y`=[2x(x+2)-x？ ]/(x？ +2x+4)

=(x？ +4x)/(x+2)^

When x=0 or x=-4 y`=0

-4 & lt； x & lt0y ` & lt； 0

X & lt-4 or x & gt0y` > 0

Because x is between-1 and 1, you can't get 0.

So when x=- 1, y= 1 x= 1, y= 1/3.

So the range is (0, 1)

It doesn't matter if you don't learn derivatives.

Divide x2 by the bottom and replace 1/x+2/x2 with t =1/x.

So the range of t is less than or equal to-1 or greater than or equal to 1.

1/2t^+t

1/2(t+ 1/4)^- 1/8

That is to say, when t=- 1, it is closest to the lowest point, that is, the extreme maximum point 1, and when t tends to infinity or

When it is infinite, the extreme value tends to 0.

You can also get (0, 1)