<s>[INST] Algebra question: Finding inverse function 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.
 [/INST] \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}
</s>