$f(x) = x^2 + 2x$ , domain ${x ≥ 1}$
Question: find the inverse
The inverse is $f(x) = 1 + \sqrt(1+x)$ (taking the positive square root only)
As $f^{-1}(5) = 2$ and as 2 is an element from the range of $f^{-1}$ meaning that it's an element from the domain from $f(x)$ (as the range of the inverse = the domain of $ f(x) $) but since $ f(x) $ has the restriction $x < 0$ this means that the inverse must be the negative square root?
How would you figure out whether to take the positive or negative square root as the inverse? Would it be by recognition? Is there another way besides recognition? Would you sub in $ x = 0 $ or any specific values to check?
Your inverse function is incorrect.
Let $y=f(x)=x^2+2x$
By completing the squares, you get:
$y=x^2+2x+1-1=(x+1)^2-1$
$(x+1)^2 = y+1$
$x=-1 + \sqrt{y+1}$
You don't take the negative square root in this case, because we must have $x\ge1$