Functions, Domain and Range: PECTAA Class 11 Maths, Punjab Board

A square of side 3 cm has one area, 9 cm². It can never have two. That "one input, one output" idea is the whole definition of a function. Once it is clear, domain, range and the vertical line test are just ways of asking the same question.

Unit 2 of the PECTAA Mathematics 11 book is Functions and Graphs. Section 2.1 builds the concept of a function: definition, notation, the vertical line test and the main types. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Where the word comes from. The book credits Leibniz with using "function" for the way one quantity depends on another. Its examples: the area of a square depends on its side (A = x²), and the volume of a sphere depends on its radius.

Definition (section 2.1.1). A function f from a set X to a set Y is a rule that assigns to each element x in X a unique element y in Y.

• Domain: the set X of inputs.
• Co-domain: the set Y in which outputs lie.
• Range: the set of outputs actually produced. The range is always a subset of the co-domain.

Unless stated otherwise, X and Y are sets of real numbers.

Notation (section 2.1.2). The book credits Euler with the symbol f(x). Think of f as a machine: put in x, get out exactly one value f(x), called the value of f at x or the image of x. Writing y = f(x), x is the independent variable and y the dependent variable.

Vertical line test (section 2.1.3). A graph is a function if and only if no vertical line meets it more than once. If some vertical line cuts it twice, one x has two outputs, so it is not a function.

Types of function (section 2.1.4).

• One-to-one (injective): different inputs give different outputs. Example: f(x) = 5x + 7, since 5x₁ + 7 = 5x₂ + 7 forces x₁ = x₂.
• Onto (surjective): every element of the co-domain has at least one pre-image. Example: f(x) = 2x + 3 on the real numbers, since any y comes from x = (y − 3)/2.
• Bijective: both one-to-one and onto.
• Piecewise: defined by different expressions on different intervals of the domain.

Key points

• Finding a domain: remove values that make a denominator zero, and values that make an expression under a square root negative.
• Finding a range: ask which outputs actually occur. Solving y = f(x) for x often helps.
• Range ⊆ co-domain. When they are equal, the function is onto.

Worked examples

Example 1 (book Example 1). f(x) = x³ − 2x² + 4x − 1. Find f(0), f(1), f(−2) and f(1 + x).

1. f(0) = −1.
2. f(1) = 1 − 2 + 4 − 1 = 2.
3. f(−2) = −8 − 8 − 8 − 1 = −25.
4. f(1 + x) = (1 + 3x + 3x² + x³) − 2(1 + 2x + x²) + 4(1 + x) − 1 = x³ + x² + 3x + 2.

Example 2 (book Example 2). Domain and range of f(x) = x².

1. Every real x can be squared, so the domain is all real numbers.
2. A square is never negative, so the range is all non-negative real numbers, [0, ∞[.

Example 3 (book Example 3). Domain of f(x) = x / (x² − 4).

1. The denominator is zero at x = 2 and x = −2.
2. Domain: all real numbers except −2 and 2, written ℝ − {−2, 2}.
3. The book solves y = x/(x² − 4) for x and shows every real y is reached, so the range is all real numbers.

Example 4 (book Example 4). Domain and range of f(x) = √(x² − 9).

1. Need x² − 9 ≥ 0. The critical points are x = ±3.
2. Test x = −4 (true), x = 0 (false), x = 4 (true).
3. Domain: ]−∞, −3] ∪ [3, ∞[.
4. The smallest value of x² − 9 here is 0, and it grows without limit, so the range is [0, ∞[.

Example 5 (book Example 6). Show that f(x) = x² − 2 on the real numbers is neither one-to-one nor onto.

1. f(2) = 2 and f(−2) = 2, so two inputs share an output. Not one-to-one.
2. The least value is −2 (at x = 0). A value such as −3 is never reached. Not onto.

Common mistakes in the exam

1. Treating co-domain and range as the same. The range can be smaller.
2. Forgetting the square root rule. Inside a real square root, the expression must be zero or positive.
3. Writing the domain of x/(x² − 4) as all reals. Remove x = ±2.
4. Using a horizontal line for the function test. The vertical line test decides if a graph is a function.
5. Expanding (1 + x)³ wrongly. It is 1 + 3x + 3x² + x³.
6. Proving onto with one example. You must show every y in the co-domain is reached.

Exam technique

• Write domain and range in interval form, and use ∪ for two pieces.
• For square-root domains, list the critical points and test one value in each region, as the book does.
• For one-to-one proofs, start from f(x₁) = f(x₂) and reach x₁ = x₂.
• Check your board's notice for the sections in your paper.

Practice questions

1. Find f(−1) for f(x) = x³ − 2x² + 4x − 1. Answer: −1 − 2 − 4 − 1 = −8.
2. State the domain of f(x) = 1/(x − 5). Answer: All real numbers except 5.
3. State the domain of f(x) = √(x − 4). Answer: x ≥ 4, that is [4, ∞[.
4. Is f(x) = 3x − 1 one-to-one? Answer: Yes. 3x₁ − 1 = 3x₂ − 1 gives x₁ = x₂.
5. A circle is drawn on graph paper. Is it the graph of a function? Answer: No. A vertical line through its middle cuts it twice.

Ready to practise this? Continue in 1st Year Mathematics.