domain for $f$, your function would have an inverse.
In either case, the inverse function would have $(-\infty,1]$ as a domain, since the range of $f$ is $(-\infty,1]$. 
A: Extending Gregory Grant's Comment
The range of $f$ is $(-\infty,1]$, while the maximal domain is $\Bbb R$.  For an inverse function $f^{-1}$ to be defined, we may choose either $x\le0$ or $x\ge0$ as the domain of $f$.  Therefore, either


*

*$$\begin{cases}\mathrm{dom}(f) &= (-\infty,0] \\ \mathrm {range}(f) &= (-\infty,1] \end{cases}, \text{ or}$$

*$$\begin{cases}\mathrm{dom}(f) &= [0,\infty) \\ \mathrm {range}(f) &= (-\infty,1] \end{cases}, \text{ or}$$


In both cases, $\mathrm{dom}(f^{-1}) = \mathrm{range}(f) = (-\infty,1]$.
