Composite Functions: Definition and Operations | 复合函数的定义与运算

📚 Composite Functions: Definition and Operations | 复合函数的定义与运算

In IB Mathematics, a composite function is formed when the output of one function becomes the input of another. Understanding how to define, evaluate, and manipulate composite functions is essential for both Analysis and Approaches and Applications and Interpretation courses.

在 IB 数学中，当一个函数的输出成为另一个函数的输入时，就形成了复合函数。理解如何定义、求值以及运筹复合函数，对于数学分析与方法（AA）以及应用与解释（AI）课程都至关重要。

1. What Is a Composite Function? | 什么是复合函数？

A composite function is written as f∘g (read as “f composed with g”) and defined by (f∘g)(x) = f(g(x)). This means we apply g first, then apply f to the result. The symbol “∘” is the composition operator, and it should not be confused with multiplication.

复合函数记作 f∘g（读作“f 与 g 的复合”），定义为 (f∘g)(x) = f(g(x))。也就是说，我们先对 x 施加 g，再将 g(x) 的结果交给 f。符号“∘”是复合运算符，不能与乘法混淆。

For example, if f(x) = x² and g(x) = x + 1, then (f∘g)(x) = f(g(x)) = (x + 1)². Notice that the order matters: (g∘f)(x) = g(f(x)) = x² + 1, which is a completely different expression.

例如，若 f(x) = x²，g(x) = x + 1，则 (f∘g)(x) = f(g(x)) = (x + 1)²。注意顺序很重要：(g∘f)(x) = g(f(x)) = x² + 1，这是完全不同的表达式。

2. Evaluating Composite Functions: Right to Left | 复合函数求值：从右到左

When evaluating (f∘g)(a) for a specific value a, always work from the inside out. First compute g(a), then substitute that result into f. This is the single most important rule in composite function calculations.

当计算 (f∘g)(a) 时，必须从内向外进行。先计算 g(a)，再将所得结果代入 f。这是复合函数运算中最重要的一条规则。

Worked example: Let f(x) = 2x + 3 and g(x) = x². Then (f∘g)(2) = f(g(2)) = f(4) = 2(4) + 3 = 11. Alternatively, we can form the composite rule first: (f∘g)(x) = 2x² + 3, then substitute x = 2 to obtain 11.

例如：设 f(x) = 2x + 3，g(x) = x²。则 (f∘g)(2) = f(g(2)) = f(4) = 2(4) + 3 = 11。也可以先写出复合规则 (f∘g)(x) = 2x² + 3，再代入 x = 2 得到 11。

x	g(x) = x²	f(g(x)) = 2x² + 3
-1	1	5
0	0	3
3	9	21

3. Domain of Composite Functions | 复合函数的定义域

For (f∘g)(x) to be defined, two conditions must hold. First, x must be in the domain of g, so g(x) exists. Second, g(x) must lie in the domain of f, so that f can act on it. Therefore the domain of f∘g is: D_f∘g = { x ∈ D_g | g(x) ∈ D_f }.

要使 (f∘g)(x) 有意义，必须满足两个条件。第一，x 必须在 g 的定义域内，使得 g(x) 存在；第二，g(x) 必须落在 f 的定义域内，使得 f 可以作用于它。因此 f∘g 的定义域为：D_f∘g = { x ∈ D_g | g(x) ∈ D_f }。

Consider f(x) = √x and g(x) = x – 2. Then (f∘g)(x) = √(x – 2), so the domain is x ≥ 2. In contrast, (g∘f)(x) = √x – 2, whose domain is x ≥ 0. This shows that the order of composition changes not only the rule but also the domain.

设 f(x) = √x，g(x) = x – 2。则 (f∘g)(x) = √(x – 2)，其定义域为 x ≥ 2。相比之下，(g∘f)(x) = √x – 2，定义域为 x ≥ 0。这说明复合顺序不仅改变表达式，也改变定义域。

4. Order Matters: Non-Commutativity | 顺序至关重要：不可交换性

In general, f∘g ≠ g∘f. Composition is not commutative. A simple counterexample: f(x) = x + 1, g(x) = 3x. Then (f∘g)(x) = 3x + 1, while (g∘f)(x) = 3x + 3. These differ by a constant, showing that order fundamentally changes the result.

一般情况下，f∘g ≠ g∘f。复合运算不满足交换律。一个简单的反例：f(x) = x + 1，g(x) = 3x。则 (f∘g)(x) = 3x + 1，而 (g∘f)(x) = 3x + 3。两者相差一个常数，说明顺序从根本上改变了结果。

In real-world terms, “add 1 then double” is not the same as “double then add 1.” If f(x) = x + 1 and g(x) = 2x, then f∘g means “double first, add 1 second,” giving 2x + 1, whereas g∘f means “add 1 first, double second,” giving 2x + 2.

从现实意义看，“先加 1 再乘 2”与“先乘 2 再加 1”并不相同。若 f(x) = x + 1，g(x) = 2x，则 f∘g 表示“先加倍、后加 1”，得到 2x + 1；而 g∘f 表示“先加 1、后加倍”，得到 2x + 2。

5. Worked Examples with Algebraic Functions | 代数函数复合实例

Let f(x) = 2x – 5 and g(x) = x² + 1. Then (f∘g)(x) = 2(x² + 1) – 5 = 2x² – 3. Similarly, (g∘f)(x) = (2x – 5)² + 1 = 4x² – 20x + 26. Notice that expanding the square is necessary because composition substitutes the entire expression.

