Dr. Matthew M. Conroy - University of Washington 1
Domain and Range Examples
The domain of a function fis the set of all values xfor which f(x)is defined.
The range of a function fis the set of all values that f(x)takes on as xruns through the domain
of f. That is, it is the set of all yvalues for which there is an xvalue such that
y=f(x).
For many functions, the domain is easy to determine. Often, all we have to do is look for which
xcause a problem when evaluating f(x); those xare not in the domain of f, and the domain is
everything else.
For example, suppose f(x) = 5
x(x−1) . To compute f(x)we have to multiply xtimes x−1. This
part causes no trouble: we can multiply any two number together. Then, we divide 5 by the
product just calculated. This goes well, unless that product is zero. That is, the only xvalues
that cause trouble are those for which
x(x−1) = 0.
Since the left hand side of this equation is factored, we see that the only xfor which x(x−1) = 0
are x= 0 and x= 1. Thus, those are the only two values not in the domain of f, and so the
domain of fis everything else. We might say: the domain of fis all x6= 0,x6= 1.
Square roots can cause us problems as well. The square root of xis only defined for x≥0. In
other words, the domain of √xis all x≥0.
Functions made up from the square root function inherit this restriction. For example, let
g(x) = √x−1 + √x+ 5.
In order for g(x)to be defined, x−1and x+ 5 must both be greater than 0. So, if xis in the
domain,
x−1≥0and x+ 5 ≥0
or, simplified,
x≥1and x≥ −5.
Now, if x≥1then xis automatically greater than or equal to −5, so the first condition is all we
need: the domain of gis all x≥1.
It is worth knowing how to evaluate the domain of a function made from another function
whose domain we know. For instance, suppose we know fhas the domain 2≤x≤11. Then,
let g(x) = f(5x−8). For g(x)to be defined, f(5x−8) must be defined. In order for f(5x−8) to
be defined, 5x−8must be in the domain of f. That is,
2≤5x−8≤11.
Solving this, we find the domain of gis
2≤x≤19
5.
