If $f:R\to R$ is defined by $f(x)=x^2-3$, find $f^{-1} (-1)$

Question

If $f:R\to R$ is defined by $f(x)=x^2-3$, find $f^{-1} (-1)$

My Attempt : $$f(x)=x^2-3$$ $$y=x^2-3$$

Then how to proceed further?

Answer 1

$$ f\left( x \right) = x^{2} - 3 $$ The function has two $x$ values that map to $y = -1$.

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$$ \color{blue}{f^{-1}(y) = \sqrt{x+3}} \qquad \Rightarrow \qquad f^{-1}\left( -1 \right) = \sqrt{2} $$ $$ \color{red}{f^{-1}(y) = -\sqrt{x+3}}\qquad \Rightarrow \qquad f^{-1}\left(-1 \right) = -\sqrt{2} $$

Answer 2

Without rigorous proof, we assume that the inverse of $f(x)$ exists, otherwise the problem would be meaningless. Let $g$ be the inverse of $f$. By definition then, $g(f(x)) = x \forall x \in D$ (D is the domain). We must find $g(-1)$ note that $f(2) = -1$. Thus $g(f(2))$ should give the required answer, which we find to be simply 2.

Hence answer is 2.