Definitions & notation
This section will refine our view of functions just a bit. In working with functions, it's important to know just what values can be put into a function and just what values the function can give back. These sets of numbers are known as domain and range, respectively.
The domain of a function is the set of all values that the independent variable (usually $x$, in these notes) can take on.
In the examples below, you'll see examples of functions that can take any value of $x$ as input, and some that have restrictions.
Likewise, some functions will only give back numbers in a certain range. For example, a quadratic function with a positive leading coefficient can give back ($y$) values greater than or equal to the y-coordinate of the vertex. There simply are no values of $f(x)$ below that.
Domain and range are key properties of functions. They are also important in understanding inverses of functions.
Note: This is important material for you to know, but the last few functions discussed on this page might not be familiar to you. Just skip those and you can come back to this section once in a while to refresh your memory with new functions.
Set notation [Bracket notation)
There is a compact notation for writing out the sets that will be the domain and range of a function, and you should learn it. Once you do, it's actually handy for writing all kinds of sets that you'll encounter in higher math, and in a lot of math and science literature. Take the time now to learn it.
We use square or round brackets to mark the boundaries of sets. Square brackets are used to indicate that the endpoint is part of the set, and round brackets mean that the set extends to the endpoint, but doesn't include it. Examples are probably the best way to learn, so take a close look at the table below.
| This notation | Means this: |
|---|---|
| $[a, b]$ | The set of all numbers between a and b, including a & b |
| $(a, b)$ | The set of all numbers between a and b, but not including a & b |
| $(a, b]$ | The set of all numbers between a and b, including b, but not including a |
| $[a, b)$ | The set of all numbers between a and b, including a, but not including b |
| $[a, \infty)$ | The set of all numbers from a (including a) to infinity, an infinitely-large set. |
| $(-\infty, b)$ | The set of all numbers from negative infinity, up to b, and not including b |
| $[a, b] \cup [c, d]$ | The union of the sets of all numbers between and including a & b, and between and including c & d. The union symbol ( ∪ ) is like the plus sign for sets. |
| $(-\infty, a) \cup (a, \infty]$ | The set of all numbers except for a |
| $(-\infty, \infty)$ | The set of all real numbers, sometimes written as $\{\mathbb{R}\}$ |
Domain & range of linear functions
The domain of any linear functions is all real numbers, or $(-\infty, \infty)$. There is simply no value of $x$ we can't put into a linear function. Another way to look at it is that the graph of a linear function, no matter now steeply sloped, extends infinitely to the right and left. The graph of
$$f(x) = x - 4$$
is shown on the right. The same is true of the range of a linear function. Provided we put in the right $x$, we should be able to obtain any $y$ in the set $y = (-\infty, \infty)$. We can write the domain and range of any linear function as
$$D: \, (-\infty, \, \infty), \: \: R: \, (-\infty, \, \infty)$$
Domain & range of linear functions
The domain and range of any linear function are $x \in (-\infty, \infty)$ and $y \in (-\infty, \infty).$
Domain & range of quadratic functions
Quadratic functions always have a vertex. That point is either the minimum or maximum of the function. The function
$$f(x) = \frac{1}{2} (x - 2)^2 - 4$$
is plotted on the left. It's vertex is (2, -4), which means that the lowest value this function can have is $y = -4$, so the range is $[-4, \infty)$.
The domain of a quadratic function is always $x = (-\infty, \infty)$. The domain and range of this function are
$$D: \, (-\infty, \, \infty), \: \: R: \, [-4, \, \infty)$$
Domain & range of polynomial functions
The domain of any polynomial function (including quadratic functions) is $x \in (-\infty, \infty).$ Functions of even degree will have a bounded range (from below if the leading coefficient is positive, from above if it's negative), and functions of odd degree will have range $y \in (-\infty, \infty).$
Domain & range of cubic functions
The cubic function plotted to the right is
$$f(x) = x^3 - 3x^2 + 3x - 2$$
Because of the odd exponent, one end of a cubic function tends toward $+\infty$ and the other toward $-\infty$. That's actually true of any polynomial function of odd degree. The domain and range of such functions are
$$D: \, (-\infty, \, \infty), \: \: R: \, (-\infty, \, \infty)$$
Polynomial functions of even degree (n = 2, 4, 6 ...)
