<s>[INST] Inverse of half a polynomial I am given the assignment of finding the inverse of the following function: 
$$f(x) = x^2 + 4x + 5$$
where $x > -2$. 
I can get rid of the $5$ obviously, but then I get stuck. Hints are appreciated.
 [/INST] You want to solve $y = x^2 + 4x + 5$ when $x > -2$, or $x^2 + 4x + (5 - y) = 0$. Solving for $x$ using the quadratic formula, we have
$$ x = \frac{-4 \pm \sqrt{16 - 4(5 - y)}}{2} = \frac{-4 \pm \sqrt{4(y - 1)}}{2} = -2 \pm \sqrt{y - 1}.$$
Since $x > -2$, we need to take the solution with the $+$ and so $x = -2 + \sqrt{y - 1}$.
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