# Range of rational function

i got that from punching in random x values on my graphing calc. (10, 20, 30) and as they got bigger the y values got smaller but never reached zero.

jbunniii
Homework Helper
Gold Member
i got that from punching in random x values on my graphing calc. (10, 20, 30) and as they got bigger the y values got smaller but never reached zero.

That's right. You can make y as close to 0 as you like, but it never quite gets there. You can also make y as negative as you like.

So that means that for $x < -1$, the range of possible y values is $y < 0$.

Other ways of expressing the same thing are

$$\{y | y < 0\}$$

or

$$(-\infty,0)$$

Now let's consider the other half of the domain: $x > -1$. What range of y values are possible here?

positve infinity or y > 0

jbunniii
Homework Helper
Gold Member
positve infinity or y > 0

OK, so if $x > 1$ then the possible range of y values is $y > 0$.

Now put the two halves together to get the total range of the function.

HallsofIvy
Homework Helper
The "natural domain" of a function where you are given a formula for it is just the set of all x values to which the formula can be applied. Here, the only arithmetic operation is "divide by x+ 1". You can divide any number except 0 so you can do that calculation for any number except x= -1. The domain is "all x except -1".

In interval notation that would be $(-\infty, -1)\cup (-1, \infty)$.

The "range" is the set of possible y values.

One way of finding the range is to try to invert the function. If y= 1/(x+1) then x+1= 1/y and x= (1/y)- 1. Since we can divide by any number except 0, y can take any value except 0. The range is "all y except 0".

For this simple function, you could also have thought, since a/b= 0 gives immediately a= 0 by multiplying both sides by b, "a fraction is 0 if and only if its numerator is 0". Here the numerator is the constant 1 which is never 0. y can never be 0.

Thanks for the help, and just to be sure, the range is always expressed as y $$\neq$$ ....

Char. Limit
Gold Member
No, only sometimes, like here.