mathmath333
  • mathmath333
find the range of the function \(f\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • 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
  • xapproachesinfinity
seems to me all real numbers?
anonymous
  • anonymous
oh no

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anonymous
  • anonymous
a fraction is only 1 if the numerator is equal to the denominator
welshfella
  • welshfella
no it cxannot have negative values
anonymous
  • anonymous
and it is unlikely that \(x^2+1=x^2\)
xapproachesinfinity
  • xapproachesinfinity
oh miss reading the function
anonymous
  • anonymous
also you notice that everything in sight is positive right? there is an actual way to do this
anonymous
  • 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
  • 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
  • mathmath333
i was looking for a way which would be different from just observation
anonymous
  • anonymous
there is nothing wrong with thinking, not matter what you math teacher tells you
anonymous
  • anonymous
but yes, there is a way solve \[y=\frac{x^2}{1+x^2}\] for \(x\)
anonymous
  • 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
  • mathmath333
i mean \(y(y-1)\geq 0\)
mathmath333
  • mathmath333
is that correct
xapproachesinfinity
  • xapproachesinfinity
is it not y(1-y)>=0
anonymous
  • anonymous
looks right
mathmath333
  • mathmath333
ya this one ^
anonymous
  • anonymous
i think if you solve for \(x\)you get \[x=\sqrt{\frac{y}{1-y}}\]
anonymous
  • anonymous
domain of that one is \[[0,1)\] so range of your function is the same
anonymous
  • anonymous
really i think this is a butt load more work, than necessary common sense often works in math too
mathmath333
  • mathmath333
but why this wont work \(y(1-y)>=0 \)
anonymous
  • anonymous
why strictly greater than?
anonymous
  • anonymous
oh i see you have \(\geq\)
mathmath333
  • mathmath333
it will be \(1\geq y\geq 0\)
anonymous
  • anonymous
but that is not right, because \(\geq\) includes 1
anonymous
  • anonymous
and there is no way for your function to be 1
mathmath333
  • mathmath333
thats why was asking why that method backstabbed
anonymous
  • anonymous
whereas solving \[\frac{y}{1-y}\geq 0\] does not include 1
mathmath333
  • mathmath333
ok thnx
anonymous
  • anonymous
ok i see my mistake i mislead you
anonymous
  • 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
  • anonymous
then solve \[\frac{y}{1-y}\geq0\] you get \[[0,1)\]
anonymous
  • anonymous
sorry about that i could never win against a king's gambit either, not matter whether i accepted or declined
mathmath333
  • mathmath333
haha
mathmath333
  • mathmath333
i have good record against kings gambit by black

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