How do you find the domain and range of # sqrt(x+4)/(x+2)#?

1 Answer
Jul 1, 2018

Answer:

The domain is #x in [-4,2) uu (2,+oo)#. The range is #y in RR#

Explanation:

For the domain, there are #2# conditions :

The denominator must be #!=0#

#=>#, #x+2!=0#

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

And

What's under the #sqrt# sign must be #>=0#

#=>#, #x+4>=0#

#=>#, #x>=-4#

Therefore,

The domain is #x in [-4,-2) uu (-2,+oo)#

For the range, let #y=sqrt(x+4)/(x+2)#

When #x=-4#, #=>#, #y=0#

And

#lim_(x->-2^-)y=-oo#

#lim_(x->-2^+)y=+oo#

#lim_(x->+oo)y=0^+#

The range is #y in RR#

graph{sqrt(x+4)/(x+2) [-11.44, 24.6, -8.1, 9.92]}