[Bob]

Actually, the range of y=f(x)=x+2/x is the set of all y whose absolute value is less than or equal to the square root of 8.

You just set 2+2/x=y, turn it into a quadratic equation in x and then compute discriminant, y^2-8.

Sorry, you lost me at "turn it into a quadratic equation in x". How would you do that? (As I said earlier, I'm no mathematician.)

[Catphish]

You just multiply through by x, so x + 2/x = y becomes x^2+2=yx. Then you get -x^2+yx-2=0. The discriminant of that is y^2-8. Since you are looking for real values that y can be y^2-8 has to be greater than or equal to 0, i.e., the absolute value of y has to be greater than or equal to the square root of 8.