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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.

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  • $\begingroup$ Welcome to MSE. Take y=√(x+1) as domain of f is positive. $\endgroup$ – Nitin Uniyal Nov 19 '15 at 17:50
  • $\begingroup$ Note that $Range(f^{-1})=Domain(f)$ $\endgroup$ – Nitin Uniyal Nov 19 '15 at 17:52
  • $\begingroup$ Thanks guys, guess im just having a weird day, don't know why i didn't get rid of the $$y^2 $$ . $\endgroup$ – Matt G Nov 19 '15 at 17:54
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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 $x \ge 0$.

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

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