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

1 Answer
Sep 20, 2015

Answer:

Domain: #(-oo, -1) uu (-1, 1) uu (1, oo)#
Range: #(-oo, -4] uu (0, oo)#

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 #4/((-1)^2-1)# which equals #4/(1-1)# or #4/0#.

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 #4/0# so we have to exclude that from the domain.

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

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 #-1/3#, which is impossible because squaring a number always has a positive result.

That means #y="-"3# is undefined and so is not part of our range. The same is true for all y-values between 4 and 0.

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, #(-oo, -4] uu (0, oo)#.

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.