Lesson 1-3: Domain and Range

Domain vs. Range

Domain: the set of all possible inputs (x-values) for a function
Range: the set of all possible outputs (y-values / f(x)-values) that the function can produce

Domain \to [Function] \to Range (inputs) (outputs)

For most functions, the domain is "all real numbers." But two situations create restrictions:

1. Division by Zero

Any value of x that makes the denominator = 0 must be excluded.

f(x) = 1/(x - 3) x \neq 3

2. Even Roots of Negatives

You can't take a square root (or 4th root, etc.) of a negative number.

f(x) = \sqrt(x - 4) x \geq 4

Find the domain of $f(x) = 5 / (x² - 9)$ .

1. Set the denominator \neq 0: 2. x² - 9 \neq 0 3. x² \neq 9 4. x \neq 3 and x \neq -3

Domain: all real numbers except 3 and −3.

In interval notation: $(-\infty, -3) \cup (-3, 3) \cup (3, +\infty)$

Find the domain of $g(x) = \sqrt(2x - 6)$ .

1. The expression inside \sqrt must be \geq 0: 2. 2x - 6 \geq 0 3. 2x \geq 6 4. x \geq 3

Domain: $x \geq 3$ , or in interval notation: $[3, +\infty)$

The range can be trickier to find analytically. For simpler functions, think about what outputs are possible:

Linear functions like $f(x) = 2x + 1$ : domain and range are both all real numbers.

Quadratic functions like $f(x) = x²$ : since squaring always gives a non-negative result, the range is $f(x) \geq 0$ , or $[0, +\infty)$ .

Square root functions like $f(x) = \sqrtx$ : the output is always ≥ 0, so range is $[0, +\infty)$ .

For complex functions, graphing is often the best way to see the range visually.

Find the range of $f(x) = (x - 2)² + 3$ .

1. (x - 2)² is always \geq 0 (squaring anything is non-negative) 2. So the minimum value of (x - 2)² is 0, when x = 2 3. Therefore the minimum of f(x) = 0 + 3 = 3 4. f(x) can grow arbitrarily large as x moves away from 2

Range: $f(x) \geq 3$ , or $[3, +\infty)$

Confusing domain (inputs) and range (outputs)
Forgetting that $\sqrt(x)$ requires the inside to be ≥ 0 (not just > 0 — 0 is fine!)
Only looking for one restriction when there might be multiple (e.g., both a square root AND a denominator)
Domain = restrictions on x. Range = what f(x) can produce.
The inside of an even root must be ≥ 0. The denominator must be ≠ 0.

Q1. Find the domain of $f(x) = 1/(x + 7)$ .

Show Answer

Set denominator ≠ 0: x + 7 ≠ 0, so x ≠ −7. Domain: all reals except −7.

Q2. Find the domain of $g(x) = \sqrt(x + 5)$ .

Show Answer

x + 5 ≥ 0, so x ≥ −5. Domain: [−5, +∞).

Q3. State the range of $f(x) = x² + 1$ .

Show Answer

x² ≥ 0 always, so f(x) ≥ 0 + 1 = 1. Range: f(x) ≥ 1, or [1, +∞).