Domain Restrictions

A domain restriction tells us which input values are not allowed in a function.

Remember:

Domain = the set of allowed input values

Usually, the input is represented by:

x

So whenever you are asked to find the domain of a function, the important question is:

“Are there any values of x that would make this function impossible or undefined?”

For Cambridge O Level Mathematics D, the two restrictions you should check first are:

division by zero and square roots of negative numbers.

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Concept

Imagine a function as a machine:

Input x → Function rule → Output y

Normally, you can put many different values of x into the machine.

But sometimes a particular value causes a mathematical problem.

For example:

f(x) = 1 / (x – 3)

If:

x = 3

then:

f(3) = 1 / (3 – 3)

= 1 / 0

But division by zero is undefined.

Therefore:

x = 3 is not allowed.

We write:

x ≠ 3

That is a domain restriction.

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What Does “Restriction” Mean?

A restriction simply means:

a condition placed on the possible values of x

For example:

x ≠ 4

means:

x can be any real number except 4.

Similarly:

x ≥ 2

means:

x can be 2 or any number greater than 2.

And:

x ≤ 7

means:

x can be 7 or any number less than 7.

So a domain restriction tells us exactly which x-values the function can accept.

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Why Do Functions Need Domain Restrictions?

Some mathematical operations are not defined for every real number.

For O Level functions, the two most important situations are:

A denominator cannot equal zero.

The expression inside a square root cannot be negative.

If neither of these problems occurs, the domain is often simply:

all real numbers

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Rule 1: The Denominator Cannot Be Zero

Consider:

f(x) = 1 / (x – 3)

The denominator is:

x – 3

We cannot allow:

x – 3 = 0

So:

x ≠ 3

Therefore:

Domain: all real numbers except 3

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Why Is Division by Zero Not Allowed?

Division asks how many times one number fits into another.

For example:

6 ÷ 2 = 3

because:

2 × 3 = 6

But:

6 ÷ 0

would ask:

“What number multiplied by 0 gives 6?”

There is no such number.

Therefore, division by zero is undefined.

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Example Math 1

Find the domain restriction of:

f(x) = 5 / (x + 2)

The denominator cannot equal zero.

So:

x + 2 ≠ 0

Therefore:

x ≠ -2

Answer:

Domain: x ≠ -2

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Example Math 2

Find the domain restriction of:

g(x) = 7 / (2x – 6)

Set the denominator not equal to zero:

2x – 6 ≠ 0

2x ≠ 6

x ≠ 3

Therefore:

Domain: x ≠ 3

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Example Math 3

Find the domain restriction of:

h(x) = (x + 4) / (x – 5)

Only the denominator causes a restriction.

So:

x – 5 ≠ 0

Therefore:

x ≠ 5

Answer:

Domain: x ≠ 5

The numerator:

x + 4

is allowed to equal zero.

That would simply make the whole fraction equal to zero.

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Important A/A* Point: Numerator Can Be Zero

Consider:

f(x) = (x – 2) / (x + 7)

If:

x = 2

then the numerator becomes:

0

So:

f(2) = 0 / 9

= 0

That is perfectly valid.

The problem happens only if the denominator becomes zero.

Therefore:

x + 7 ≠ 0

x ≠ -7

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Rule 2: Inside a Square Root Must Be Non-Negative

Consider:

f(x) = √(x – 4)

For real-number functions, you cannot take the square root of a negative number.

So the expression inside the square root must satisfy:

x – 4 ≥ 0

Therefore:

x ≥ 4

So:

Domain: x ≥ 4

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Why Do We Use ≥ and Not >?

Because:

√0 = 0

So zero is allowed inside a square root.

Therefore:

inside square root ≥ 0

not:

inside square root > 0

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Example Math 4

Find the domain of:

f(x) = √(x + 3)

Require:

x + 3 ≥ 0

Subtract 3:

x ≥ -3

Therefore:

Domain: x ≥ -3

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Example Math 5

Find the domain of:

g(x) = √(10 – x)

Require:

10 – x ≥ 0

Subtract 10:

-x ≥ -10

Multiply by -1 and reverse the inequality:

x ≤ 10

Therefore:

Domain: x ≤ 10

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Example Math 6

Find the domain of:

h(x) = √(12 – 3x)

Require:

12 – 3x ≥ 0

-3x ≥ -12

Divide by -3.

Remember: dividing an inequality by a negative number reverses the sign.

So:

x ≤ 4

Therefore:

Domain: x ≤ 4

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Why Does the Inequality Sign Reverse?

