# What is the domain and range of #y= 4 / (x^2-1)#?

##### 1 Answer

#### Answer:

Domain:

Range:

#### Explanation:

Best explained through the graph.

graph{4/(x^2-1) [-5, 5, -10, 10]}

We can see that for the domain, the graph starts at negative infinity. It then hits a vertical asymptote at x = -1.

That's fancy math-talk for the graph is not defined at x = -1, because at that value we have

Since you can't divide by zero, you can't have a point at x = -1, so we keep it out of the domain (recall that the domain of a function is the collection of all the x-values that produce a y-value).

Then, between -1 and 1, everything's fine, so we have to include it in the domain.

Things start getting funky at x = 1 again. Once more, when you plug in 1 for x, the result is

To sum it up, the function's domain is from negative infinity to -1, then from -1 to 1, and then to infinity. The mathy way of expressing that is

The range follows the same idea: it's the set of all y-values of the function. We can see from the graph that from negative infinity to -4, all is well.

Then things start going south. At y=-4, x=0; but then, if you try y=-3, you won't get an x. Watch:

#-3 = 4/(x^2-1)#

#-3(x^2-1) = 4#

#x^2-1 = -4/3#

#x^2 = -4/3+1 = -1/3#

#x = sqrt(-1/3)#

There is no such thing as the square root of a negative number. That's saying some number squared equals

That means

From 0 above, everything is good all the way to infinity. Our range is then negative infinity to -4, then 0 to infinity; in math terms,

In general, to find domain and range, you have to look for places where things are suspicious. That usually involves stuff like dividing by zero, taking the square root of a negative number, etc.

Whenever you find a point like this, remove it from the domain/range and build up your interval notation.