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anonymous
 4 years ago
Find a rational function f:R> with range f(R)=[1,1]. (Thus f(x)=P(x)/Q(x) for all xeR for suitable polynomials P and Q where Q has no real root.
anonymous
 4 years ago
Find a rational function f:R> with range f(R)=[1,1]. (Thus f(x)=P(x)/Q(x) for all xeR for suitable polynomials P and Q where Q has no real root.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0could you explain it please? thankyou!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you could try something like \[f(x)=\frac{x}{x^2+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i mean to say something "like" it. that one doesn't work because the range of \[f(x)=\frac{x}{x^2+1}\] is \([\frac{1}{2},\frac{1}{2}]\) you will have to adjust it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the first response isn't a rational function. They have to be polynomials.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0only problem with @mahmit answer is \(x+2\) is not a polynomial

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh what @scarydoor said

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@satellite73 's hint is on the money... easy to convert that to the right function.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@scarydoor how can i convert it to the right function? i dont understand

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=\frac{ x+1 }{ x^2+1 }\] can anyone confirm this answer? i think its right...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Satellite's function is almost right, in that the range is [1/2, 1/2]. But you want it [1,1]. So you want to stretch it out to that. If you multiply the function by 2, then if you think about it a bit, you'll see that the range will be [1,1].

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh thankyou! yes it makes sense
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