Suppose:

3 < 5

Multiply both sides by -1:

-3 > -5

The direction changes.

So whenever you multiply or divide an inequality by a negative number, reverse the inequality sign.

For example:

-2x ≥ -8

becomes:

x ≤ 4

This is very important in square-root domain questions.

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Functions with No Restrictions

Consider:

f(x) = 4x + 7

There is:

no denominator involving x

and:

no square root involving x

So any real value of x can be used.

Therefore:

Domain = all real numbers

The same is true for ordinary polynomial functions such as:

f(x) = x² + 3x – 2

g(x) = 2x³ – 5x

h(x) = x⁴ + 7

Their domain is:

all real numbers

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Quick Recognition Table

Function type	What to check	Typical domain
Linear / polynomial	Usually no restriction	All real numbers
Fraction	Denominator cannot be 0	Exclude value(s) making denominator 0
Square root	Inside root must be ≥ 0	Solve an inequality

This is an excellent first check in an exam.

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A Function Can Have More Than One Restriction

Consider:

f(x) = 1 / [(x – 2)(x + 5)]

The denominator cannot equal zero.

So:

(x – 2)(x + 5) ≠ 0

This means:

x ≠ 2

and:

x ≠ -5

Therefore:

Domain: x ≠ 2 and x ≠ -5

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Example with a Quadratic Denominator

Consider:

f(x) = 1 / (x² – 9)

Factorise:

x² – 9

= (x – 3)(x + 3)

The denominator is zero when:

x = 3

or:

x = -3

Therefore:

Domain: x ≠ 3 and x ≠ -3

This is a useful A/A* extension because domain restrictions can require algebra first.

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Domain Restrictions with Square Roots and Fractions

In harder questions, a function may contain both.

For example:

f(x) = √(x – 1) / (x – 4)

Now we need to satisfy both conditions.

For the square root:

x – 1 ≥ 0

so:

x ≥ 1

For the denominator:

x – 4 ≠ 0

so:

x ≠ 4

Therefore the domain is:

x ≥ 1, x ≠ 4

Both restrictions must be obeyed.

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A/A* Example

Find the domain of:

f(x) = √(8 – 2x) / (x + 1)

First check the square root:

8 – 2x ≥ 0

-2x ≥ -8

x ≤ 4

Now check the denominator:

x + 1 ≠ 0

x ≠ -1

Therefore:

Domain: x ≤ 4, x ≠ -1

This means all numbers up to and including 4 are allowed, except -1.

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Domain of Composite Functions

Domain restrictions can also appear in composite functions.

Suppose:

f(x) = 1/x

and:

g(x) = x + 3

Find:

f(g(x))

Since:

f(x) = 1/x

replace x with:

g(x) = x + 3

So:

f(g(x)) = 1 / (x + 3)

The denominator cannot equal zero.

Therefore:

x + 3 ≠ 0

x ≠ -3

So the composite function has the restriction:

x ≠ -3

This is why understanding domain restrictions is important before studying harder composite-function questions.

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Domain Restrictions and Inverse Functions

Domain restrictions are also important when working with inverse functions.

Consider:

f(x) = x²

Over all real numbers:

f(2) = 4

and:

f(-2) = 4

So the function is not one-to-one.

If we restrict the domain to:

x ≥ 0

then the function becomes one-to-one and has inverse:

f⁻¹(x) = √x

This shows that sometimes a domain is restricted not because an operation is undefined, but because we need a function to satisfy a particular property.

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Example Math

Example 7

Find the domain of:

f(x) = 3 / (x – 8)

Denominator:

x – 8 ≠ 0

So:

x ≠ 8

Answer:

x ≠ 8

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Example 8

Find the domain of:

f(x) = √(2x + 6)

Require:

2x + 6 ≥ 0

2x ≥ -6

x ≥ -3

Answer:

x ≥ -3

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Example 9

Find the domain of:

f(x) = √(5 – x)

Require:

5 – x ≥ 0

-x ≥ -5

Therefore:

x ≤ 5

Answer:

x ≤ 5

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Example 10

Find the domain of:

f(x) = 1 / (x² – 16)

Factorise:

x² – 16

= (x – 4)(x + 4)

Therefore:

x ≠ 4

and:

x ≠ -4

Answer:

x ≠ ±4

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Exam Tip

Before doing anything else with a function, look for:

a fraction

and:

a square root

If you see a fraction:

denominator ≠ 0

If you see a square root:

inside ≥ 0

This quick check can immediately reveal the domain restriction.

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A/A* Exam Technique

When asked to find the domain, use this routine:

Step 1: Identify any denominator.

