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

Question

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

Eva Abramson · Accepted Answer

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

Isaiah Anthony · Answer

undefined To find the domain and range of $ f(x) = \frac{x^2}{1 - x^2} $:

1. **Domain**: Identify values of $ x $ that make the function undefined. In this case, the function is undefined when the denominator, $ 1 - x^2 $, equals zero. Thus, $ 1 - x^2 \neq 0 $. Solving $ 1 - x^2 = 0 $, we get $ x = \pm 1 $. So, the domain of $ f(x) $ is all real numbers except $ x = \pm 1 $.

2. **Range**: To find the range, observe the behavior of the function as $ x $ approaches positive and negative infinity. As $ x $ approaches positive or negative infinity, $ f(x) $ approaches positive infinity. Also, the function is symmetric about the y-axis. Hence, the range of $ f(x) $ is all real numbers greater than or equal to zero, excluding zero.