Composite Functions: How to Combine and Deconstruct Functions | 复合函数：如何组合与分解函数

📚 Composite Functions: How to Combine and Deconstruct Functions | 复合函数：如何组合与分解函数

Composite functions are a fundamental concept in A-Level Mathematics. They allow us to combine two or more functions into one, and they appear across pure mathematics, mechanics, and statistics. Understanding them is essential for solving problems involving chains of operations.

复合函数是A-Level数学中的一个基本概念。它允许我们将两个或多个函数组合成一个函数，出现在纯数学、力学和统计学中。理解复合函数对于解决涉及连续操作的问题至关重要。

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

A composite function is formed when one function is applied to the result of another function. If f(x) and g(x) are functions, then the composite function f(g(x)) is written as (f ∘ g)(x). We first apply g to x, then apply f to the output of g.

复合函数是当一个函数作用于另一个函数的结果时形成的。若f(x)和g(x)是函数，则复合函数f(g(x))记为(f ∘ g)(x)。我们先对x应用g，然后将f作用于g的输出。

For example, let f(x) = x² and g(x) = x + 1. Then the composite function (f ∘ g)(x) is found by replacing every x in f(x) with g(x):

(f ∘ g)(x) = f(g(x)) = (x + 1)²

It is important to note that (f ∘ g)(x) is usually different from (g ∘ f)(x). In our example, (g ∘ f)(x) = g(x²) = x² + 1, which is not the same as (x + 1)².

注意：(f ∘ g)(x)通常不同于(g ∘ f)(x)。在我们的例子中，(g ∘ f)(x) = g(x²) = x² + 1，这与(x + 1)²不同。

2. Notation and Order of Operations | 记号与运算顺序

The notation f ∘ g is read as “f composed with g” or “f after g”. The operation order is crucial: the function on the right of the ∘ symbol is always applied first.

记号f ∘ g读作“f与g的复合”或“在g之后的f”。运算顺序至关重要：∘符号右侧的函数总是先应用。

Consider the following table to clarify the difference:

Expression	Meaning	Apply order
(f ∘ g)(x)	f after g: f(g(x))	First g, then f
(g ∘ f)(x)	g after f: g(f(x))	First f, then g

When asked to evaluate a composite function at a specific value, you can work step by step. For instance, if f(x) = 2x and g(x) = x − 3, then to find (f ∘ g)(5): first compute g(5) = 5 − 3 = 2, then f(2) = 2 × 2 = 4. So (f ∘ g)(5) = 4.

当被要求计算复合函数在特定值的值时，你可以逐步操作。例如，若f(x) = 2x，g(x) = x − 3，要求(f ∘ g)(5)：先计算g(5) = 5 − 3 = 2，然后f(2) = 2 × 2 = 4。因此(f ∘ g)(5) = 4。

3. Finding Composite Functions | 求复合函数

To find the expression for a composite function, you substitute the inner function into the outer function. The key is to pay attention to brackets and simplification.

要求复合函数的表达式，你需要将内层函数代入外层函数。关键在于注意括号和化简。

Example: Let f(x) = 3x + 2 and g(x) = x² − 1. Find (f ∘ g)(x).

示例：设f(x) = 3x + 2，g(x) = x² − 1。求(f ∘ g)(x)。

Substitute g(x) into f(x):

(f ∘ g)(x) = f(g(x)) = 3(x² − 1) + 2 = 3x² − 3 + 2 = 3x² − 1

Now find (g ∘ f)(x):

现在求(g ∘ f)(x)：

(g ∘ f)(x) = g(f(x)) = (3x + 2)² − 1 = 9x² + 12x + 4 − 1 = 9x² + 12x + 3

Notice how different the two results are. Always double-check which function is applied first.

注意这两个结果是多么不同。始终检查哪个函数先应用。

4. Domain and Range of Composite Functions | 复合函数的定义域与值域

The domain of a composite function (f ∘ g)(x) is the set of all x values that can be input into g and then produce an output that is acceptable for f. In other words, x must be in the domain of g, and g(x) must be in the domain of f.

