Is $\{(1, 4), (2, 6), (3, 4), (2, 8)\}$ a function? State its domain and range.

Not a function; domain $\{1, 2, 3\}$, range $\{4, 6, 8\}$.

For $f(x) = 2x^2 - 5$, find $f(3)$, $f(-1)$ and $f(a + 1)$.

$f(3) = 13$, $f(-1) = -3$ and $f(a + 1) = 2a^2 + 4a - 3$.

State the domain and range of $f(x) = \sqrt{x + 2} - 3$.

Domain $\{x \in \mathbb{R} \mid x \ge -2\}$, range $\{y \in \mathbb{R} \mid y \ge -3\}$.

State the domain and range of $g(x) = -2(x - 1)^2 + 5$, then solve $g(x) = -3$.

Domain $\{x \in \mathbb{R}\}$, range $\{y \in \mathbb{R} \mid y \le 5\}$; $g(x) = -3$ at $x = -1$ and $x = 3$.

For $h(x) = \frac{3}{x - 4} + 1$, state the domain and range, find $h(5)$, and solve $h(x) = -2$.

Domain $\{x \in \mathbb{R} \mid x \ne 4\}$, range $\{y \in \mathbb{R} \mid y \ne 1\}$; $h(5) = 4$; $x = 3$.

Grade 11 · MCR3U · Functions

MCR3U: Domain, Range and Function Notation

A function sends each input to exactly one output. Function notation names the rule, $f(x)$ , and the domain and range say which inputs it accepts and which outputs it can produce. Almost every question in this topic asks for one of those three things.

How It Works

Function or relation. A relation is any set of ordered pairs. It is a function when no input appears with two different outputs. On a graph, use the vertical line test: if some vertical line hits the graph twice, it is not a function.

Function notation is substitution. $f(3)$ means replace every $x$ in the rule with $3$ . For $f(x) = 2x^2 - 5$ , $f(3) = 2(9) - 5 = 13$ . The input can be an expression too: $f(a + 1) = 2(a + 1)^2 - 5$ . Keep the brackets, or the square lands on the wrong term.

Solving runs the other way. $f(x) = 13$ asks which inputs give an output of $13$ . Set the rule equal to $13$ and solve. A quadratic rule can give two answers, and both count.

Domain from an equation. Start with all real numbers and remove whatever breaks the rule. A denominator cannot be $0$ , and the expression under a square root cannot be negative. So $\frac{1}{x - 4}$ excludes $x = 4$ , and $\sqrt{x + 2}$ needs $x \ge -2$ .

Range from an equation. Ask which outputs the rule can reach. A square root is never negative. A parabola stops at its vertex: $y = (x - 1)^2 + 3$ has range $y \ge 3$ , and a negative $a$ flips the inequality. $\frac{1}{x - 4}$ is never $0$ , so its range leaves out $0$ .

Domain and range from a graph. Read the domain left to right along the $x$ -axis and the range bottom to top along the $y$ -axis. A closed dot includes the endpoint, an open dot leaves it out, and an arrow means the graph keeps going.

Set notation. Tests expect answers like $\{x \in \mathbb{R} \mid x \ge -2\}$ . Check every boundary before you write it: is the endpoint in or out?

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Is $\{(1, 4), (2, 6), (3, 4), (2, 8)\}$ a function? State its domain and range.
Show solution
1. The input $2$ goes to both $6$ and $8$ , so the relation is not a function. Two inputs sharing the output $4$ is fine.
2. Domain: the first coordinates, $\{1, 2, 3\}$ . Range: the second coordinates, $\{4, 6, 8\}$ .
Answer: Not a function; domain $\{1, 2, 3\}$ , range $\{4, 6, 8\}$ .
Easy
For $f(x) = 2x^2 - 5$ , find $f(3)$ , $f(-1)$ and $f(a + 1)$ .
Show solution
1. $f(3) = 2(3)^2 - 5 = 18 - 5 = 13$ .
2. $f(-1) = 2(-1)^2 - 5 = 2 - 5 = -3$ .
3. $f(a + 1) = 2(a + 1)^2 - 5 = 2(a^2 + 2a + 1) - 5 = 2a^2 + 4a - 3$ .
Answer: $f(3) = 13$ , $f(-1) = -3$ and $f(a + 1) = 2a^2 + 4a - 3$ .
Medium
State the domain and range of $f(x) = \sqrt{x + 2} - 3$ .
Show solution
1. The expression under the root cannot be negative: $x + 2 \ge 0$ , so $x \ge -2$ .
2. $\sqrt{x + 2} \ge 0$ , so $f(x) \ge -3$ . The value $-3$ is reached at $x = -2$ , so it is included.
Answer: Domain $\{x \in \mathbb{R} \mid x \ge -2\}$ , range $\{y \in \mathbb{R} \mid y \ge -3\}$ .
Medium
State the domain and range of $g(x) = -2(x - 1)^2 + 5$ , then solve $g(x) = -3$ .
Show solution
1. A polynomial accepts every real $x$ , so the domain is all real numbers.
2. $a = -2$ , so the parabola opens down and its vertex $(1, 5)$ is a maximum. The range is $y \le 5$ .
3. $-2(x - 1)^2 + 5 = -3$ gives $(x - 1)^2 = 4$ , so $x - 1 = \pm 2$ .
Answer: Domain $\{x \in \mathbb{R}\}$ , range $\{y \in \mathbb{R} \mid y \le 5\}$ ; $g(x) = -3$ at $x = -1$ and $x = 3$ .
Test level
For $h(x) = \frac{3}{x - 4} + 1$ , state the domain and range, find $h(5)$ , and solve $h(x) = -2$ .
Show solution
1. The denominator is $0$ at $x = 4$ , so the domain leaves out $4$ .
2. $\frac{3}{x - 4}$ is never $0$ , so $h(x)$ is never $1$ . Every other output is reached, so the range leaves out only $1$ .
3. $h(5) = \frac{3}{1} + 1 = 4$ .
4. $\frac{3}{x - 4} + 1 = -2$ gives $\frac{3}{x - 4} = -3$ , so $x - 4 = -1$ and $x = 3$ .
Answer: Domain $\{x \in \mathbb{R} \mid x \ne 4\}$ , range $\{y \in \mathbb{R} \mid y \ne 1\}$ ; $h(5) = 4$ ; $x = 3$ .

Learn It on Martian Lab

Martian Lab teaches this in the Function Notation, Domain & Range lesson of its MCR3U Functions & Transformations unit. Watch it, then take the Functions & Transformations practice tests to see if it stuck.

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More Functions & Transformations topics: Inverse Functions · Transformations of Functions

More MCR3U topics: Sine Law and the Ambiguous Case · Simplifying Rational Expressions · Operations with Rational Expressions · Max/Min and Vertex Form · Linear-Quadratic Systems · Quadratic Word Problems · Simplifying Radicals · Exponent Laws and Rational Exponents · Graphing Exponential Functions · Exponential Growth and Decay · Special Angles and the CAST Rule · Cosine Law and 3D Problems · Transformations of Sine and Cosine · Sinusoidal Word Problems · Arithmetic Sequences and Series · Geometric Sequences and Series · Pascal's Triangle and the Binomial Theorem · Simple and Compound Interest · Annuities: Future and Present Value

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