Inverse on a restricted domain

Text

Let $f(x)=x^{2}-6x$ restricted to $x\ge 3$. Show it is invertible on this domain, find the inverse and compute $f^{-1}(0)$.

Solution

Completing the square, $f(x)=(x-3)^{2}-9$: the vertex of the parabola is at $x=3$, so for $x\ge 3$ the function is strictly increasing, hence injective and invertible. Set $y=(x-3)^{2}-9$: with $x\ge 3$ we have $x-3\ge 0$, so $x-3=\sqrt{y+9}\;\Rightarrow\; f^{-1}(x)=3+\sqrt{x+9}.$ $f^{-1}(0)=3+\sqrt{9}=6.$ $\boxed{\,f^{-1}(x)=3+\sqrt{x+9},\quad f^{-1}(0)=6\,}$