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Section 10.2 Domain and Range of Functions
When working with functions, we need to understand what inputs are allowed and what outputs are possible. The
domain of a function is the set of all possible input values (typically
\(x\) -values), and the
range is the set of all possible output values (typically
\(y\) -values or
\(f(x)\) -values).
Think of a function as a machine: the domain is all the items you’re allowed to put into the machine, and the range is all the possible items that can come out of the machine.
Definition 10.2.1 . Domain and Range.
For a function
\(f\text{:}\)
The domain of \(f\) is the set of all possible input values: \(\set{x : f(x) \text{ is defined}}\)
The range of \(f\) is the set of all possible output values: \(\set{y : y = f(x) \text{ for some } x \text{ in the domain}}\)
We will use interval notation to express domains and ranges. Recall from the previous section that interval notation provides a concise way to describe sets of numbers.
Subsection 10.2.1 Finding the Domain of a Function
To find the domain of a function, we need to identify which values of
\(x\) make the function undefined. The most common restrictions are:
Division by zero: We cannot divide by zero, so any value that makes a denominator equal to zero must be excluded from the domain.
Even roots of negative numbers: We cannot take the square root (or any even root) of a negative number in the real number system, so the expression under an even root must be non-negative.
Logarithms of non-positive numbers: We can only take logarithms of positive numbers (covered in a later chapter).
Let’s look at examples of finding domains for different types of functions.
Example 10.2.2 . Domain of Polynomial Functions.
Find the domain of
\(f(x) = x^2 + 3x - 5\text{.}\)
Solution .
Polynomial functions have no restrictions. We can substitute any real number for
\(x\) and get a valid output.
Therefore, the domain is
\((-\infty, \infty)\) or
\(\R\) (all real numbers).
Example 10.2.3 . Domain of Another Polynomial.
Find the domain of
\(p(x) = 2x^3 - 4x + 1\text{.}\)
Solution .
Again, this is a polynomial function with no restrictions.
Domain:
\((-\infty, \infty)\)
Example 10.2.4 . Domain of a Simple Rational Function.
Find the domain of
\(f(x) = \frac{1}{x-3}\text{.}\)
Solution .
This function has a variable in the denominator. We need to find where the denominator equals zero:
\begin{align*}
x - 3 \amp= 0\\
x \amp= 3
\end{align*}
The function is undefined when
\(x = 3\text{,}\) so we must exclude this value from the domain.
Domain:
\((-\infty, 3) \cup (3, \infty)\)
In words: all real numbers except 3.
Example 10.2.5 . Domain with Two Restrictions.
Find the domain of
\(g(x) = \frac{x+2}{x^2-4}\text{.}\)
Solution .
First, we factor the denominator:
\begin{equation*}
x^2 - 4 = \brac{x-2}\brac{x+2}
\end{equation*}
The denominator equals zero when:
\begin{align*}
x - 2 = 0 \quad \amp\text{or} \quad x + 2 = 0\\
x = 2 \quad \amp\text{or} \quad x = -2
\end{align*}
We must exclude both values from the domain.
Domain:
\((-\infty, -2) \cup (-2, 2) \cup (2, \infty)\)
In words: all real numbers except
\(-2\) and
\(2\text{.}\)
Example 10.2.6 . Domain with No Restrictions.
Find the domain of
\(h(x) = \frac{5}{x^2+1}\text{.}\)
Solution .
We check if the denominator can equal zero:
\begin{align*}
x^2 + 1 \amp= 0\\
x^2 \amp= -1
\end{align*}
Since
\(x^2\) is always non-negative for real numbers,
\(x^2 + 1\) is always at least 1. The denominator never equals zero.
Domain:
\((-\infty, \infty)\)
Example 10.2.7 . Domain of a Square Root Function.
Find the domain of
\(f(x) = \sqrt{x-2}\text{.}\)
Solution .
The expression under the square root must be non-negative:
\begin{align*}
x - 2 \amp\geq 0\\
x \amp\geq 2
\end{align*}
Example 10.2.8 . Domain of Another Radical Function.
Find the domain of
\(g(x) = \sqrt{5-2x}\text{.}\)
Solution .
