The domain was given to be $\mathbb{R}$. I calculated that the range was $(0,1]$ and that for every $y$ I could find an $x$ in that domain since $x = \sqrt{\frac{1}{y^{2}}-1}$. I therefore thought $f(x)$ was surjective, but when I asked Wolfram Alpha it said that it isn't surjective, because it isn't surjective on $\mathbb{R}$.
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6Of course, surjectivity will depend on what codomain the function has. – Parcly Taxel Oct 20 '18 at 07:20
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Yep. How can $f(x)$ be $0$? – Shervin Sorouri Oct 20 '18 at 07:21
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You are using $Range$ as the codomain. – Avinash N Oct 20 '18 at 07:39
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It depends upon the codomain we are assuming for $f$, indeed if we consider
$$f: \mathbb{R}\to \mathbb{R}$$
it is not surjective since for example $\not \exists x$ such that $f(x)=-1$ but if we assume
$$f: \mathbb{R}\to (0,1]$$
it is surjective and for a suitable restriction of the ("natural") domain
$$f: [0,\infty) \to (0,1]$$
it is also injective and thus bijective, that is invertible.
Refer also to the related
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