Finding Functions | 求函数

📚 Finding Functions | 求函数

In A-Level Mathematics, the ability to find unknown functions from given conditions is a fundamental skill. Whether you are decomposing composite functions, determining inverses, or solving functional equations, understanding how to manipulate and derive function expressions is essential for success in the Edexcel specification.

在A-Level数学中，根据给定条件求出未知函数是一项基本技能。无论是分解复合函数、求反函数还是解函数方程，掌握如何操作并推导函数表达式对 Edexcel 考试大纲的成功至关重要。

1. Understanding Functions | 理解函数

A function is a rule that assigns each input value exactly one output value. It can be represented by an equation, a graph, or a mapping diagram.

函数是一种规则，它为每个输入值分配唯一一个输出值。它可以用方程、图形或映射图来表示。

The domain of a function is the set of all possible input values, while the range (or image) is the set of all output values that actually occur.

函数的定义域是所有可能输入值的集合，而值域（或像集）是所有实际出现的输出值的集合。

xample, f(x) = 2x + 1 has domain all real numbers, and its range is also all real numbers.

例如，f(x) = 2x + 1 的定义域为全体实数，其值域也是全体实数。

When 'finding functions', we are often given partial information and need to deduce the complete rule f(x).

在“求函数”时，我们通常被提供部分信息，需要推导出完整的规则 f(x)。

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2. Composite Functions and Finding Inner/Outer Functions | 复合函数与求内/外函数

Given two functions f and g, the composite function f(g(x)) means apply g first, then f. We denote this as fg(x).

给定两个函数 f 和 g，复合函数 f(g(x)) 表示先应用 g，再应用 f。我们记作 fg(x)。

sking you to find the other. For instance, if fg(x) = 6x - 5 and g(x) = 2x + 1, find f(x).

常见的考试题是给出 fg(x) 的表达式和其中一个函数，要求找出另一个函数。例如，若 fg(x) = 6x - 5 且 g(x) = 2x + 1，求 f(x)。

x = (u-1)/2. Substitute: f(u) = 6((u-1)/2) - 5 = 3(u-1) - 5 = 3u - 8. Hence f(x) = 3x - 8.

设 f(g(x)) = f(2x+1) = 6x - 5。为求 f(x)，用变量表示输入，令 u = 2x+1，则 x = (u-1)/2。代入得：f(u) = 6((u-1)/2) - 5 = 3(u-1) - 5 = 3u - 8。因此 f(x) = 3x - 8。

s given and fg(x) is known, you may need to find g(x). Since f(g(x)) = expression, you can 'undo' f by applying the inverse of f, provided it exists, to get g(x) = f⁻¹(fg(x)).

类似地，如果已知 f(x) 和 fg(x)，可能需要求 g(x)。因为 f(g(x)) = 表达式，你可以通过施加 f 的反函数（如果存在）来“撤销” f，得到 g(x) = f⁻¹(fg(x))。

Always check the domain restrictions to ensure the operations are valid.

始终检查定义域限制，以确保操作有效。

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3. Inverse Functions: Finding and Verifying | 反函数的求法与验证

inverse, rewrite y = f(x), then rearrange to make x the subject, and finally swap x and y.

反函数 f⁻¹(x) 逆转了 f(x) 的效果。只有当 f 在其定义域上是一一映射时，反函数才存在。要求反函数，改写 y = f(x)，然后重新排列使 x 成为主项，最后交换 x 和 y。

x+3 => yx - 2x = y+3 => x(y-2) = y+3 => x = (y+3)/(y-2). Swap to get f⁻¹(x) = (x+3)/(x-2).

例如，若 f(x) = (2x+3)/(x-1)，x ≠ 1，令 y = (2x+3)/(x-1)。两边同乘：y(x-1)=2x+3 => yx - y = 2x+3 => yx - 2x = y+3 => x(y-2)=y+3 => x = (y+3)/(y-2)。交换得 f⁻¹(x) = (x+3)/(x-2)。

domain of f. Diagrammatically, the graphs of f and f⁻¹ are reflections in the line y = x.

f⁻¹ 的定义域是 f 的值域，f⁻¹ 的值域是 f 的定义域。从图形上看，f 和 f⁻¹ 的图像关于直线 y = x 对称。

ying inverses: check f(f⁻¹(x)) = x and f⁻¹(f(x)) = x for all x in the appropriate domains.

验证反函数：检查在适当定义域内，对所有 x 有 f(f⁻¹(x)) = x 且 f⁻¹(f(x)) = x。

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4. Domain and Range Considerations | 定义域与值域考量

When finding functions, it is crucial to specify the domain. For example, if a composite function involves a square root, the inner expression must be non-negative.

在求函数时，明确规定义域至关重要。例如，如果复合函数涉及平方根，内部表达式必须非负。

Suppose we have fg(x) = √x² - 9 and g(x) = x². To find f(x), we set u = x², then f(u) = √u - 9. Thus f(x) = √x - 9 fore f is only defined for x ≥ 9, and the domain of the composite will be x such that x² ≥ 9, i.e., |x| ≥ 3.

假设有 fg(x) = √x² - 9 且 g(x) = x²。求 f(x)，令 u = x²，则 f(u) = √u - 9。因此 f(x) = √x - 9。但我们必须考虑 g 的定义域和值域，以确定 f 的定义域。对于实数 x，g(x) ≥ 0，但 f 要求 x - 9 ≥ 0 => x ≥ 9。因此 f 只在 x ≥ 9 上有定义，而复合函数的定义域将是满足 x² ≥ 9 的 x，即 |x| ≥ 3。

Always pay attention to whether the function is one-to-one over its natural domain. If not, you may need to restrict the domain to find an inverse.

始终注意函数在其自然定义域上是否为一对一的。如果不是，你可能需要限制定义域以求出反函数。

For instance, f(x) = x² is not one-to-one over ℝ, but if we restrict the domain to x ≥ 0, then f⁻¹(x) = √x exists.

例如，f(x) = x² 在 ℝ 上不是一对一的，但如果将定义域限制为 x ≥ 0，则存在反函数 f⁻¹(x) = √x。

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5. Solving Functional Equations | 解函数方程

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