We need
\(5 - 2x \geq 0\text{:}\)
\begin{align*}
5 - 2x \amp\geq 0\\
5 \amp\geq 2x\\
\frac{5}{2} \amp\geq x
\end{align*}
Domain:
\((-\infty, \frac{5}{2}]\)
Example 10.2.9 . Domain of a Cube Root Function.
Find the domain of
\(h(x) = \sqrt[3]{x+4}\text{.}\)
Solution .
Cube roots (odd roots) of negative numbers are defined in the real number system. For example,
\(\sqrt[3]{-8} = -2\text{.}\)
Therefore, there are no restrictions on the domain.
Domain:
\((-\infty, \infty)\)
Example 10.2.10 . Domain with Combined Restrictions.
Find the domain of
\(f(x) = \frac{\sqrt{x+1}}{x-2}\text{.}\)
Solution .
This function has two types of restrictions:
From the square root: We need
\(x + 1 \geq 0\text{,}\) so
\(x \geq -1\text{.}\)
From the denominator: We need
\(x - 2 \neq 0\text{,}\) so
\(x \neq 2\text{.}\)
Combining both restrictions:
\(x \geq -1\) AND
\(x \neq 2\text{.}\)
Domain:
\([-1, 2) \cup (2, \infty)\)
Exercise Group 10.2.1 . Finding Domain I: Basic Functions.
Find the domain of each function. Express your answer in interval notation.
(a) \(f(x) = 3x - 7\)
(b) \(g(x) = x^2 - 5x + 6\)
(c) \(h(x) = \frac{1}{x+4}\) Answer .
\((-\infty, -4) \cup (-4, \infty)\)
(d) \(p(x) = \frac{2}{x-5}\) Answer .
\((-\infty, 5) \cup (5, \infty)\)
(e) \(f(x) = \frac{x+1}{2x-6}\) Answer .
\((-\infty, 3) \cup (3, \infty)\)
(f) \(g(x) = \frac{3x}{x^2-9}\) Answer .
\((-\infty, -3) \cup (-3, 3) \cup (3, \infty)\)
(g) \(h(x) = \frac{1}{x^2+4}\)
(h) \(f(x) = \frac{x-1}{x^2+1}\)
(i) \(g(x) = \frac{2x+3}{x^2-x-6}\) Answer .
\((-\infty, -2) \cup (-2, 3) \cup (3, \infty)\)
(j) \(h(x) = \frac{5}{x^2-7x+10}\) Answer .
\((-\infty, 2) \cup (2, 5) \cup (5, \infty)\)
Exercise Group 10.2.2 . Finding Domain II: Radical Functions.
Find the domain of each function. Express your answer in interval notation.
(a) \(f(x) = \sqrt{x}\)
(b) \(g(x) = \sqrt{x+3}\)
(c) \(h(x) = \sqrt{2x-8}\)
(d) \(f(x) = \sqrt{10-x}\)
(e) \(g(x) = \sqrt{4-3x}\) Answer .
\((-\infty, \frac{4}{3}]\)
(f) \(h(x) = \sqrt[3]{x-5}\)
(g) \(f(x) = \sqrt{x^2-16}\) Answer .
\((-\infty, -4] \cup [4, \infty)\)
(h) \(g(x) = \sqrt{9-x^2}\)
(i) \(h(x) = \sqrt{x^2+4}\)
(j) \(f(x) = \sqrt{2x+5}\) Answer .
\([-\frac{5}{2}, \infty)\)
Exercise Group 10.2.3 . Finding Domain III: Mixed Functions.
Find the domain of each function. Express your answer in interval notation.
(a) \(f(x) = \frac{\sqrt{x}}{x-1}\) Answer .
\([0, 1) \cup (1, \infty)\)
(b) \(g(x) = \frac{\sqrt{x+2}}{x+5}\) Answer .
\([-2, -5) \cup (-5, \infty)\)
(c) \(h(x) = \sqrt{\frac{1}{x}}\)
(d) \(f(x) = \frac{1}{\sqrt{x-3}}\)
(e) \(g(x) = \frac{\sqrt{x-1}}{x-4}\) Answer .
