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?
 A: First of all, this is an excellent example of an OP providing enough details of their attempt at solving a problem so that it's possible to give answers tailored to the OP's error or confusion. Kudos to Bibliophile for doing so.
The first step is perfectly correct. It's the second step where things go awry. It's always tempting to plug the coefficients of a quadratic polynomial, which $2y^2+8y-7$ certainly is, into the quadratic formula $-b\pm\sqrt{b^2-4ac})\over2a$.  The thing to remember is that the quadratic formula gives the solution to a quadratic equation in which the polynomial is being set equal to $0$.  That's not what we have here. Instead, we have $2y^2+8y-7$ set equal to $x$. So we first need to move the $x$ over, and only then apply the quadratic formula:
$$2y^2+8y-7=x\implies2y^2+8y-(7+x)=0\implies y={-8\pm\sqrt{64+8(7+x)}\over4}={-8\pm\sqrt{120+8x}\over4}$$
Finally, note that the problem only asks for the domain of the inverse function.  In this case, that amounts to asking where $\sqrt{120+8x}$ is a real number.  It's easy to see that the domain is $x\in[-15,\infty)$.  
Remark:  The function $f(x)=2x^2+8x-7$ has an inverse only in the sense that $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$ (i.e., for all $x\in[-15,\infty)$).  Indeed, there really is no such thing as "the" inverse here: because of the $\pm$ sign in the quadratic formula, there are two natural possibilities:
$$f_+^{-1}(x)={-8+\sqrt{120+8x}\over4}\quad\text{and}\quad f_-^{-1}(x)={-8-\sqrt{120+8x}\over4}$$
Additional remark: Another way to get at the domain for the inverse here is to rewrite the quadratic $2x^2+8x-7=2(x^2+4x)-7=2(x^2+4x+4)-8-7$ to get
$$f(x)=2(x+2)^2-15$$
from which it's clear that $f(x)$ takes all values greater than or equal to $-15$, so that the domain of its inverse is $[-15,\infty)$.
A: So that a function admits an inverse it should be monotone in the range. Let's find where your function is monotone: 
$$f'(x) = 4x + 8 = 0$$
then at $x = -2$, you have a local extrema. So, at $]-\infty,-2]$ we have an inverse. And at $]-2,\infty[$ we also have an inverse. It happens to be that the range of $f(x)$ (consequently the domain of the inverse) is the same in both domains of $f(x)$, which is $$[f(-2),f(+\infty)[=[-15,+\infty[$$
A: First, your calculations are incorrect. We have
$$
f(x) = 2x^2 + 8x -7
$$
set $y = f(x)$ and solve for $x$.
$$
y = 2x^2 + 8x - 7 \Leftrightarrow
2(x+2)^2 = y + 15 \Leftrightarrow
x = -2 \pm \sqrt{\frac{y+15}{2}}
$$
So the inverse is $f^{-1}(x) = -2 \pm \sqrt{\frac{x+15}{2}}$
In order for this to be a function, we must have that:
for every $x$ we can only obtain one $y$. With the $\pm$ this is not the case. 
It turns out that the $-2 + \sqrt{\frac{x+15}{2}}$ corresponds to those $x$ which are to the right of the extrem point in the range of $f(x)$ and $-2 - \sqrt{\frac{x+15}{2}}$ corresponds to those $x$ which are to the left of the extrem point in the range of $f(x)$. I think the easiest way to see this is by looking at the mirror image of $f(x)$ in the line $y=x$ graphically.
A: 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 
$$
\psi(y)=-\frac{b}{2a}+\sqrt{\frac{1}{a}\left(y+\frac{\Delta}{4a}\right)}
$$
