Question

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

Answer 1

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]}