This question is about finding the inverse function of $f(x)=-\sqrt{9-x^2}$
I seem to be making an error with one of the manipulations. Here is my attempt.
$$x=-\sqrt{9-y^2}$$ $$x^2=(-\sqrt{9-y^2})^2$$ $$x^2= -(9-y^2)$$ $$x^2=y^2-9$$ $$x^2+9=y^2$$ $$\sqrt{x^2+9}=\sqrt{y^2}$$ $$y=\sqrt{x^2+9}$$
The answer is $f^{-1} (x)=-\sqrt{9-x^2},-3≤x ≤0$
In which step did I make a mistake?
Thank you for your time.
\begin{align} y=-\sqrt{9-x^2} \end{align} Change $x$ to $y$ and vice versa. \begin{align} x&=-\sqrt{9-y^2}\\ -x&=\sqrt{9-y^2}\\ (-x)^2&=\left(\sqrt{9-y^2}\right)^2\\ x^2&=9-y^2\\ y^2&=9-x^2\\ y&=\pm\sqrt{9-x^2} \end{align} The last step, change $y$ to $f^{-1}(x)$. \begin{align} f^{-1}(x)&=\pm\sqrt{9-x^2} \end{align} Since there is a symmetry between a function and its inverse. Specifically, if $f(x)$ is an invertible function with domain $x$ and range $y$, then its inverse $f^{-1}(x)$ has domain $y$ and range $x$. Therefore, the inverse function of $f(x)=-\sqrt{9-x^2}$ is \begin{align} f^{-1}(x)&=-\sqrt{9-x^2} \end{align}