Step 2: Set the denominator ≠ 0.

Step 3: Identify any square root.

Step 4: Set the expression inside ≥ 0.

Step 5: Solve the equation or inequality.

Step 6: Combine all restrictions.

Step 7: Check that your answer actually makes the function valid.

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Common Mistakes

Mistake 1: Setting the denominator equal to zero

To find the forbidden value, you may solve:

x – 3 = 0

which gives:

x = 3

But your final restriction is:

x ≠ 3

because 3 is the value you must exclude.

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Mistake 2: Writing the Square Root Condition as > 0

Incorrect:

x > 4

for:

√(x – 4)

Correct:

x ≥ 4

because:

√0

is allowed.

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Mistake 3: Restricting the Numerator

For:

(x – 2)/(x + 5)

the numerator may equal zero.

Only the denominator causes the restriction.

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Mistake 4: Forgetting to Reverse an Inequality

From:

-2x ≥ -10

the correct result is:

x ≤ 5

not:

x ≥ 5

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Mistake 5: Checking Only One Restriction

For:

√(x – 2)/(x – 5)

you need both:

x ≥ 2

and:

x ≠ 5

Do not stop after checking only the square root.

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Question & Answer

1. What is a domain restriction?

A condition showing which input values are not allowed or which values are permitted.

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2. What does domain mean?

The set of allowed input values.

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3. Why can the denominator not equal zero?

Because division by zero is undefined.

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4. What condition do we use for a square root?

The expression inside the square root must be:

≥ 0

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5. What is the domain of:

f(x) = 1/(x – 4)

x ≠ 4

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6. What is the domain of:

f(x) = √(x – 4)

x ≥ 4

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7. Why is x = 4 allowed in √(x – 4)?

Because:

√0 = 0

which is defined.

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8. What is the domain of a normal linear function?

Usually all real numbers.

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9. Can a function have more than one excluded value?

Yes.

For example:

1/(x² – 9)

has:

x ≠ 3 and x ≠ -3

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10. What two things should I check first?

Denominators and square roots.

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Practice Questions

1

Find the domain of:

f(x) = 4x + 3

2

Find the domain of:

f(x) = 5/(x – 7)

3

Find the domain of:

f(x) = 2/(x + 4)

4

Find the domain of:

f(x) = √(x – 6)

5

Find the domain of:

f(x) = √(x + 5)

6

Find the domain of:

f(x) = √(12 – x)

7

Find the domain of:

f(x) = √(10 – 2x)

8

Find the domain of:

f(x) = 1/(x² – 25)

9

Find the domain of:

f(x) = √(x – 2)/(x – 5)

10

Find the domain of:

f(x) = √(6 – x)/(x + 3)

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Answers

1

There are no restrictions.

Domain = all real numbers

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2

x – 7 ≠ 0

x ≠ 7

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3

x + 4 ≠ 0

x ≠ -4

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4

x – 6 ≥ 0

x ≥ 6

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5

x + 5 ≥ 0

x ≥ -5

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6

12 – x ≥ 0

x ≤ 12

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7

10 – 2x ≥ 0

-2x ≥ -10

x ≤ 5

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8

x² – 25 ≠ 0

(x – 5)(x + 5) ≠ 0

Therefore:

x ≠ 5 and x ≠ -5

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9

Square-root condition:

x – 2 ≥ 0

x ≥ 2

Denominator condition:

x – 5 ≠ 0

x ≠ 5

Therefore:

x ≥ 2, x ≠ 5

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10

Square-root condition:

6 – x ≥ 0

x ≤ 6

Denominator condition:

x + 3 ≠ 0

x ≠ -3

Therefore:

x ≤ 6, x ≠ -3

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Want to Know More?

After mastering domain restrictions, continue with:

• Domain of a Function
• Range of a Function
• Function Notation
• Composite Functions
• Inverse Functions
• One-to-One Functions
• Functions and Graphs
• Domain and Range from Graphs

Domain restrictions are particularly important when functions involve fractions, square roots, composite functions and inverses.

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Final Exam Reminder

When you see a function, ask:

Which values of x would make this function invalid?

For fractions:

denominator ≠ 0

For square roots:

inside the square root ≥ 0

For ordinary polynomial functions:

domain is usually all real numbers

For example:

f(x) = 1/(x – 3)

gives:

x ≠ 3

while:

g(x) = √(x – 3)

gives:

x ≥ 3

And if both appear in the same function, both restrictions must be satisfied.

That simple checking habit will help an A/A* Math D student avoid many of the most common domain mistakes.