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Functions...bijections?
Let f(x) = (3x-1)/(2x+3).
I am able to find the domain...was unsure how to go about finding the range. and also I dont know how to find subsets X and Y of R such that f: X --> Y is a bijection.
Thanks!
2 Answers
- suesysgoddessLv 61 decade agoFavourite answer
The domain is all reals except -3/2
The range is all reals except 3/2 (horizontal asymptote at y = 3/2)
A bijection is onto and one to one. For every x there is one and only one value for y
.. this function is onto and one to one, which means it is a bijection. X is the domain of the function and Y is the range of the function.
- MathTutorLv 61 decade ago
Domain = R - {-3/2}
Range= R - {3/2}
You can never reach the value y = 3/2. This is the functions asymptote
X = Domain and Y = Range in this case. Since f: Domain -> R is an injective function, then you only have to consider f: X->Y to have a bijective one.
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