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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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littmo12Best ResponseYou've already chosen the best response.0
could you explain it please? thankyou!
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
you could try something like \[f(x)=\frac{x}{x^2+1}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
i 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
 one year ago

scarydoorBest ResponseYou've already chosen the best response.0
the first response isn't a rational function. They have to be polynomials.
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
only problem with @mahmit answer is \(x+2\) is not a polynomial
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
oh what @scarydoor said
 one year ago

scarydoorBest ResponseYou've already chosen the best response.0
@satellite73 's hint is on the money... easy to convert that to the right function.
 one year ago

littmo12Best ResponseYou've already chosen the best response.0
@scarydoor how can i convert it to the right function? i dont understand
 one year ago

littmo12Best 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...
 one year ago

scarydoorBest ResponseYou've already chosen the best response.0
Satellite'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].
 one year ago

littmo12Best ResponseYou've already chosen the best response.0
ahh thankyou! yes it makes sense
 one year ago
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