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

1 Answer
May 15, 2017

Answer:

The domain is #D_f(x)=RR-{-1,1}#
The range is #f(x) in (-oo,-1) uu [0,+oo) #

Explanation:

#f(x)=x^2/(1-x^2)=x^2/((1-x)(1+x))#

As we cannot divide by #O#, #x!=1# and #x!=-1#

The domain of #f(x)# is #D_f(x)=RR-{-1,1}#

To calculate the range, we need to calculate #f^-1(x)#

Let #y=x^2/(1-x^2)#

We interchange #y# and #x#

#x=y^2/(1-y^2)#

Now, we calculate #y# in terms of #x#

#x(1-y^2)=y^2#

#x-xy^2=y^2#

#y^2(x+1)=x#

#y^2=x/(x+1)#

#y=sqrt(x/(x+1))#

The domain of #y# is the range of #f(x)#

What is underneath the #sqrt# sign is #>=0#

Therefore,

#x/(1+x)>=0#

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##-1##color(white)(aaaaaaaa)##0##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##x+1##color(white)(aaaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##f^-1(x)##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f^-1(x)>=0# when #x in (-oo,-1) uu [0,+oo)#