Additional Mathematics Synthesis Guide 37: Domains, Ranges, Restrictions and Admissible Solutions Across A-Math

Identify the mathematical object → record its legal inputs → transform carefully → generate candidates → return every candidate to the original conditions.

1. Domain is part of the definition

The domain of a function is the set of inputs for which the function is defined in the stated number system. If a function is written without a domain, its natural real domain is usually inferred from the expression.

For example, the real function

f(x)=1/(x−3)

has natural domain x≠3. The excluded value is not a later inconvenience. It belongs to the function from the start.

2. Range is the set of possible outputs

For f(x)=x² with real x, the range is y≥0. For f(x)=eˣ, the range is y>0. For f(x)=sinx, the range is −1≤y≤1.

Range matters when solving equations because a target outside the range cannot be attained. The equation sinx=2 has no real solution before any inverse trigonometric calculation is attempted.

3. Rational expressions: denominator restrictions

For

g(x)=(x+1)/(x²−4),

factor the denominator:

x²−4=(x−2)(x+2).

Therefore x≠2 and x≠−2.

Those exclusions remain even if later algebra cancels a common factor in a related expression. Cancellation can simplify a formula without retroactively changing where the original expression was defined.

4. Cancellation can leave a hole

Consider

h(x)=(x²−1)/(x−1).

For x≠1, this simplifies to x+1. But the original function is still undefined at x=1.

So h behaves like the line y=x+1 with a missing point at (1,2). The simplified expression and the original function agree on the original domain; they are not identical as domain-specified functions.

5. Square roots: radicand conditions

For real-valued

f(x)=√(x−2),

we require

x−2≥0, so x≥2.

If the square root is in a denominator, the inequality becomes strict because zero is also forbidden. Thus 1/√(x−2) requires x>2.

6. Logarithms: positive arguments

For

f(x)=ln(3x−5),

the argument must satisfy

3x−5>0, so x>5/3.

Zero is not allowed. Negative inputs are not allowed for the real logarithm. These conditions must be checked before and after solving logarithmic equations.

7. Exponential functions reverse the domain-range pattern

For y=eˣ, the domain is all real x and the range is y>0. Its inverse y=lnx therefore has domain x>0 and range all real numbers.

Inverse functions exchange domain and range. This is one reason domain-range control matters before finding an inverse.

8. Composite functions inherit hidden restrictions

Suppose

f(x)=lnx and g(x)=x²−4.

Then

(f∘g)(x)=ln(x²−4).

It is not enough that g is defined. We also need the output of g to lie inside the domain of f:

x²−4>0.

Hence x<−2 or x>2.

9. Composite domain is an input-output compatibility problem

For f(g(x)) to exist, two things must be true:

• x must lie in the domain of g;
• g(x) must lie in the domain of f.

This is a useful way to think about composites: the output contract of the inner function must satisfy the input contract of the outer function.

10. Inverse functions require one-to-one control

The function f(x)=x² on all real x is not one-to-one because f(2)=f(−2). Therefore it has no inverse function on all real numbers.

If its domain is restricted to x≥0, then f is one-to-one and its inverse is

f⁻¹(x)=√x, x≥0.

The inverse domain is the original range, and the inverse range is the original restricted domain.

11. Trigonometric domains

Sine and cosine are defined for every real angle. Tangent is undefined where cosx=0. Secant is also undefined where cosx=0. Cotangent and cosecant are undefined where sinx=0.

For tangent, the excluded angles are

x=π/2+kπ

in radians, or 90°+180°k in degrees.

12. Identity simplification does not restore excluded angles

If an expression simplifies after cancelling sinx or cosx, any angle excluded by the original denominator remains excluded from the original expression.

This is the trigonometric version of the removable-hole idea in rational functions.

13. Squaring creates candidate solutions

Solve

√(x+1)=x−1.

Because the square root is non-negative, the right side must satisfy x−1≥0, so x≥1.

Squaring gives x+1=x²−2x+1, hence x(x−3)=0. Candidates are x=0 and3, but x=0 violates the sign condition and the original equation. Only x=3 is admissible.

14. Multiplying by an expression can introduce forbidden cases

Suppose an equation contains 1/(x−2). Multiplying through by x−2 may be a legal transformation only after recording x≠2.

The multiplied equation may be algebraically meaningful at x=2 even though the original equation was not. The original restriction must survive the transformation.

15. Dividing can destroy solution branches

From

sinx(2cosx−1)=0,

dividing by sinx assumes sinx≠0 and would lose all solutions from the branch sinx=0.

Factorised equations should usually be solved by setting each factor to zero rather than dividing by a factor that may vanish.

16. Range can reject equation targets immediately

For y=3sinx+5, the range is [2,8]. Therefore the equation

3sinx+5=10

has no real solution because 10 lies outside the range.

This avoids unnecessary inverse-trigonometric work and provides an immediate graph-based check.

17. Calculus inherits the original function domain

If f(x)=1/x, then f′(x)=−1/x². Both exclude x=0.

Sometimes a derivative formula can algebraically simplify to something defined at a point where the original function was not. That does not create differentiability of the original function at the missing point.

18. Stationary points must belong to the domain

A solution of f′(x)=0 is only a stationary-point candidate if f is defined there and the point lies in the requested interval.

Likewise, optimisation on a restricted domain requires checking only admissible stationary points and the relevant endpoints.