复合函数(f ∘ g)(x)的定义域是能够输入到g中，并且g的输出能被f接受的所有x值的集合。换句话说，x必须在g的定义域内，并且g(x)必须在f的定义域内。

For example, let f(x) = 1/x and g(x) = x + 1. To form (f ∘ g)(x) = 1/(x + 1), we must exclude x = −1 because the denominator becomes zero. Also, since g(x) is defined for all real numbers, the domain of the composite is all real numbers except x = −1.

例如，设f(x) = 1/x，g(x) = x + 1。形成(f ∘ g)(x) = 1/(x + 1)时，必须排除x = −1，因为分母变为零。另外，由于g(x)对所有实数都有定义，复合函数的定义域是除去x = −1的所有实数。

The range of a composite function is often found by considering the range of the inner function, then the effect of the outer function. A common method is to sketch the graph or use algebraic reasoning. Always state the domain before drawing conclusions about the range.

复合函数的值域通常通过考虑内层函数的值域，再考虑外层函数的效应来找到。一个常用方法是画出图像或使用代数推理。在得出值域结论之前，始终先说明定义域。

Worked domain problem: If f(x) = √x and g(x) = 2x − 5, find the domain of (f ∘ g)(x).

定义域问题示例：若f(x) = √x，g(x) = 2x − 5，求(f ∘ g)(x)的定义域。

We have (f ∘ g)(x) = f(g(x)) = √(2x − 5). The square root is defined only when 2x − 5 ≥ 0, so 2x ≥ 5, hence x ≥ 5/2. Therefore the domain is x ∈ [5/2, ∞).

我们有(f ∘ g)(x) = f(g(x)) = √(2x − 5)。平方根仅在2x − 5 ≥ 0时有定义，因此2x ≥ 5，即x ≥ 5/2。因此定义域为x ∈ [5/2, ∞)。

5. Inverse of a Composite Function | 复合函数的反函数

An important property is that the inverse of a composite function reverses the order of the original functions. If both f and g are one-to-one functions, then:

一个重要的性质是：复合函数的反函数会逆转原函数的顺序。如果f和g都是一一对应函数，则：

(f ∘ g)⁻¹ = g⁻¹ ∘ f⁻¹

This means to undo (f ∘ g), we first undo f, then undo g. Let us verify with an example.

这意味着要撤销(f ∘ g)，我们先撤销f，然后撤销g。让我们用一个例子验证。

Let f(x) = 2x and g(x) = x + 3. Then (f ∘ g)(x) = 2(x + 3) = 2x + 6. Its inverse is found by solving y = 2x + 6 for x: x = (y − 6)/2, so (f ∘ g)⁻¹(x) = (x − 6)/2.

设f(x) = 2x，g(x) = x + 3。则(f ∘ g)(x) = 2(x + 3) = 2x + 6。其反函数通过解y = 2x + 6得到x：x = (y − 6)/2，所以(f ∘ g)⁻¹(x) = (x − 6)/2。

Now compute g⁻¹(x) = x − 3 and f⁻¹(x) = x/2. Then (g⁻¹ ∘ f⁻¹)(x) = g⁻¹(f⁻¹(x)) = g⁻¹(x/2) = x/2 − 3 = (x − 6)/2. This matches the inverse we found directly.

现在计算g⁻¹(x) = x − 3，f⁻¹(x) = x/2。则(g⁻¹ ∘ f⁻¹)(x) = g⁻¹(f⁻¹(x)) = g⁻¹(x/2) = x/2 − 3 = (x − 6)/2。这与我们直接求得的反函数一致。

In exam questions, you may be asked to find the inverse of a composite function without first combining them. In such cases, apply the formula above directly, but always verify that the domain of the inverse matches the range of the original.

