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

1 Answer
Jul 11, 2017

Answer:

See Below.

Explanation:

Before we do anything, let's see if we can simplify the function by factoring the numerator and denominator.

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

#((x)(x-1))/(x+1)#

The domain of a function is all of the #x#values (horizontal axis) that will give you a valid y-value (vertical axis) output.

Since the function given is a fraction, dividing by #0# will not yield a valid #y# value. To find the domain, let's set the denominator equal to zero and solve for #x#. The value(s) found will be excluded from the range of the function.

#x+1=0#

#x=-1#

So, the domain is all real numbers EXCEPT #-1#. In set notation, the domain would be written as follows:

#(-oo,-1)uu(-1,oo)#

The range of a function is all of the #y#-values that it can take on. Let's graph the function and see what the range is.

graph{(x^2-x)/(x+1) [-10, 10, -5, 5]}