Polynomial functions of even degree always have a global maximum or (in this case) minimum, depending on the sign of the leading term.
Therefore the ranges of polynomials of even degree (where both ends either go to positive or negative infinity) are determined by the location of that local min. or max.
We won't always know the exact location of that max. or min. That's a problem that calculus will help you with later. In such cases, it's OK just to give that max. or min. a label like we did in the graph, $(a, b)$. The minimum value of the function is just $b$ (a negative number in this example), and we write the domain and range like this:
$$D: \; (-\infty, \; \infty) \; \; R: \; (b, \; \infty)$$
Domain & range of polynomial functions
The range of any polynomial function of odd degree is $y = (-\infty, \infty)$.
The range of a polynomial function of even degree is bounded either from below or above because the function will always have a maximum or minimum value.
Rational functions
Because rational functions have discontinuities like asymptotes and holes, their domains and ranges can be pretty broken up.
The function
$$f(x) = \frac{x^2 - 1}{x^2 + x - 3} = \frac{(x + 1)(x - 1)}{(x - 2)(x + 3)}$$
plotted below has both vertical and horizontal asymptotes.
A function has no actual value at a vertical asymptote. It's easy to see why. When $x = -3$ in our function above, the denominator goes to zero, so the function is infinite. That's represented by the dashed vertical line at $x = -3$. It's the same when $x = 2$. So the domain is interrupted twice by values that the function just can't take.
Furthermore, for large values of $x$, in either direction, $f(x)$ approaches (but never quite reaches) $y = 1$. The range is also disrupted. As in the example above, we don't actually know how high that upside-down U-shaped bit in the middle gets, so we'll call that value $a$. Calculus (or trial and error with your calculator) would help us to find that, but for now it's enough to recognize that there's a hard number above which the function can't have a value (until the plus side of $y = 1$).
The domain and range of this function are
$$ \begin{align} &D: \: (-\infty, -3) \cup (-3, 2) \cup (2, \infty) \\[5pt] &R: \: (-\infty, a] \cup (1, \infty) \end{align}$$
Root functions
The function
$$f(x) = \sqrt{x}$$
is plotted on the right. The obvious problem with the domains of root functions is that the expression under the radical can't be negative.
That means that root functions just begin somewhere (in this case at $x = 0$) and move off to the right. The domain and range of this function are
$$D: \: [0, \infty) \phantom{0000} R: \: [0, \infty)$$
Here's a second example of a root function,
$$f(x) = \sqrt{x - 3}$$
You will recognize that this function is just the root function on the left translated three units to the right.
This function has a different domain because the expression under the radical is less than zero for values of $x \lt 3$. Its domain and range are
$$D: \: [3, \infty) \phantom{0000} R: \: [0, \infty)$$
What restricts the domain?
If the domain of a function is restricted, it's usually because of one of two things: (1) some value(s) of x make the denominator go to zero (mathematicians often say that the function "blows up" in that case), or (2) some value(s) of x cause an even root (like a square root) of a negative number.
A logarithmic function: $f(x) = ln(x)$
Logarithmic functions tend toward infinity as the independent variable grows, and tend (rapidly) toward negative infinity as it approaches zero.
Log functions have no value for x < 0 because a log expression asks "to what power must we raise a positive base (in this case the base, e) to get x?" For a positive base, there is no number that will do that. We can't raise a positive number to any power and get a negative number. The domain and range of this function are
$$D: \: (0, \infty) \phantom{0000} R: \: (-\infty, \infty)$$
An Exponential function: $f(x) = 2^x$
This exponential function can take any value of $x$ we wish to put into it, but it can only return positive values.
Remember that negative exponents don't mean negative function values. The minus in the exponent just means "take the reciprocal. The reciprocal of a large number is a small number, thus the appearance of the graph.
