What is the domain and range of #ƒ(x)= (5x+15)/ ((x^2) +1)#?

1 Answer
Konstantinos Michailidis
Sep 24, 2015

Refer to explanation

Explanation:

The range is the set of real numbers hence #D(f)=R#.

For the range we set #y=f(x)# and we solve with respect to #x#

Hence

#y=(5x+5)/(x^2+1)=>y*(x^2+1)=5x+5=> x^2*(y)-5x+(y-5)=0#

The last equation is a trinomial with respect to x.In order to have a meaning in real numbers its discriminant must be equal or greater than zero.Hence

#(-5)^2-4*y*(y-5)>=0=>-4y^2+20y+25>=0#

The last is always true for the following values of #y#

#-5/2(sqrt2-1)<=y<=5/2(sqrt2+1)#

Hence the range is

#R(f)=[-5/2(sqrt2-1),5/2(sqrt2+1)]#

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