在考试题中，你可能会被要求在不先合并函数的情况下求复合函数的反函数。在这种情况下，直接应用上述公式，但始终验证反函数的定义域与原函数的值域相匹配。

6. Common Exam Pitfalls | 常见考试陷阱

Reversing order: Confusing f(g(x)) with g(f(x)). Remember that the function on the right acts first.
Order 颠倒：混淆f(g(x))与g(f(x))。记住右侧的函数先起作用。
Ignoring domain: Forgetting to exclude values that make denominators zero or square roots negative.
忽略定义域：忘记排除使分母为零或使平方根为负的值。
Poor expansion: Failing to expand brackets correctly when simplifying composite expressions.
展开错误：在化简复合表达式时未能正确展开括号。
Not simplifying: Leaving expressions in a form that is not fully simplified, which can lose marks.
不化简：将表达式留在未完全化简的形式，可能会丢分。
Misunderstanding inverse order: Applying the inverse functions in the wrong order.
误解反函数顺序：以错误的顺序应用反函数。

To avoid these pitfalls, always write down the expression before simplifying, and check your work by substituting a simple value such as x = 1.

为了避免这些陷阱，在化简前先写出表达式，并通过代入一个简单值（如x = 1）来检查你的答案。

7. Worked Example | 例题解析

Let us solve a typical exam question step by step.

让我们逐步解答一道典型考试题。

Question: The functions f and g are defined by f(x) = x² − 3x and g(x) = 2 − x, for all real x.

题目：函数f和g定义为f(x) = x² − 3x，g(x) = 2 − x，对所有实数x。

(a) Find (f ∘ g)(x).

(a) 求(f ∘ g)(x)。

(b) Find (g ∘ f)(x).

(b) 求(g ∘ f)(x)。

Solution:

解答：

(a) (f ∘ g)(x) = f(g(x)) = f(2 − x) = (2 − x)² − 3(2 − x)

= (4 − 4x + x²) − 6 + 3x = x² − x − 2

(b) (g ∘ f)(x) = g(f(x)) = g(x² − 3x) = 2 − (x² − 3x) = −x² + 3x + 2

2x² − 4x − 4 = 0 ⟹ x² − 2x − 2 = 0

Using the quadratic formula:

使用二次公式：

x = (2 ± √(4 + 8)) / 2 = (2 ± √12) / 2 = 1 ± √3

So the solutions are x = 1 + √3 and x = 1 − √3.

所以解为x = 1 + √3和x = 1 − √3。

8. Practice Questions | 练习题

Test your understanding with these questions.

用以下问题测试你的理解。

Question 1: Given f(x) = 5x − 4 and g(x) = x², find (f ∘ g)(x) and (g ∘ f)(x).

题目1：已知f(x) = 5x − 4，g(x) = x²，求(f ∘ g)(x)和(g ∘ f)(x)。

Question 2: The functions p(x) = 1/(x − 2) and q(x) = 3x are defined for suitable domains. Find the domain of (p ∘ q)(x).

题目2：函数p(x) = 1/(x − 2)和q(x) = 3x在适当定义域上有定义。求(p ∘ q)(x)的定义域。

Question 3: If f(x) = 3x − 1 and g(x) = x/2, find an expression for (f ∘ g)⁻¹(x).

题目3：若f(x) = 3x − 1，g(x) = x/2，求(f ∘ g)⁻¹(x)的表达式。

Answers: 1) (f ∘ g)(x) = 5x² − 4; (g ∘ f)(x) = (5x − 4)². 2) x ≠ 2/3. 3) (f ∘ g)⁻¹(x) = 2(x + 1)/3.

答案：1) (f ∘ g)(x) = 5x² − 4；(g ∘ f)(x) = (5x − 4)²。2) x ≠ 2/3。3) (f ∘ g)⁻¹(x) = 2(x + 1)/3。

9. Conclusion | 总结

Composite functions are a powerful tool that appears repeatedly throughout the A-Level syllabus. Master the order of operations, be careful with domains, and remember that the inverse of a composite reverses the function order. With consistent practice, these problems become straightforward.

复合函数是A-Level课程中反复出现的强大工具。掌握运算顺序，注意定义域，记住复合函数的反函数会逆转函数顺序。通过持续练习，这类问题将变得简单容易。

Always show your working clearly and check whether your final answer is in its simplest form. Using a step-by-step approach will help you avoid careless errors.

始终清晰展示你的解题步骤，并检查最终答案是否已化为最简形式。采用分步方法将帮助你避免粗心错误。

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Composite Functions: How to Combine and Deconstruct Functions | 复合函数：如何组合与分解函数