<s>[INST] Determine the domain of the inverse of $f(x) = 2x^2 + 8x - 7$, which is a function. Determine the domain where the inverse of $f(x) = 2x^2 + 8x - 7$ is a function.


*

*So, I started off switching the $y$- (the $f(x)$) and $x$-value, like so:
$x = 2y^2 + 8y - 7$

*Then, I plugged these values into the quadratic formula, getting:
$f^{-1}(x) = (-8 \pm \sqrt{120})/4$
First off, is the inverse written correctly (this problem is different from the others I worked on)? And second, since I have a ($\pm$) sign, does that mean there will be two different domains?
 [/INST] Let $a>0$ and $f(x)=ax^2+bx+c$. There is 
$$
\varphi:  \left[\frac{-\Delta}{4a},+\infty)\right) \to\left(-\infty, -\frac{b}{2a}\right] 
\qquad \mbox{and} \qquad 
\psi:\left[ \frac{-\Delta}{4a},+\infty)\right) \to \left[ -\frac{b}{2a}, +\infty,\right) 
$$
such that 
$$
\begin{matrix}
\varphi ( f(x))=x \qquad f(\varphi(y))=y\\
\psi(f(x))=x      \qquad f(\psi(y))=y
\end{matrix}
$$
Here $\Delta=b^2-4ac$. Let $y=ax^2+bx+c$. Check for yourself
$$
y=\left[a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a} \right]
$$
Then
$$
\left( x+\frac{b}{2a}\right)^2=\frac{1}{a}\left( y+\frac{\Delta}{4a}\right)
$$
and 
$$
x=-\frac{b}{2a} \pm\sqrt{\frac{1}{a}\left(y+\frac{\Delta}{4a}\right)}
$$
Finally we have
$$
\varphi(y)=-\frac{b}{2a}-\sqrt{\frac{1}{a}\left(y+\frac{\Delta}{4a}\right)}
$$
and 
$$
\psi(y)=-\frac{b}{2a}+\sqrt{\frac{1}{a}\left(y+\frac{\Delta}{4a}\right)}
$$
</s>