Find the inverse of $x^2+4x+1, [-2,0)$

Question

I am to find the inverse of $x^2+4x+1, [-2,0)$.

I am unsure how to 'untangle' the equation in a certain form. I made one step:

$$x=y^2+4y+1$$ $$x-1=y^2+4y$$

I realize this should probably be simple. Because on the right side I have $y$ appear in two pieces, $y^2$ and $4y$ I am uncertain on my next step.

If I square root both sides: $$\sqrt{x-1}=\sqrt{y^2-4y}$$

Can I write the right side as $\sqrt{y^2} + \sqrt{4y}$ which would be $y$ and $2y$ respectivly? This would give me:

$$\sqrt{x-1}=y-2y$$

But I feel like I'm adrift and uncertain here.

I'm seeking hand holding and baby steps. If I am here: $$x-1=y^2+4y$$

What is the prescribed next step to isolate y on it's own (Solve for y)?

Answer 1

Using the quadratic formula, you get that$$x=y^2+4y+1\iff y=-2\pm\sqrt{4-(1-x)}=-2\pm\sqrt{x+3}.$$You will have to take the $+$ sign here; otherwise, $x<-2$. So, $f^{-1}(x)=-2+\sqrt{x+3}$ when $x\in[-2,0)$.

Answer 2

Take $$y^2+4y+1-x=0$$ and solve the quadratic: $$y=\frac{-4\pm\sqrt{16-4(1-x)}}{2}=-2\pm\sqrt{x+3}$$ Now decide which sign you take the $\pm$ as in your interval.

Answer 3

Given $[E_1]~ y = x^2 + 4x + 1$
and $[E_2]~x = y^2 + 4y + 1$,
I don't understand how you go from $E_1$ to $E_2$.
Therefore, I can't critique your overall method until this question is resolved.

Anyway, the approach that I favor is to use $E_1$ to conclude that $[E_3]~ (y+3) = x^2 + 4x + 4 = (x + 2)^2 ~\Rightarrow $
$[E_4]~ (x+2) = \sqrt{y+3}.$

Assuming that you agree with my logic, it seems to me that $E_4$
constitutes expressing $x$ as a function of $y$.