19. Definite integrals require interval awareness

An antiderivative formula should not be used blindly across a discontinuity of the original integrand. For example, 1/x is not continuous across x=0.

At school level, if a denominator or logarithmic restriction splits an interval, that split must be recognised before interpreting a definite integral or area.

20. Modelling domains are often narrower than algebraic domains

A formula may be mathematically defined for many x-values but physically meaningful only for a smaller set. Time may require t≥0. A length may require x>0. A population model may be meaningful only over the stated observation period.

Mathematical admissibility therefore has two layers:

• the expression must be mathematically defined;
• the value must also satisfy the context or question restrictions.

21. Simultaneous restrictions must be intersected

For

f(x)=ln(x−1)/√(5−x),

we require

• x−1>0 → x>1;
• 5−x>0 because the square root is in a denominator → x<5.

Therefore the domain is

1<x<5.

Restrictions combine by intersection, not by choosing whichever one is more convenient.

22. A complete admissibility checklist

• Denominator non-zero?
• Logarithm argument positive?
• Even-root radicand non-negative, or positive if in a denominator?
• Inverse function based on a one-to-one domain?
• Composite output entering the next function legally?
• Trigonometric denominator defined?
• Candidate created by squaring, multiplying or another non-equivalent transformation?
• Candidate within the stated interval or modelling domain?
• Final value meaningful in context?

23. Common failure patterns

• Finding restrictions only after completing the algebra.
• Forgetting that a cancelled factor leaves the original excluded input excluded.
• Using ≥0 for a logarithm argument instead of >0.
• Forgetting strict positivity when a square root appears in a denominator.
• Finding an inverse without restricting a non-one-to-one function.
• Checking only the inner function’s domain in a composite.
• Accepting every root after squaring.
• Dividing by a factor that may be zero and losing solutions.
• Ignoring range before solving an equation.
• Keeping mathematically valid roots that violate the physical time, length or interval condition.

24. A reliable domain-and-admissibility routine

1. Scan the original expression for denominators, logarithms, roots, inverse functions and trigonometric denominators.
2. Record all restrictions before transforming.
3. For composites, require the inner output to belong to the outer domain.
4. For equations, mark any non-reversible step such as squaring or division by a variable expression.
5. Generate candidate solutions algebraically.
6. Intersect them with every original restriction and stated interval.
7. Substitute surviving candidates into the original equation when practical.
8. Return to context and reject values that are mathematically defined but physically inadmissible.

25. Practice set

1. Find the domain of 1/(x−4).
2. Find the domain of √(2x−3).
3. Find the domain of 1/√(2x−3).
4. Find the domain of ln(5−x).
5. Find the domain of ln(x−1)/√(5−x).
6. State the range of eˣ.
7. State the range of 2sinx+3.
8. Explain why x=1 remains excluded from (x²−1)/(x−1).
9. If f(x)=lnx and g(x)=x²−9, find the domain of f(g(x)).
10. Why must x² be restricted before defining an inverse function?
11. On x≥0, find the inverse of f(x)=x².
12. Where is tanx undefined?
13. Where is secx undefined?
14. Solve √(x+1)=x−1 and state the admissible answer.
15. Why should sinx(2cosx−1)=0 not be solved by immediately dividing by sinx?
16. Can 4sinx+1=7 have a real solution? Explain by range.
17. If f′(a)=0 but f(a) is undefined, is a a stationary point of f?
18. A motion model is stated only for 0≤t≤6. A solution gives t=8. Is it admissible?
19. Find the domain of 1/(x²−5x+6).
20. Give the two layers of admissibility in a modelling problem.

Answers

1. x≠4.
2. x≥3/2.
3. x>3/2.
4. x<5.
5. 1<x<5.
6. y>0.
7. [1,5].
8. The original denominator is zero at x=1 even though the simplified formula x+1 is defined there.
9. x²−9>0, so x<−3 or x>3.
10. Because x² gives the same output for x and −x, so it is not one-to-one on all reals.
11. f⁻¹(x)=√x, x≥0.
12. x=π/2+kπ.
13. The same: cosx=0, so x=π/2+kπ.
14. Domain/sign control gives x≥1; squaring gives x=0 or3; only x=3 is admissible.
15. Because sinx=0 is itself a full solution branch and division would erase it.
16. No. The range of 4sinx+1 is [−3,5], so 7 is impossible.
17. No. A stationary point must be a point on the function.
18. No. It lies outside the modelled time interval.
19. x≠2,3.
20. The expression must be mathematically defined, and the value must satisfy the stated physical/contextual restrictions.

26. What mastery looks like

Mastery means the learner treats restrictions as part of the mathematical object, not as a last-minute filter. They can combine denominator, root and logarithmic conditions, control composite domains and inverse functions, recognise when a transformation creates candidates or destroys branches, and return every final answer to the original domain and context.

The transfer test is to remove all explicit prompts about domain or validity. If the learner still records restrictions before manipulation and can explain why an algebraically obtained value is inadmissible, domain control has become part of mathematical judgement.

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Continue through Batch 10

• Guide 38: Parameter Thresholds and Number of Solutions — Discriminants, Tangency, Range and Intersections
• Guide 39: Proof and Justification — Algebra, Geometry, Trigonometry and Calculus
• Guide 40: Exactness, Approximation, Calculator State, Rounding and Verification
• Additional Mathematics Directory
• BTT Mathematics Hub