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Find the inverse of $f(x)=\sqrt{3x^2 +1}$.

We don't know if $f$ is invertible so we have to prove it, but how?

$f$ bijective $\Leftrightarrow $ $f$ invertible

Is $f$ injective? $ x \neq y \Rightarrow f(x) \neq f(y) $

$f(x) = f(y) $

$\sqrt{3x^2 +1} = \sqrt{3y^2 +1}$

$ x^2 = y^2 $

$x = y$

No injective $\Rightarrow$ No bijective $\Rightarrow$ No invertible.

Is that right?

Spectree
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    Right, it is not injective on $\mathbb R$ and is has no inverse. If you take the domain as $(0,\infty)$ you can find the inverse. – Kavi Rama Murthy Jan 23 '21 at 00:37
  • Continuing Kavi Rama Murthy's comment, when the domain of $f(x)$ is $(0, \infty)$, the property that $f(x)$ is well-defined will depend on the convention that in the original equation, $f(x)$ (i.e. the range of $f$) is non-negative. Note the relationship between $f$ being well defined and $f^{(-1)}$ being well defined : $f^{(-1)}$ is well defined because $f(x)$ is injective, which has happened because the domain of $f$ has been restricted. – user2661923 Jan 23 '21 at 00:54

2 Answers2

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If you are looking at the "natural domain" of $f$, then it is not invertible since $f(1)=f(-1)$.

If you look at a subset of the natural domain, say $(0,\infty)$, then setting $y=\sqrt{3x^2+1}$ and solve for $x$, you get $$ x=\sqrt{\frac{y^2-1}{3}},\quad y>1 $$ which gives you the inverse of $f:(0,\infty)\to\mathbb{R}$ where $$ f^{-1}(y)=\sqrt{\frac{y^2-1}{3}},\quad y>1 $$

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When we talk about functions and inverses, we should be careful about domains and codomains. Function $f$ can be defined on whole $\mathbb R$, but then it will not be injective, because $f(x)=f(-x)$ for all $x\in\mathbb R$. If we restrict function to $[0,\infty)$ then $f$ will be injective and its image will be $[1,\infty)$. Similar thing happens if we restrict $f$ to $(-\infty,0]$.

Nikola Ubavić
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