Composite functions and intermediate domains
Введено| English | Значение |
|---|---|
| composition/ˌkɒmpəˈzɪʃn/ | составная функция |
An allowed starting input can still produce an invalid intermediate value.
- One machine’s output becomes another machine’s input.
- This lesson studies composition 复合函数: Applying one function to the output of another, in a specified order.
The method
- Here fg means f composed with g: fg(x)=f(g(x)), not multiplication.
- Apply g first.
- A composite input must belong to the inner function’s domain, and its output must belong to the outer function’s domain.
- In general fg and gf have different formulas and different domains.
For f(x)=√(x-2) and g(x)=1/(x-1), why is fg(6) undefined even though g(6) exists?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
A composite input must survive BOTH stages: the inner function’s domain and then its output must lie in the outer function’s domain. Checking only the inner domain misses the second condition.
A worked case
Let f(x)=√(x−2), with x≥2, and g(x)=1/(x−1), with x≠1. Then fg(x)=√(1/(x−1)−2). It requires 1/(x−1)≥2. A negative denominator cannot meet this positive lower bound; for x>1, multiplying by x−1 preserves order and gives x≤3/2. Thus the domain is 1<x≤3/2. In the reverse order, gf(x)=1/(√(x−2)−1). It requires x≥2 and √(x−2)≠1, so x≠3. Its domain is [2,3)∪(3,∞). At x=6, gf(6)=1, while fg(6) is not real.
Composite functions and intermediate domains
Here fg means f composed with g: fg(x)=f(g(x)), not multiplication
Use domains to justify inverse and composite function steps.
Find gf(6) for the worked functions.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
f(6)=√4=2, then g(2)=1/(2−1)=1.
Where it goes wrong
- Checking only the inner domain is insufficient.
- Do not multiply the inequality by x−1 before deciding its sign.
- An excluded intermediate denominator remains excluded even if another form appears simpler.
- A formula and domain together define the composite function.
Not true: A composite is defined wherever its inner function is defined. The inner output must also lie in the outer function's domain.
Find the included upper endpoint of the domain of fg.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
For x>1, 1≥2(x−1) gives x≤3/2=1.5.
The inner output must also lie in the outer function's domain.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. The inner output must also lie in the outer function's domain.
Applying it
- Trace the intermediate value: gf(6) uses f(6)=2 then g(2)=1.
- For fg(5/4), g(5/4)=4 then f(4)=√2.
- At x=3, f(3)=1 is valid, but g(1) is undefined, so gf(3) is undefined.
- Mark every retained or excluded endpoint in interval notation.
Find the excluded input within x≥2 for gf.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The outer denominator is zero when √(x−2)=1, giving x=3.
Match each composition condition to the stage it constrains.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
fg means apply g first; the full domain is the set of inputs passing both tests, here 1<x≤3/2.
In your course
- AQA A-level Mathematics (7357), specification section B.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Composition: Applying one function to the output of another, in a specified order.
- Here fg means f composed with g: fg(x)=f(g(x)), not multiplication.
- Watch for: checking only the inner domain is insufficient.