The domain and range of this exponential function are
$$D: \: (-\infty, \infty) \phantom{0000} R: \: (0, \infty)$$
Practice problems
Determine the domain and range of the following functions, and express them in proper bracket notation.
| 1. |
$$f(x) = \frac{1}{x^2 + 1}$$ |
| 2. |
$$f(x) = \frac{x^2 - 1}{x^2 + 3x + 2}$$ |
| 3. |
$$f(x) = \frac{x^2 - x - 2}{x^2 - x - 6}$$ |
| 4. |
$$f(x) = \sqrt{x - 3}$$ |
| 5. |
$$f(x) = \sqrt{-x - 1}$$ |
| 6. |
$$f(x) = 2 \, ln(x - 3)$$ |
Problem 2 solution
This function can be factored to
$$\frac{x^2 - 1}{x^2 + 3x + 2} = \frac{(x + 1)(x - 1)}{(x + 1)(x + 2)}$$
The binomial (x + 1) is in both the numerator and denominator, so this function has a hole at x = -1; we take that point out of the domain. The remaining function is:
$$f(x) = \frac{x - 1}{x + 2}$$
The denominator has a zero at x = -2, therefore there is a vertical asymptote there. The domain is:
$$D: \; (-\infty, -2) \cup (-2, -1) \cup (-1, \infty)$$
With the hole removed, our function is:
$$f(x) = \frac{x - 1}{x + 2},$$
which has a single zero at x = 1. The inequality
$$\frac{x - 1}{x + 2} \gt 0 $$
shows that the function is positive for all x > 0, for x < -2, and negative otherwise. Here is a sketch of the function graph:
Finally, the location of the hole is determined from the version of f(x) without it, f(-1) = -2. From the graph, we can write the range:
$$R: \; (-\infty, -2) \cup (-2, 1) \cup (1, \infty)$$
Problem 3 solution
This function can be factored like this:
$$f(x) = \frac{(x + 1)(x - 2)}{(x - 3)(x + 2)}$$
It has zeros at x = -1, 2, and vertical asymptotes (the zeros of the denominator) at x = -2, 3. It has no holes. The location of the zeros and VAs give us a pretty clear sketch of the graph:
The domain is: $D: \; (-\infty, -2) \cup (-2, 3) \cup (3, \infty)$
The range is a bit more difficult. The location of the horizontal asymptote is clear, y = 1, but the maximum value of the center section of the graph needs work. We can find it numerically or using calculus. If you do, you'll find that there's a local maximum at x = 1/2, at the point $(\frac{1}{2}, \frac{9}{25}).$ Thus the range is:
$$ R: \; \left(-\infty, \frac{9}{25}\right) \cup (1, \infty)$$
Problem 4 solution
The graph of this function is just the square root function, $f(x) = \sqrt{x},$ shifted to the right by 3 units. In this sketch, the parent function is in gray and our function is in black:
We can read the domain and range right from the graph:
$$ \begin{align} D&: \; [3, \infty) \\ R&: \; [0, \infty) \end{align}$$
You could also check the domain by insisting that the function under the radical be greater than zero:
$$x - 3 \ge 0 \; \longrightarrow \; x \ge 3$$
Problem 5 solution
The graph of this function is just the square root function, $f(x) = \sqrt{x},$ shifted to the left by 1 unit, and reflected to the left. In this sketch, the parent function is in gray and our function is in black:
We can read the domain and range right from the graph:
$$ \begin{align} D&: \; (-\infty, -1] \\ R&: \; [0, \infty) \end{align}$$
You could also check the domain by insisting that the function under the radical be greater than zero:
$$-x - 1 \ge 0 \; \rightarrow \; -x \ge 1 \; \rightarrow \; x \le -1$$
Problem 6 solution
The graph of this function is just the the ln(x) function shifted to the right by 3 units, and stretched vertically by a factor of 2. The latter doesn't affect the domain or range. In this sketch, the parent function is in gray and our function is in black:
Notice that the vertical asymptote of the function is translated 3 units to the right as well. We can read the domain and range right from the graph:
$$ \begin{align} D&: \; (3, \infty) \\ R&: \; (-\infty, \infty) \end{align}$$
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