Inverse of $f(x)=4x-x^2; x≥2$

Question

I just asked a question a few minutes ago, but I stumbled upon another one.

This is the question:

Find the inverse of $f(x)=4x-x^2,$ where the domain is $x≥2. $ I did:

$x=4y-y^2$

$y^2-4y+x=0$

I am stuck here.

Answer 1

I'll not provide the whole answer, but a slight hint as you prefer.

The equation $y^2-4y+x=0$ is in quadratic form, which you probably know if you are taking Algebra 2, so think of $y$ as a constant term.

Spoiler (don't hover unless you're stuck):

You obtain $y=2±\sqrt{4-x}$ after you use the formula $\frac{-b/2±\sqrt{(-b/2)^2-ac}}{a}.$

I think you can figure out the domain of $f^{-1}(x)$, as the $x$ and $y$ values of f(x) become the $y$ and $x$ values of $f^{-1}(x)$.

Answer 2

You can find the inverse by completing the square for y $$y^2-4y=-x$$ $$y^2-4y+4=4-x$$ $$(y-2)^2=4-x$$ $$y=\sqrt{4-x} +2$$ So we get $f^{-1}(x)=\sqrt{4-x}+2$. Note, it has to be positive square root, not negative, since the restriction on x of the original function is $x\geq2$.