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

1 Answer
Jan 1, 2018

Answer:

Domain: #{x|x!=+-1}#
Range: #{y|y<=-(2+sqrt3)/2,y>=-(2-sqrt3)/2}#

Explanation:

Finding the Domain
The domain is where the function is defined in terms of real numbers.

The only thing that might make this function undefined is when the denominator equals zero. So let's solve for when that is the case:

#x^2-1=0#

#x^2=1#

#x=+-sqrt1#

#x=+-1#

This means that our domain is:
#{x|x!=+-1}#

Finding the Range
You can usually reason your way to finding the range, but this function has some quite odd behaviour, so we're going to use a bit of a classic method.

The range of the function is the same as the domain of the inverse of the function. So let us find the inverse. We can do this by setting the function equal to #y# and solving for #x#:

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

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

#yx^2-y=x+2#

#yx^2-x-y-2=0#

#y/yx^2-1/yx-y/y-2/y=0#

#x^2-1/yx-1-2/y=0#

This is a quadratic in #x#, so we can solve using the quadratic formula:
#x=1/(2y)+-sqrt(1/(4y^2)+1+2/y)#

#x=1/(2y)+-sqrt(1/(4y^2)+(4y^2)/(4y^2)+(8y)/(4y^2))#

#x=1/(2y)+-sqrt((4y^2+8y+1)/(4y^2))#

#x=1/(2y)+-sqrt(4y^2+8y+1)/(2y)#

#x=(1+-sqrt(4y^2+8y+1))/(2y)#

The range of our original function is the same as the domain of this inverse function, so we want to look at when this might be undefined. The function is not defined if the bit in the square root is negative, so we can solve the following inequality to find out when that's the case:
#4y^2+8y+1<0#

We can use the quadratic formula to solve for the zeroes and factor like so:
#(x+(2+sqrt3)/2)(x+(2-sqrt3)/2)<0#

For the product to be negative, only one of the factors may be negative, so we want to find where that's the case. There are three possible intervals where that could be the case, and they are between the zeroes (since that is where the function could go negative). The only one of these intervals that make one of the factors negative is:
#-(2+sqrt3)/2< y< -(2-sqrt3)/2#

You might think that the denominator equalling zero would make the function undefined - and it does, but that doesn't mean the range doesn't include those values.

This is because the division by #2y# was something the quadratic formula arbitrarily introduced, and that excludes any solutions where #y=0#. We can also back up this reasoning by showing that #x=-2# in the original function produces a value of #0#.

All this knowledge allows us to state the range of the function as follows:
#{y|y<=-(2+sqrt3)/2,y>=-(2-sqrt3)/2}#