# 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?

• The domain of the original function is all of $\Bbb R$. The range is all $y\leq1$. For example plug in $x=0$ and you get one. Anything else will give you something less than one. The domain and range of the inverse are respectively the range and domain of the original function, Dec 29 '15 at 2:53
• Oh thanks! Why is it that plugging in only gives a number less than or equal to 1, if the range of the inverse clearly shows that ALL positive and negative numbers of the number line are included? Dec 29 '15 at 18:47
• Because the range of the inverse is the domain of $f$ which is all of $\Bbb R$. It's the range of $f$ (and therefore the domain of $f^{-1}$ that are restricted to $(-\infty,1]$. Dec 29 '15 at 18:51

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:

1. We could define $f$ on $(-\infty,0]$
2. 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 $(-\infty,1]$.

• Thank you for this graphical illustration! If we were to use the negative infinity side of the parabola, would the line of reflection for its inverse function still be y=x or would it be y= -x ? Dec 29 '15 at 18:50
• @user152810 it is always y=x Dec 29 '15 at 18:52

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