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

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.

• Welcome to MSE. Take y=√(x+1) as domain of f is positive. Nov 19 '15 at 17:50
• Note that $Range(f^{-1})=Domain(f)$ Nov 19 '15 at 17:52
• Thanks guys, guess im just having a weird day, don't know why i didn't get rid of the $$y^2$$ . Nov 19 '15 at 17:54

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