What is the domain and range of #y = (2x^2)/( x^2 - 1)#?

1 Answer
Aug 7, 2017

Answer:

The domain is #x in (-oo,-1) uu (-1,1) uu (1,+oo)#
The range is #y in (-oo,0] uu (2,+oo)#

Explanation:

The function is

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

We factorise the denominator

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

Therefore,

#x!=1# and #x!=-1#

The domain of y is #x in (-oo,-1) uu (-1,1) uu (1,+oo)#

Let's rearrage the function

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

#yx^2-y=2x^2#

#yx^2-2x^2=y#

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

#x=sqrt(y/(y-2))#

For #x# to a solution, #y/(y-2)>=0#

Let #f(y)=y/(y-2)#

We need a sign chart

#color(white)(aaaa)##y##color(white)(aaaa)##-oo##color(white)(aaaaaa)##0##color(white)(aaaaaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##y##color(white)(aaaaaaaa)##-##color(white)(aaa)##0##color(white)(aaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##y-2##color(white)(aaaaa)##-##color(white)(aaa)##color(white)(aaa)##-##color(white)(aa)##||##color(white)(aa)##+#

#color(white)(aaaa)##f(y)##color(white)(aaaaaa)##+##color(white)(aaa)##0##color(white)(aa)##-##color(white)(aa)##||##color(white)(aa)##+#

Therefore,

#f(y)>=0# when #y in (-oo,0] uu (2,+oo)#

graph{2(x^2)/(x^2-1) [-16.02, 16.02, -8.01, 8.01]}