Domain Of A Function Practice Problems

Finding the domain of a function is one of the foundational skills in algebra and precalculus. It requires identifying all possible input values (usually x) for which a function produces a real, defined output. While the concept is straightforward—avoid division by zero and avoid even roots of negative numbers—the application across different function types can be tricky.

This guide provides a structured set of domain of a function practice problems, ranging from basic polynomials to complex piecewise and composite functions. Each section includes the problem, a step-by-step solution, and the reasoning behind the restrictions Nothing fancy..

Understanding the Core Restrictions

Before diving into the practice problems, recall the two primary "deal-breakers" for real-valued functions:

1. Denominators cannot be zero. If the function is a rational expression (a fraction), set the denominator equal to zero and solve for x. Those values are excluded.
2. Radicands of even roots must be non-negative. For square roots, fourth roots, etc., the expression inside the radical (the radicand) must be greater than or equal to zero ($\ge 0$).
3. Arguments of logarithms must be positive. For $\log(f(x))$ or $\ln(f(x))$, the inside expression must be strictly greater than zero (${content}gt; 0$).

Polynomials, exponential functions, sine, and cosine generally have a domain of all real numbers $(-\infty, \infty)$ unless restricted by context.

Level 1: Rational Functions (Division by Zero)

These problems focus solely on denominator restrictions.

Problem 1

Find the domain of $f(x) = \frac{2x + 5}{x^2 - 9}$ The details matter here..

Solution: The numerator is a polynomial (defined everywhere). The restriction comes from the denominator.

1. Set denominator $\neq 0$: $x^2 - 9 \neq 0$.
2. Factor: $(x - 3)(x + 3) \neq 0$.
3. Solve for excluded values: $x \neq 3$ and $x \neq -3$.

Domain in Interval Notation: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$. Domain in Set Builder Notation: ${x \in \mathbb{R} \mid x \neq -3, x \neq 3}$ Most people skip this — try not to..

Problem 2

Find the domain of $g(x) = \frac{x - 4}{x^2 + 4}$.

Solution:

1. Set denominator $\neq 0$: $x^2 + 4 \neq 0$.
2. Solve: $x^2 \neq -4$.
3. Since $x^2$ is always non-negative for real numbers, it can never equal $-4$. There are no restrictions.

Domain: $(-\infty, \infty)$ or $\mathbb{R}$. Key Takeaway: Always check if the denominator actually has real roots.

Level 2: Radical Functions (Even Roots)

These problems require solving inequalities to ensure the radicand is non-negative.

Problem 3

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

Solution:

1. Radicand $\ge 0$: $5 - 2x \ge 0$.
2. Subtract 5: $-2x \ge -5$.
3. Divide by -2 (flip inequality sign): $x \le \frac{5}{2}$.

Domain: $(-\infty, \frac{5}{2}]$ or ${x \mid x \le 2.5}$.

Problem 4

Find the domain of $k(x) = \sqrt{x^2 - 4x - 5}$.

Solution:

1. Radicand $\ge 0$: $x^2 - 4x - 5 \ge 0$.
2. Factor quadratic: $(x - 5)(x + 1) \ge 0$.
3. Find critical points (zeros): $x = 5, x = -1$.
4. Test intervals on a number line:
   • $x < -1$: Test $x = -2 \rightarrow (-)(-) = +$ (True).
   • $-1 < x < 5$: Test $x = 0 \rightarrow (-)(+) = -$ (False).
   • $x > 5$: Test $x = 6 \rightarrow (+)(+) = +$ (True).
5. Include endpoints because of "or equal to".

Domain: $(-\infty, -1] \cup [5, \infty)$ That's the whole idea..

Problem 5 (Cube Root Exception)

Find the domain of $m(x) = \sqrt[3]{x - 7}$.

Solution: Cube roots (odd roots) are defined for all real numbers. Negative inputs yield negative outputs. Domain: $(-\infty, \infty)$. Remember: Only even roots (square, 4th, 6th) restrict the domain.

Level 3: Combined Rational and Radical Functions

These require satisfying both restrictions simultaneously (intersection of domains).

Problem 6

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

Solution: We have two conditions:

1. Radical (Numerator): $x + 3 \ge 0 \implies x \ge -3$.
2. Denominator: $x - 1 \neq 0 \implies x \neq 1$.

Intersection: $x$ must be $\ge -3$ AND $x \neq 1$.

Domain: $[-3, 1) \cup (1, \infty)$. Note the bracket at -3 (included) and parenthesis at 1 (excluded).

Problem 7

Find the domain of $q(x) = \frac{1}{\sqrt{x - 4}}$.

Solution: This is a rational function with a radical in the denominator That's the part that actually makes a difference..

1. Radicand must be positive (strictly ${content}gt; 0$) because it is in the denominator. If radicand $= 0$, denominator $= 0$. $x - 4 > 0 \implies x > 4$.
2. No separate denominator check needed; it's covered by the strict inequality.

Domain: $(4, \infty)$. Crucial Distinction: $\sqrt{u}$ in numerator $\rightarrow u \ge 0$. $\sqrt{u}$ in denominator $\rightarrow u > 0$.

Problem 8

Find the domain of $r(x) = \sqrt{\frac{x + 2}{x - 5}}$ The details matter here..

Solution: The entire fraction must be $\ge 0$. We solve the rational inequality $\frac{x + 2}{x - 5} \ge 0$ Not complicated — just consistent. Turns out it matters..

1. Critical points (zeros of num/den): $x = -2$ (num zero), $x = 5$ (den zero, excluded).
2. Sign chart intervals: $(-\infty, -2], [-2, 5), (5, \infty)$.
3. Test signs:
   • $x < -2$: $(-)/(-) = +$ (Valid).
   • $-2 < x < 5$: $(+)/(-) = -$ (Invalid).
   • $x > 5$: $(+)/(+) = +$ (Valid).
4. Check endpoints: $x = -2$ makes fraction 0 (Valid). $x = 5$ undefined (Invalid).

Domain: $(-\infty, -2] \cup (5, \infty)$ The details matter here..

Level 4: Logarithmic Functions

Logarithms require strictly positive arguments Turns out it matters..

Problem 9

Find the domain of $f(x) = \ln(2x - 6)$.

Solution:

1. Argument ${content}gt;