设 f(x) = 2x – 5，g(x) = x² + 1。则 (f∘g)(x) = 2(x² + 1) – 5 = 2x² – 3。类似地，(g∘f)(x) = (2x – 5)² + 1 = 4x² – 20x + 26。注意展开平方是必要的，因为复合时需要替换整个表达式。

For rational functions, composition can create restrictions. If f(x) = 1/x and g(x) = x + 1, then (f∘g)(x) = 1/(x + 1) with domain x ≠ -1. Meanwhile (g∘f)(x) = 1/x + 1 with domain x ≠ 0. Always state the domain after simplifying.

对于有理函数，复合可能引入额外限制。若 f(x) = 1/x，g(x) = x + 1，则 (f∘g)(x) = 1/(x + 1)，定义域为 x ≠ -1。而 (g∘f)(x) = 1/x + 1，定义域为 x ≠ 0。化简后务必写出定义域。

6. Decomposing a Composite Function | 分解复合函数

In IB exams, you may be asked to “decompose” a function into two simpler functions. For example, h(x) = sin(3x) can be written as h = f∘g, where g(x) = 3x and f(x) = sin(x). The clue is to identify which operation is performed on x first.

在 IB 考试中，有时会让你“分解”一个函数为两个更简单的函数。例如，h(x) = sin(3x) 可以写成 h = f∘g，其中 g(x) = 3x，f(x) = sin(x)。关键线索是判断哪个运算先作用于 x。

More generally, h(x) = (2x + 1)⁵ can be decomposed as g(x) = 2x + 1 and f(x) = x⁵. A common technique is to ask: “What is the inner operation?” Then let that be g(x), and let f(x) represent whatever remains to be applied to the whole result.

更一般地，h(x) = (2x + 1)⁵ 可分解为 g(x) = 2x + 1 与 f(x) = x⁵。常用技巧是问自己：“哪一步是内层运算？”令该内层运算为 g(x)，再令 f(x) 表示对整体结果施加的剩余运算。

7. Composite Functions and Graph Transformations | 复合函数与图像变换

Graph transformations are a special application of composition. If y = f(x) and we define g(x) = x + k, then (f∘g)(x) = f(x + k) shifts the graph of f horizontally by k units. For k > 0, the graph moves left because the input is increased before f is applied.

图像变换是复合运算的特殊应用。若 y = f(x)，令 g(x) = x + k，则 (f∘g)(x) = f(x + k) 将 f 的图像水平平移 k 个单位。当 k > 0 时，图像向左移动，因为输入先被增大再交给 f 处理。

Similarly, if g(x) = kx, then (f∘g)(x) = f(kx) produces a horizontal stretch by factor 1/k. If k is negative, the graph also reflects across the y-axis. Understanding composition helps you remember why f(2x) compresses the graph, while f(x/2) stretches it.

类似地，若 g(x) = kx，则 (f∘g)(x) = f(kx) 产生水平伸缩，伸缩倍数为 1/k。若 k 为负，图像还会关于 y 轴反射。理解复合运算有助于记住为什么 f(2x) 压缩图像，而 f(x/2) 拉伸图像。

8. Inverse Functions and Composition | 反函数与复合

A function f has an inverse f⁻¹ if f is one-to-one. A key property is that f⁻¹ ∘ f (x) = x for all x in the domain of f, and f ∘ f⁻¹ (y) = y for all y in the range of f. The composite of a function with its inverse is the identity function.

函数 f 在单射条件下存在反函数 f⁻¹。重要性质是：对于 f 定义域内所有 x，有 f⁻¹ ∘ f (x) = x；对于 f 值域内所有 y，有 f ∘ f⁻¹ (y) = y。函数与其反函数的复合是恒等函数。

For example, if f(x) = 2x + 3, then f⁻¹(x) = (x – 3)/2. Computing f⁻¹(f(x)) = ((2x + 3) – 3)/2 = x. This is a powerful way to check whether two functions are truly inverses of each other.

例如，若 f(x) = 2x + 3，则 f⁻¹(x) = (x – 3)/2。计算 f⁻¹(f(x)) = ((2x + 3) – 3)/2 = x。这是检验两个函数是否互为反函数的强有力方法。

9. Real-World Applications | 实际应用

Composite functions naturally model multi-step processes. Consider a currency converter: if x is the amount in US dollars, g(x) = 0.85x converts to euros, and f(e) = e – 2 adds a transaction fee. Then (f∘g)(x) = 0.85x – 2 gives the final amount after both steps.

复合函数天然适用于多步骤过程的建模。以货币兑换为例：若 x 为美元金额，g(x) = 0.85x 将其兑换为欧元，f(e) = e – 2 表示扣除手续费。则 (f∘g)(x) = 0.85x – 2 给出最终金额。

In physics, if g(t) represents the position of an object at time t, and f(s) converts position to potential energy, then (f∘g)(t) tracks energy as a function of time. IB problems often use such compositions to connect measurement, calculation, and interpretation.

在物理中，若 g(t) 表示物体在时间 t 的位置，f(s) 将位置转化为势能，则 (f∘g)(t) 描述能量随时间的变化。IB 题目常利用这种复合结构将测量、计算与解释联系起来。

10. Common Mistakes and Exam Tips | 常见错误与应试技巧

Do not read (f∘g)(x) as multiplication: it means f(g(x)), never f(x) × g(x). | 不要把 (f∘g)(x) 理解为乘法：它表示 f(g(x))，绝非 f(x) × g(x)。
Always evaluate from the inside out. Skipping this step often leads to reversed compositions. | 务必从内向外求值。跳过这一步常导致复合顺序颠倒。
Never simplify the expression without keeping track of its domain. Squaring or taking roots may hide restrictions. | 化简表达式的过程中不要忽略定义域。平方或开方可能隐藏限制条件。
Check whether f and g are one-to-one before attempting inverse composition. | 在尝试反函数复合前，先确认 f 与 g 是否为单射。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Composite Functions: Definition and Operations | 复合函数的定义与运算