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How to find the domain of a function

The rule: look for a denominator and for an even root

The domain of a function is the set of all $x$ you are allowed to put in. Instead of hunting for what is allowed, look for what is forbidden - in school algebra there are only two common traps. A denominator may not be zero: set it equal to zero, solve, and throw those points out. An even root (square root, fourth root) may not contain a negative number: set the expression inside $\geq 0$ and solve the inequality. If the formula has neither - a polynomial, a linear function - every real number works and the domain is $(- \infty, \infty)$ . The answer is usually written in interval notation: a square bracket where the endpoint is included, a round one where it is not, and $\cup$ to join the pieces.

Problem

Find the domain of each function. Example 1. $f (x) = - 6 x + 6$ Example 2. $g (x) = \frac{x}{x ^{2} - 25}$ Example 3. $f (x) = 5 x - 20$ (answer in interval notation) Example 4. $h (x) = 6 x - 7 x^{2}$ .

Answer

f (x) = - 6 x + 6

: domain

(- \infty, \infty)

g (x) = \frac{x}{x ^{2} - 25}

: domain

(- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)

f (x) = 5 x - 20

: domain

[4, \infty)

. For

h (x) = 6 x - 7 x^{2}

, the domain is

[0, 6/7]

All four domains were recomputed by hand; zero under a square root is allowed.

Step-by-step solution

Example 1: a linear function

f (x) = - 6 x + 6

There is no denominator and no root - nothing can go wrong for any $x$ . Multiplying and adding work for every real number, so

D (f) = (- \infty, \infty)

The same holds for any polynomial: $x^{2} - 7$ , $3 x^{4} + x$ , and so on.

Example 2: a fraction

g (x) = \frac{x}{x ^{2} - 25}

The only danger is division by zero, so find where the denominator vanishes:

x^{2} - 25 = 0

x^{2} = 25

x = 5 or x = - 5

Every other number is fine, so the number line is cut at two points and the domain falls into three pieces:

D (g) = (- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)

Note that the numerator plays no part: $x = 0$ is perfectly allowed, it simply gives $g (0) = 0$ .

Example 3: a square root

f (x) = 5 x - 20

A square root needs its inside to be zero or positive:

5 x - 20 \geq 0

5 x \geq 20

x \geq 4

The endpoint $4$ itself is allowed, because $0 = 0$ is a perfectly good number - so the bracket on the left is square:

D (f) = [4, \infty)

Example 4: a quadratic under a square root

For $h (x) = 6 x - 7 x^{2}$ , the radicand must satisfy $6 x - 7 x^{2} \geq 0$ . Factor it as $x (6 - 7 x) \geq 0$ . The zeros are $0$ and $6/7$ . Between them both factors are nonnegative; outside, their product is negative. Include both endpoints because the radicand can equal zero: $D (h) = [0, 6/7] .$

How to check yourself

Pick a number just outside your answer and substitute it. For Example 3, $x = 3$ gives $- 5$ - not a real number, so $3$ is rightly excluded. Then pick one just inside: $x = 4$ gives $0 = 0$ , which exists, so the endpoint really does belong to the domain.

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