Finding the inverse function to $f(x) = x^2 - 6x$

Question

I'm having trouble with this inverse function. I know how to invert some basic functions but this beats me this time... $y= x^2 - 6x$ and $x \in (-\infty, +3 ]$

Step by step solution would be really helpful. Thanks!

PS: I tried to look for other posts about inverse functions, but I couldn't find any of them treating the same case.

Answer 1

One my solve as follows:

$$y=x^2-6x$$

$$0=x^2-6x-y$$

Apply the quadratic formula:

$$x=3\pm\sqrt{9+y}$$

Appropriately matching the domain and range:

$$x=3-\sqrt{9+y}$$

Answer 2

Hint: you shall need a completion of square $$x^2-6x=(x-3)^2-9$$

Answer 3

$f(x)=x^2-6x$

$f(x)=(x-3)^2-9$ (equivalent equation just in a different form)

$x=(f^{-1}(x)-3)^2-9$ (switch x's and $f^{-1}(x)$ to invert)

$x+9=(f^{-1}(x)-3)^2$

$\pm\sqrt {x+9}=f^{-1}(x)-3$

$\pm\sqrt {x+9}+3=f^{-1}(x)$

$f^{-1}(x)=3\pm\sqrt {x+9}$ (rearrange for readability)

$f^{-1}(x)=3-\sqrt {x+9}$ (adjust for domain)