\([1, 4) \cup (4, \infty)\)
(f) \(h(x) = \sqrt{x} + \frac{1}{x}\)
(g) \(f(x) = \frac{x}{\sqrt{x+4}}\)
(h) \(g(x) = \sqrt{\frac{x+2}{x-3}}\) Answer .
\((-\infty, -2] \cup (3, \infty)\)
(i) \(h(x) = \frac{2}{\sqrt{5-x}}\)
(j) \(f(x) = \sqrt{x^2-5x+6}\) Answer .
\((-\infty, 2] \cup [3, \infty)\)
(k) \(g(x) = \frac{\sqrt{4-x^2}}{x+1}\) Answer .
\([-2, -1) \cup (-1, 2]\)
(l) \(h(x) = \sqrt{6x-x^2}\)
Subsection 10.2.2 Finding the Range of a Function
Finding the range of a function is generally more challenging than finding the domain. We need to determine all possible output values. There are several strategies:
For simple functions, analyze the behavior algebraically
For transformed functions, use knowledge of transformations
For complex functions, use graphing or calculus techniques
Example 10.2.11 . Range of a Linear Function.
Find the range of
\(f(x) = 2x + 3\text{.}\)
Solution .
Linear functions (with non-zero slope) take on all real number values. As
\(x\) goes from
\(-\infty\) to
\(\infty\text{,}\) the function values also go from
\(-\infty\) to
\(\infty\text{.}\)
Range:
\((-\infty, \infty)\)
Example 10.2.12 . Range of a Quadratic Function (Opens Up).
Find the range of
\(f(x) = x^2\text{.}\)
Solution .
The function
\(f(x) = x^2\) has a minimum value at
\(x = 0\text{,}\) where
\(f(0) = 0\text{.}\)
Since
\(x^2 \geq 0\) for all real
\(x\text{,}\) the function values are never negative.
As
\(x\) gets larger (in either direction),
\(f(x)\) increases without bound.
Example 10.2.13 . Range of a Quadratic Function (Opens Down).
Find the range of
\(g(x) = -\brac{x-2}^2 + 5\text{.}\)
Solution .
This is a quadratic function in vertex form with vertex
\((2, 5)\text{.}\)
Since the coefficient of the squared term is negative (
\(-1\) ), the parabola opens downward.
The maximum value occurs at the vertex:
\(g(2) = 5\text{.}\)
As
\(x\) moves away from 2 (in either direction),
\(g(x)\) decreases without bound.
Example 10.2.14 . Range of a Simple Rational Function.
Find the range of
\(f(x) = \frac{1}{x}\text{.}\)
Solution .
For
\(f(x) = \frac{1}{x}\text{:}\)
When \(x \gt 0\text{,}\) we have \(f(x) \gt 0\)
When \(x \lt 0\text{,}\) we have \(f(x) \lt 0\)
As \(x\) approaches 0, \(\abs{f(x)}\) grows without bound
As \(\abs{x}\) grows large, \(f(x)\) approaches 0 but never equals 0
The function can output any non-zero value, but never 0.
Range:
\((-\infty, 0) \cup (0, \infty)\)
Example 10.2.15 . Range of a Transformed Rational Function.
Find the range of
\(g(x) = 2 + \frac{3}{x-1}\text{.}\)
Solution .
This function is a transformation of
\(\frac{1}{x}\text{:}\)
If
\(\frac{3}{x-1}\) can be any value except 0, then
\(2 + \frac{3}{x-1}\) can be any value except
\(2 + 0 = 2\text{.}\)
Range:
\((-\infty, 2) \cup (2, \infty)\)
Example 10.2.16 . Range of a Square Root Function.
Find the range of
\(f(x) = \sqrt{x}\text{.}\)
Solution .
The square root function produces only non-negative outputs.
The minimum value is
\(f(0) = 0\text{,}\) and as
\(x\) increases,
\(f(x)\) increases without bound.
Example 10.2.17 . Range of a Transformed Radical Function.
Find the range of
\(g(x) = 3 - \sqrt{x+2}\text{.}\)
Solution .
First, note the domain is
\(x \geq -2\) (so the square root is defined).
The expression
\(\sqrt{x+2}\) has range
\([0, \infty)\) over this domain.
