Finding The Domain and Range of a Function

Date: 01/27/2001 at 00:04:57
From: Angela Armstrong
Subject: Finding The Domain and Range of a function

How do I find the domain and range of a function? My problem is 
f(x)= x squared -2.

I do not understand how you determine the domain and range of a given 
function.

Date: 01/27/2001 at 10:32:05

Hi Angela,

Thanks for writing to Dr. Math. 

The domain of a function f(x) is usually fairly easy to find: it is 
the set of all the numbers x that can be put into the formula f(x) and 
have the result make sense. That is, the x values that don't make some 
expression inside a square root sign negative, or that don't make a 
denominator zero, and so on.  

If your function has sqrt(x-2) in the formula, you can only put in 
x values greater than or equal to 2, otherwise you will have a 
negative number under a square root sign. If your function has 
1/(x^2-4) [here x^2 means x squared], then you can't let x = 2 or -2, 
because that would require a division by 0, which isn't allowed.  

For your example, is there any value of x that would make the right 
side of the formula meaningless? If not, then the domain is the set of 
all numbers, unless for some reason the domain is explicitly 
restricted. For example, a problem might define f(x) = x^2 - 2 
for x > 0. In this case, with the explicit restriction x > 0 given, 
the domain is {x | x > 0}, even though the formula makes sense for 
more values of x.

The range is often harder to determine, but for your example, it isn't
too hard. If I write y = f(x), then

     y = x^2 - 2.

The range is the set of all values y can take, as x takes every value 
in the domain. In this problem, you know that the square of a number 
is greater than or equal to 0. Could y take the value -3? If we try to 
solve
          -3 = x^2 - 2

we get
          -1 = x^2 

which is impossible to solve, so -3 is not in the range. One way to 
find the values of y that are possible is to try solving the equation 
for x:

            y = x^2 - 2

        y + 2 = x^2

            x = +/- sqrt(y+2) .

Now you can use the logic you used for the domain: what values of y 
will let this formula make sense?

If you have further questions, please write again.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/