find the range of the function \(f\)

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find the range of the function \(f\)

Mathematics
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let \(\large \color{black}{\begin{align} f=\left\{\left(x,\ \dfrac{x^2}{1+x^2}\right):\ x\in \mathbb{R}\right\} \hspace{.33em}\\~\\ \end{align}}\) be a function from \(\mathbb{R}\) to \(\mathbb{R}\) .Determine the range of \(f\)
seems to me all real numbers?
oh no

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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
or just wing it the least it can be is zero, since everything is positive unless x is zero and it cannot reach 1 since the numerator is smaller than the denominator
making the range \[[0,1)\] otherwise you have to set \[y=\frac{x^2}{1+x^2}\] and solve for \(x\) which is a colossal pain
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 \(x\)
you will get a quadratic equation in \(y\) you then take the discriminant and make sure it is greater than or equal to zero enjoy, it is a pain in the neck
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
you do not get a quadratic, you solve \[y=\frac{x^2}{1+x^2}\] \[yx^2+y=x^2\\ yx^2-x^2=-y\\ (y-1)x^2=-y\\ x^2=\frac{y}{1-y}\\ x=\sqrt{\frac{y}{1-y}}\]
then solve \[\frac{y}{1-y}\geq0\] you get \[[0,1)\]
sorry about that i could never win against a king's gambit either, not matter whether i accepted or declined
haha
i have good record against kings gambit by black

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