find the range of the function \(f\)

- mathmath333

find the range of the function \(f\)

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- mathmath333

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 \(f\)

- xapproachesinfinity

seems to me all real numbers?

- anonymous

oh no

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## More answers

- anonymous

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

- welshfella

no it cxannot have negative values

- anonymous

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

- xapproachesinfinity

oh miss reading the function

- anonymous

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

- anonymous

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

- anonymous

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

- mathmath333

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

- anonymous

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

- anonymous

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

- anonymous

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

- mathmath333

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

- mathmath333

is that correct

- xapproachesinfinity

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

- anonymous

looks right

- mathmath333

ya this one ^

- anonymous

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

- anonymous

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

- anonymous

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

- mathmath333

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

- anonymous

why strictly greater than?

- anonymous

oh i see you have \(\geq\)

- mathmath333

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

- anonymous

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

- anonymous

and there is no way for your function to be 1

- mathmath333

thats why was asking why that method backstabbed

- anonymous

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

- mathmath333

ok thnx

- anonymous

ok i see my mistake i mislead you

- anonymous

you do not get a quadratic, you solve
\[y=\frac{x^2}{1+x^2}\]
\[yx^2+y=x^2\\
yx^2-x^2=-y\\
(y-1)x^2=-y\\
x^2=\frac{y}{1-y}\\
x=\sqrt{\frac{y}{1-y}}\]

- anonymous

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

- anonymous

sorry about that
i could never win against a king's gambit either, not matter whether i accepted or declined

- mathmath333

haha

- mathmath333

i have good record against kings gambit by black

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