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How do you find the domain and range of #x/(x^2+1)#?

1 Answer

Konstantinos Michailidis

Sep 18, 2015

The domain is #R# and the range is #[-1/2,1/2]#

Explanation:

The domain is #R# the set of all reals because #1+x^2>0#.

To find the range we have that

#y=x/(x^2+1)=>y(x^2+1)=x=>yx^2+x+y=0=#

this is trinomial with respect to x hence

#1^2-4y*y>=0=>1-4y^2>=0=>-1/2<=y<=1/2#

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