Domain and Range (Algebraic Methods)

Finding the domain and range of a function is one of the most important skills in college algebra. In Algebra 2, you may have found domains by inspection. Here we develop systematic algebraic methods that work for every function type — rational, radical, logarithmic, and combinations.

Interval Notation Review

Before we begin, let us review interval notation — the standard way to express domains and ranges.

We use the union symbol $\cup$ to join disconnected intervals. For example, $(-\infty, 2) \cup (2, \infty)$ means all real numbers except $2$.

Set-Builder Notation

As an alternative, set-builder notation writes the domain as $\{x \mid x \neq 2\}$ (read “the set of all $x$ such that $x$ is not equal to 2”). Both notations appear in college algebra courses — you should be fluent in both.

Rule 1: Denominators Cannot Be Zero

If a function has a variable in a denominator, the denominator must not equal zero. To find the domain, set the denominator equal to zero and exclude those solutions.

Worked Example 1: Simple Rational Function

Find the domain of $f(x) = \frac{3x + 1}{x - 4}$.

Step 1: Set the denominator equal to zero: $x - 4 = 0$, so $x = 4$.

Step 2: Exclude $x = 4$ from all real numbers.

Domain: $(-\infty, 4) \cup (4, \infty)$, or equivalently $\{x \mid x \neq 4\}$.

Worked Example 2: Quadratic Denominator

Find the domain of $g(x) = \frac{x}{x^2 - 9}$.

Step 1: Factor and solve: $x^2 - 9 = (x - 3)(x + 3) = 0$, so $x = 3$ or $x = -3$.

Step 2: Exclude both values.

Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$.

Rule 2: Even-Index Radicands Must Be Non-Negative

For $\sqrt[n]{f(x)}$ where $n$ is even (square root, fourth root, etc.), the radicand $f(x)$ must satisfy $f(x) \geq 0$.

Odd-index radicals (cube root, fifth root) accept any real input — they have no domain restriction.

Worked Example 3: Square Root Function

Find the domain of $h(x) = \sqrt{5 - 2x}$.

Step 1: Set the radicand $\geq 0$:

$5 - 2x \geq 0$

Step 2: Solve:

$-2x \geq -5 \implies x \leq \frac{5}{2}$

Domain: $\left(-\infty, \frac{5}{2}\right]$.

Worked Example 4: Fourth Root

Find the domain of $p(x) = \sqrt[4]{x^2 - 1}$.

Set $x^2 - 1 \geq 0$:

$(x - 1)(x + 1) \geq 0$

Using a sign chart: the expression is non-negative when $x \leq -1$ or $x \geq 1$.

Domain: $(-\infty, -1] \cup [1, \infty)$.

Rule 3: Logarithm Arguments Must Be Positive

For $\log_b(f(x))$, the argument $f(x)$ must be strictly positive: $f(x) > 0$.

Worked Example 5: Logarithmic Domain

Find the domain of $k(x) = \ln(3x - 6)$.

Set the argument positive: $3x - 6 > 0$, so $x > 2$.

Domain: $(2, \infty)$.

Worked Example 6: Logarithm with Quadratic Argument

Find the domain of $m(x) = \log(x^2 - 4x + 3)$.

Factor: $x^2 - 4x + 3 = (x - 1)(x - 3) > 0$.

Sign chart: positive when $x < 1$ or $x > 3$.

Domain: $(-\infty, 1) \cup (3, \infty)$.

Combining Multiple Restrictions

When a function involves more than one restriction, the domain is the intersection of all individual restrictions.

Worked Example 7: Combined Restrictions

Find the domain of $f(x) = \frac{\sqrt{x + 3}}{x - 1}$.

Restriction 1 (square root): $x + 3 \geq 0 \implies x \geq -3$, giving $[-3, \infty)$.

Restriction 2 (denominator): $x - 1 \neq 0 \implies x \neq 1$.

Combined: $[-3, 1) \cup (1, \infty)$.

Worked Example 8: Three Restrictions at Once

Find the domain of $g(x) = \frac{\ln(x)}{\sqrt{4 - x}}$.

Restriction 1 (logarithm): $x > 0$, giving $(0, \infty)$.

Restriction 2 (square root): $4 - x \geq 0 \implies x \leq 4$.

Restriction 3 (denominator): $\sqrt{4 - x} \neq 0 \implies 4 - x \neq 0 \implies x \neq 4$.

Combining: $x > 0$ and $x \leq 4$ and $x \neq 4$, so $x \in (0, 4)$.

Domain: $(0, 4)$.

Finding the Range Algebraically

Finding the range is generally harder than finding the domain. The main strategy is to solve for $x$ in terms of $y$, then determine which $y$-values produce valid $x$-values.

Worked Example 9: Range of a Rational Function

Find the range of $f(x) = \frac{2x + 1}{x - 3}$.

Step 1: Set $y = \frac{2x + 1}{x - 3}$ and solve for $x$:

$y(x - 3) = 2x + 1$ $yx - 3y = 2x + 1$ $yx - 2x = 3y + 1$ $x(y - 2) = 3y + 1$ $x = \frac{3y + 1}{y - 2}$

Step 2: This expression for $x$ is defined for all $y \neq 2$. Since every such $y$ gives a valid $x$ in the domain, the range is all real numbers except $2$.

Range: $(-\infty, 2) \cup (2, \infty)$.

Worked Example 10: Range of a Square Root Function

Find the range of $h(x) = \sqrt{5 - 2x}$.

The square root function outputs values $\geq 0$. As $x$ decreases toward $-\infty$, $5 - 2x$ grows without bound, so the square root grows without bound. At the right endpoint of the domain ($x = 5/2$), the output is $\sqrt{0} = 0$.

Range: $[0, \infty)$.

Real-World Application: Structural Engineering

An engineer models the deflection of a beam as $d(x) = \frac{x\sqrt{L - x}}{EI}$, where $x$ is the position along the beam (in meters), $L$ is the beam length, and $EI$ is a stiffness constant. The domain restrictions are:

• $x$ must be non-negative (physical position): $x \geq 0$
• The radicand must be non-negative: $L - x \geq 0 \implies x \leq L$

Domain: $[0, L]$.

For a 10-meter beam with $EI = 200$, the deflection at $x = 6$ is:

$d(6) = \frac{6\sqrt{10 - 6}}{200} = \frac{6(2)}{200} = \frac{12}{200} = 0.06 \text{ m}$

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• For denominators, set them equal to zero and exclude those $x$-values
• For even-index radicals, set the radicand $\geq 0$ and solve the inequality
• For logarithms, set the argument strictly $> 0$
• When a function has multiple features, take the intersection of all restrictions
• Finding the range algebraically means solving for $x$ in terms of $y$ and determining which $y$-values yield valid results
• Interval notation and set-builder notation are interchangeable — master both
• Always present your domain with proper notation: interval notation with $\cup$ for unions

Return to College Algebra for more topics in this section.