Find the inverse of $f(x)=x^2-1 , \ x \in \mathbb{R}, \ x \geq 1$

Question

Find the inverse function $f^{-1}$ and state it's domain of:

$$ f(x)=x^2-1 ,\ x \in \mathbb{R}, \ x \geq 1.$$

I think I've got the inverse function by switching $x$ and $y$ and then making $y$ the function.

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

This is where I get stuck, what is the domain, how do I work it out without using a graph, this is probably really simple but I can't get my head around functions right now. Thanks in advance.

Answer 1

We have $$ y = x^2 - 1 \wedge x\ge 1 \iff \\ x = \sqrt{y+1} \wedge \sqrt{y+1} \ge 1 \iff \\ x = \sqrt{y+1} \wedge y + 1 \ge 1 \iff \\ x = \sqrt{y+1} \wedge y \ge 0 $$

So $f^{-1}(x) = \sqrt{x+1}$. The domain is all real values $x \ge 0$.

Answer 2

$$ f(x)=x^2-1 ,\ x \in \mathbb{R}, \ x \geq 1$$ $$f(x) \in [0,\infty )$$ $$ \implies x=\left(f^{-1}(x)\right)^2-1 ,\ x \in [0,\infty)$$ $$ \implies f^{-1}(x)=\sqrt{x+1} ,\ x \in [0,\infty) \ \ [\text{because we know that $f^{-1}(x)$ is positive}.]$$