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seems to me all real numbers?

oh no

a fraction is only 1 if the numerator is equal to the denominator

no it cxannot have negative values

and it is unlikely that \(x^2+1=x^2\)

oh miss reading the function

also you notice that everything in sight is positive right?
there is an actual way to do this

i was looking for a way which would be different from just observation

there is nothing wrong with thinking, not matter what you math teacher tells you

but yes, there is a way
solve
\[y=\frac{x^2}{1+x^2}\] for \(x\)

i mean \(y(y-1)\geq 0\)

is that correct

is it not y(1-y)>=0

looks right

ya this one ^

i think if you solve for \(x\)you get
\[x=\sqrt{\frac{y}{1-y}}\]

domain of that one is
\[[0,1)\] so range of your function is the same

really i think this is a butt load more work, than necessary
common sense often works in math too

but why this wont work \(y(1-y)>=0 \)

why strictly greater than?

oh i see you have \(\geq\)

it will be \(1\geq y\geq 0\)

but that is not right, because \(\geq\) includes 1

and there is no way for your function to be 1

thats why was asking why that method backstabbed

whereas solving
\[\frac{y}{1-y}\geq 0\] does not include 1

ok thnx

ok i see my mistake i mislead you

then solve \[\frac{y}{1-y}\geq0\] you get
\[[0,1)\]

haha

i have good record against kings gambit by black