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College Algebra

Function Domain

Lesson

The domain of a function is the set of all inputs xxfor which the function is defined. Most of the time the domain is “all real numbers” — but a few operations break at certain values, and those values must be excluded.

The two main domain restrictions:

  1. Division by zero is undefined. If xx appears in a denominator, exclude any value that makes the denominator zero.
  2. Square roots of negatives aren’t real. Inside a square root, the expression must be ≥0\geq 0. (Touched on briefly here, explored more in later topics.)

For this topic we focus on the first restriction: finding values of xx that must be excluded because they make a denominator zero. The strategy is the same every time:

  1. Identify the denominator.
  2. Set it equal to zero.
  3. Solve — factor first if needed.
  4. Each solution is excluded from the domain.

Worked example 1

f(x)=x+1x−4f(x) = \frac{x + 1}{x - 4}

Set the denominator to zero:

x−4=0  ⟹  x=4x - 4 = 0 \implies x = 4

The domain is all real numbers except x=4x = 4.

Worked example 2

f(x)=1x2−x−12f(x) = \frac{1}{x^2 - x - 12}

Factor the denominator:

x2−x−12=(x−4)(x+3)x^2 - x - 12 = (x - 4)(x + 3)

Set each factor to zero:

x−4=0orx+3=0x - 4 = 0 \quad\text{or}\quad x + 3 = 0

So x=4x = 4 and x=−3x = -3 are excluded.

How to type your answer

List every excluded xx value, separated by commas— order doesn’t matter. One value is fine too. Examples: 4, 3,-2, -1/2,2.

Practice

Work through these. Stuck? Click Get a hint.

Warm-Up

Quick problems to get going.

Problem 1

f(x)=1x−2f(x) = \frac{1}{x - 2}

Problem 2

f(x)=1x+5f(x) = \frac{1}{x + 5}

Problem 3

f(x)=x+1x−4f(x) = \frac{x + 1}{x - 4}

Problem 4

f(x)=3x−7f(x) = \frac{3}{x - 7}

Practice

Standard problems matching the lesson.

Problem 5

f(x)=1(x−3)(x+2)f(x) = \frac{1}{(x - 3)(x + 2)}

Problem 6

f(x)=x+1(x−5)(x−1)f(x) = \frac{x + 1}{(x - 5)(x - 1)}

Problem 7

f(x)=1x2−9f(x) = \frac{1}{x^2 - 9}

Problem 8

f(x)=1x2−16f(x) = \frac{1}{x^2 - 16}

Problem 9

f(x)=1x2−x−12f(x) = \frac{1}{x^2 - x - 12}

Problem 10

f(x)=1x2+5x+6f(x) = \frac{1}{x^2 + 5x + 6}

Problem 11

f(x)=x−2x2−25f(x) = \frac{x - 2}{x^2 - 25}

Problem 12

f(x)=1x2−xf(x) = \frac{1}{x^2 - x}

Problem 13

f(x)=1x2+x−6f(x) = \frac{1}{x^2 + x - 6}

Problem 14

f(x)=x+1x2+4x+3f(x) = \frac{x + 1}{x^2 + 4x + 3}

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 15

f(x)=1x2−2x−15f(x) = \frac{1}{x^2 - 2x - 15}

Problem 16

f(x)=12x2−8f(x) = \frac{1}{2x^2 - 8}

Problem 17

f(x)=1x3−xf(x) = \frac{1}{x^3 - x}

Problem 18

f(x)=x2−4x2−x−6f(x) = \frac{x^2 - 4}{x^2 - x - 6}

Problem 19

f(x)=1x2−7x+10f(x) = \frac{1}{x^2 - 7x + 10}

Problem 20

f(x)=1x2+2x−8f(x) = \frac{1}{x^2 + 2x - 8}

Problem 21

f(x)=1x3−4xf(x) = \frac{1}{x^3 - 4x}

Problem 22

f(x)=12x2−3x−2f(x) = \frac{1}{2x^2 - 3x - 2}