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$f(x)=-x^2+1$

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?

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    $\begingroup$ 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, $\endgroup$ – Gregory Grant Dec 29 '15 at 2:53
  • $\begingroup$ 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? $\endgroup$ – user152810 Dec 29 '15 at 18:47
  • $\begingroup$ 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 $(-\infty,1]$. $\endgroup$ – Gregory Grant Dec 29 '15 at 18:51
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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.

enter image description here

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 $(-\infty,1]$.

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  • $\begingroup$ 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 ? $\endgroup$ – user152810 Dec 29 '15 at 18:50
  • $\begingroup$ @user152810 it is always y=x $\endgroup$ – GaussTheBauss Dec 29 '15 at 18:52
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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, $\mathrm{dom}(f^{-1}) = \mathrm{range}(f) = (-\infty,1]$.

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