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mathmath333
 one year ago
find the range of the function \(f\)
mathmath333
 one year ago
find the range of the function \(f\)

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0let \(\large \color{black}{\begin{align} f=\left\{\left(x,\ \dfrac{x^2}{1+x^2}\right):\ x\in \mathbb{R}\right\} \hspace{.33em}\\~\\ \end{align}}\) be a function from \(\mathbb{R}\) to \(\mathbb{R}\) .Determine the range of \(f\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1seems to me all real numbers?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a fraction is only 1 if the numerator is equal to the denominator

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0no it cxannot have negative values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and it is unlikely that \(x^2+1=x^2\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1oh miss reading the function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also you notice that everything in sight is positive right? there is an actual way to do this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or 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
 one year ago
Best ResponseYou've already chosen the best response.0making 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
 one year ago
Best ResponseYou've already chosen the best response.0i was looking for a way which would be different from just observation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is nothing wrong with thinking, not matter what you math teacher tells you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but yes, there is a way solve \[y=\frac{x^2}{1+x^2}\] for \(x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you 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
 one year ago
Best ResponseYou've already chosen the best response.0i mean \(y(y1)\geq 0\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1is it not y(1y)>=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think if you solve for \(x\)you get \[x=\sqrt{\frac{y}{1y}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0domain of that one is \[[0,1)\] so range of your function is the same

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0really i think this is a butt load more work, than necessary common sense often works in math too

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but why this wont work \(y(1y)>=0 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why strictly greater than?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i see you have \(\geq\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0it will be \(1\geq y\geq 0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but that is not right, because \(\geq\) includes 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and there is no way for your function to be 1

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0thats why was asking why that method backstabbed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whereas solving \[\frac{y}{1y}\geq 0\] does not include 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i see my mistake i mislead you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you do not get a quadratic, you solve \[y=\frac{x^2}{1+x^2}\] \[yx^2+y=x^2\\ yx^2x^2=y\\ (y1)x^2=y\\ x^2=\frac{y}{1y}\\ x=\sqrt{\frac{y}{1y}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then solve \[\frac{y}{1y}\geq0\] you get \[[0,1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry about that i could never win against a king's gambit either, not matter whether i accepted or declined

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i have good record against kings gambit by black
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