How do you find the domain and range of #sqrt((13x)/((x^2)-1)#?

1 Answer
May 29, 2017

Answer:

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

Explanation:

What's uner the #sqrt# sign is #>=0#

So,

#(13x)/(x^2-1)>=0#

#(13x)/((x+1)(x-1))>=0#

Let #f(x)=(13x)/((x+1)(x-1))#

We can build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-1##color(white)(aaaaa)##0##color(white)(aaaaaaa)##+1##color(white)(aaaa)##+oo#

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

#color(white)(aaaa)##x##color(white)(aaaaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aa)##+#

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

#color(white)(aaaa)##f(x)##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+#

Therefore,

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

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

Let,

#y=sqrt((13x)/(x^2-1))#

When #x=-1#, #=>#, #y=+oo#

When #x=0#, #=>#, #y=0#

The range is #y in [0,+oo)#