Solution
We have the following equation given:
[tex]y=x^2-2x=x(x-2)[/tex]For this case we can do the following:
[tex]x=y^2-2y=y(y-2)[/tex]We can solve for the quadratic equation and we got:
[tex]y=\frac{-(-2)\pm\sqrt[]{(-2)^2-4(1\cdot-x)}}{2\cdot1}=\frac{2\pm\sqrt[]{4\cdot(1+x)}}{2}=1\pm\sqrt[]{1+x}[/tex]The two solutions are:
[tex]y_1=1-\sqrt[]{1+x},y_2=1+\sqrt[]{1+x}[/tex]Then the answer is:
Part 1
[tex]x\ge1[/tex]Part 2
[tex]f^{-1}(x)=1+\sqrt[]{1+x}[/tex]