Chapter 1

Functions

Domain and range done properly for the first time, composite and inverse functions, and the modulus function — the topic that quietly decides whether the rest of this syllabus makes sense.

หลักสูตรแกนกลาง / สสวท.: จำนวนและพีชคณิต (Number and Algebra)ตรงกับ ม.4 ความสัมพันธ์และฟังก์ชัน (เพิ่มเติม) — 0606 ลงลึกกว่าในเรื่องโดเมน เรนจ์ และฟังก์ชันค่าสัมบูรณ์
Common Core (US): F-IF, F-BF
Cambridge: IGCSE 0606 — Topic 1: Functions

1.1 Domain and range

In 0580 a function was mostly a machine you fed numbers into. Here the SET of numbers you are allowed to feed in, and the set that comes out, become part of the answer — and marks are given for them.

นิยาม

Domain and range

The domain is every input the function is defined for. The range is every output it actually produces. A question that says "state the range" is asking what the graph reaches, not what it could reach if you were allowed different inputs.

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The two things that restrict a domain
- division by zero — $f (x) = \frac{1}{x - 3}$ excludes $x = 3$
- the square root of a negative — $f (x) = x - 5$ needs $x \geq 5$
- and a domain the question simply hands you, which overrides both
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Reading a range off a shape you already know
- $f (x) = x^{2} + 3$ has a minimum of $3$ , so the range is $f (x) \geq 3$
- $f (x) = 5 - (x - 2)^{2}$ has a maximum of $5$ , so the range is $f (x) \leq 5$
- This is completing the square doing work in a topic that does not look like quadratics

ตัวอย่าง

For $f (x) = x^{2} - 6 x + 11$ with domain $x \geq 3$ , state the range.

วิธีทำ

1.Complete the square: $f (x) = (x - 3)^{2} + 2$
2.The domain starts exactly at the vertex $x = 3$
3.From there the curve only rises

ตอบ $f (x) \geq 2$

ข้อควรระวัง

The restricted domain in that example was not decoration. Without it the range is still $f (x) \geq 2$ , but the function has no inverse — two inputs share each output. Restricting the domain is how a question makes an inverse exist, and spotting that is usually the whole point of part (i).

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