How do you find the domain and range of #g(x)= (3+x^2)/(4-x^2)#?

1 Answer
Oct 7, 2017

Answer:

The domain is #x in (-oo,-2)uu(-2,2)uu(2,+oo)#
The range is #g(x) in (-oo,-1)uu [3/4,+oo)#

Explanation:

As you cannot divide by #0#, the denominator is #!=0#

#4-x^2!=0#

#(2+x)(2-x)!=0#

#=>#, #x!=-2# and #x!=2#

Therefore,

The domain is #x in (-oo,-2)uu(-2,2)uu(2,+oo)#

Let

#y=(3+x^2)/(4-x^2)#

#y(4-x^2)=3+x^2#

#4y-yx^2=3+x^2#

#x^2(1+y)=4y-3#

#x^2=(4y-3)/(1+y)#

#x=sqrt((4y-3)/(1+y))#

Therefore,

#(4y-3)/(1+y)>=0#

#=>#, #y!=-1#

Let #g(y)=(4y-3)/(1+y)#

We build a sign chart

#color(white)(aaaa)##y##color(white)(aaaa)##-oo##color(white)(aaaaaa)##-1##color(white)(aaaaaaa)##3/4##color(white)(aaaaaaa)##+oo#

#color(white)(aaaa)##1+y##color(white)(aaaaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##+##color(white)(aaaaaaa)##+#

#color(white)(aaaa)##4y-3##color(white)(aaaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##-##color(white)(aa)##0##color(white)(aaaa)##+#

#color(white)(aaaa)##g(y)##color(white)(aaaaaaa)##+##color(white)(aaa)##||##color(white)(aaaa)##-##color(white)(aa)##0##color(white)(aaaa)##+#

Therefore,

#g(y)>=0# when #y in (-oo,-1)uu [3/4,+oo)#

graph{(3+x^2)/(4-x^2) [-12.66, 12.65, -6.33, 6.33]}