Therefore,
\(-\sqrt{x+2}\) has range
\((-\infty, 0]\text{.}\)
Adding 3:
\(3 - \sqrt{x+2}\) has range
\((-\infty, 3]\text{.}\)
Maximum: when \(x = -2\text{,}\) \(g(-2) = 3 - \sqrt{0} = 3\)
As \(x \to \infty\text{,}\) \(g(x) \to -\infty\)
Exercise Group 10.2.4 . Finding Range I.
Find the range of each function. Express your answer in interval notation.
(a) \(f(x) = 5x - 2\)
(b) \(g(x) = -3x + 7\)
(c) \(h(x) = x^2 + 4\)
(d) \(f(x) = \brac{x-1}^2\)
(e) \(g(x) = -x^2 + 3\)
(f) \(h(x) = \brac{x+2}^2 - 5\)
(g) \(f(x) = -2\brac{x-3}^2 + 1\)
(h) \(g(x) = \sqrt{x-4}\)
(i) \(h(x) = \sqrt{x} + 2\)
(j) \(f(x) = -\sqrt{x+1}\)
Exercise Group 10.2.5 . Finding Both Domain and Range.
Find both the domain and range of each function. Express your answers in interval notation.
(a) \(f(x) = x^2 - 6x + 5\) Answer .
Domain:
\((-\infty, \infty)\)
Range:
\([-4, \infty)\) (vertex form:
\(\brac{x-3}^2 - 4\) )
(b) \(g(x) = \frac{2}{x+3}\) Answer .
Domain:
\((-\infty, -3) \cup (-3, \infty)\)
Range:
\((-\infty, 0) \cup (0, \infty)\)
(c) \(h(x) = \sqrt{x-5} + 1\)
(d) \(f(x) = -3\brac{x+1}^2 + 2\) Answer .
Domain:
\((-\infty, \infty)\)
(e) \(g(x) = \frac{1}{x^2+1}\) Answer .
Domain:
\((-\infty, \infty)\)
Range:
\((0, 1]\) (maximum when
\(x=0\) )
(f) \(h(x) = 2 - \sqrt{4-x}\)
(g) \(f(x) = \frac{x+1}{x-2}\) Answer .
Domain:
\((-\infty, 2) \cup (2, \infty)\)
Range:
\((-\infty, 1) \cup (1, \infty)\)
(h) \(g(x) = \abs{x} - 3\) Answer .
Domain:
\((-\infty, \infty)\)
(i) \(h(x) = 3\sqrt{x+2} - 1\)
(j) \(f(x) = \frac{5}{2x-1} + 3\) Answer .
Domain:
\((-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)\)
Range:
\((-\infty, 3) \cup (3, \infty)\)
Subsection 10.2.3 Summary and Strategies
Here are the key strategies for finding domain and range:
Strategies for Finding Domain.
Step 1: Identify the type of function.
Step 2: Check for restrictions:
Step 3: Combine all restrictions using interval notation.
Strategies for Finding Range.
For linear functions: The range is all real numbers (unless it’s a horizontal line).
For quadratic functions: Find the vertex. If the parabola opens up, the range is
\([k, \infty)\) where
\(k\) is the
\(y\) -coordinate of the vertex. If it opens down, the range is
\((-\infty, k]\text{.}\)
For rational functions of the form \(\frac{a}{x-h} + k\text{:}\) The range is all real numbers except
\(k\text{.}\)
For square root functions of the form \(a\sqrt{x-h} + k\text{:}\)
General strategy: Think about the minimum and maximum values the function can achieve, either through algebra, graphing, or calculus techniques.
Quick Reference: Common Functions.
Polynomial
\(x^2 + 3x - 5\) \((-\infty, \infty)\) Depends on degree and leading coefficient
Rational
\(\frac{1}{x-2}\) All reals except where denominator = 0
Usually all reals except horizontal asymptote value
Square Root
\(\sqrt{x-3}\) Where expression under root \(\geq 0\)
\([0, \infty)\) or \((-\infty, 0]\) (depends on sign)
Absolute Value
\(\abs{x}\) \((-\infty, \infty)\)
\([0, \infty)\) for basic form
