📚 Finding Functions | 求函数
In A-Level Mathematics, ‘finding functions’ refers to determining an unknown function from given relationships, such as composite functions, inverse functions, or functional equations. This skill is essential for solving many exam problems where you are given one or more functions and asked to find another.
在 A-Level 数学中，“求函数”指的是根据给定的关系来确定未知函数，例如复合函数、反函数或函数方程。这一技能对于解决许多考试问题至关重要，因为题目通常会给出一个或多个函数，要求你求出另一个函数。
1. Understanding the Concept of a Function | 函数的概念
A function is a rule that maps each input x to exactly one output f(x). It is often written as f : x ↦ expression, or simply f(x) = expression. For example, f(x) = 2x + 1 is a linear function.
函数是一种规则，它将每个输入 x 映射到唯一的一个输出 f(x)。通常写作 f : x ↦ 表达式，或简写为 f(x) = 表达式。例如，f(x) = 2x + 1 是一个线性函数。
To find a function, you need to identify this rule from given information. The information may involve operations such as addition, multiplication, composition, or inversion of functions.
要求出一个函数，你需要从给定信息中识别这个规则。这些信息可能涉及函数的加法、乘法、复合或反函数等运算。
2. Function Notation and Equations | 函数记号与方程
Function notation allows us to write equations involving functions clearly. For example, fg(x) means f(g(x)), i.e. first apply g, then apply f. Similarly, f²(x) means f(f(x)), not (f(x))² unless stated.
函数记号使我们能够清晰地书写涉及函数的方程。例如，fg(x) 表示 f(g(x))，即先应用 g，再应用 f。类似地，f²(x) 表示 f(f(x))，除非特别说明，否则不表示 (f(x))²。
When finding a function, you often set up an equation such as f(g(x)) = h(x) and solve for the unknown function. This requires careful substitution and algebraic manipulation.
在求函数时，你常常会建立这样的方程：f(g(x)) = h(x)，然后求解未知函数。这需要仔细的代换和代数运算。
3. Finding a Function When One Composite Is Known | 已知复合函数求其中一个函数
If you know the composite function fg(x) and one of the original functions, you can often find the other by substitution. For example, if f(x) = 2x + 1 and fg(x) = 4x + 5, then g(x) can be found by setting 2g(x) + 1 = 4x + 5.
如果你知道复合函数 fg(x) 以及其中一个原函数，通常可以通过代换求出另一个。例如，若 f(x) = 2x + 1 且 fg(x) = 4x + 5，则令 2g(x) + 1 = 4x + 5，即可求出 g(x)。
The approach differs depending on which function is known. If the outer function f is known, substitute g(x) as the input. If the inner function g is known, replace x in f(x) with g(x).
具体方法取决于已知的是哪一个函数。如果已知外函数 f，则将 g(x) 作为输入代入。如果已知内函数 g，则将 f(x) 中的 x 替换为 g(x)。
4. Example: Finding g(x) Given f(x) and fg(x) | 例：已知 f(x) 和 fg(x) 求 g(x)
Let f(x) = 3x − 2 and fg(x) = 3x² + 4. Since f(g(x)) = 3g(x) − 2, we equate:
设 f(x) = 3x − 2 且 fg(x) = 3x² + 4。因为 f(g(x)) = 3g(x) − 2，所以我们令两边相等：
3g(x) − 2 = 3x² + 4
Add 2 to both sides:
两边同时加 2：
3g(x) = 3x² + 6
Divide by 3:
两边同时除以 3：
g(x) = x² + 2
Always check by composing: f(g(x)) = 3(x² + 2) − 2 = 3x² + 6 − 2 = 3x² + 4. This confirms the answer.
务必通过复合验证：f(g(x)) = 3(x² + 2) − 2 = 3x² + 6 − 2 = 3x² + 4。这验证了答案的正确性。
5. Example: Finding f(x) Given g(x) and fg(x) | 例：已知 g(x) 和 fg(x) 求 f(x)
Now suppose g(x) = 2x − 1 and fg(x) = 4x² + 4x + 2. Then f(2x − 1) = 4x² + 4x + 2. To find f(x), we need to express the right-hand side in terms of (2x − 1).
现在假设 g(x) = 2x − 1 且 fg(x) = 4x² + 4x + 2。那么 f(2x − 1) = 4x² + 4x + 2。为了求出 f(x)，我们需要将右边用 (2x − 1) 表示。
Let u = 2x − 1, so x = (u + 1)/2. Substitute:
令 u = 2x − 1，则 x = (u + 1)/2。代入得：
f(u) = 4[(u + 1)/2]² + 4[(u + 1)/2] + 2
Simplify:
化简：
f(u) = (u + 1)² + 2(u + 1) + 2 = u² + 4u + 5
Therefore f(x) = x² + 4x + 5. This method of substitution is a powerful tool when the inner function is linear.
因此 f(x) = x² + 4x + 5。当内函数为线性函数时，这种换元法非常有用。
6. Using Inverse Functions | 利用反函数求原函数
The inverse function f⁻¹ reverses the effect of f. If f(x) = y, then f⁻¹(y) = x. To find f(x) from its inverse, you can rearrange the equation f⁻¹(y) = x to y = f(x).
反函数 f⁻¹ 能逆反函数 f 的作用。若 f(x) = y，则 f⁻¹(y) = x。要从反函数求 f(x)，可以将方程 f⁻¹(y) = x 整理为 y = f(x)。
For example, if f⁻¹(x) = 2x − 3, then let y = f⁻¹(x) = 2x − 3. But this means x = f(y), so we need to solve for y in terms of x. Here y = (x + 3)/2, hence f(x) = (x + 3)/2.
例如，若 f⁻¹(x) = 2x − 3，令 y = f⁻¹(x) = 2x − 3。这意味着 x = f(y)，因此我们需要解出用 x 表示的 y。这里 y = (x + 3)/2，故 f(x) = (x + 3)/2。
f(x) = (x + 3) / 2
Always verify that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x for the relevant domain.
务必在相关定义域内验证 f(f⁻¹(x)) = x 和 f⁻¹(f(x)) = x。
7. Substitution Methods for Functional Equations | 换元法解函数方程
Some problems present a functional equation directly, such as f(x + 1) = 2x + 5. To find f(x), let u = x + 1, so x = u − 1. Then f(u) = 2(u − 1) + 5 = 2u + 3, so f(x) = 2x + 3.
有些题目直接给出函数方程，例如 f(x + 1) = 2x + 5。要求 f(x)，令 u = x + 1，则 x = u − 1。于是 f(u) = 2(u − 1) + 5 = 2u + 3，所以 f(x) = 2x + 3。
This substitution technique is especially useful when the argument is a linear expression. If the argument is non-linear, you may need a different substitution or may need to impose extra conditions.
当变量是线性表达式时，这种换元技术特别有用。如果变量是非线性的，则可能需要换一种代换方式，或者需要附加额外条件。
8. Domain and Range Restrictions | 定义域与值域的限制
When finding a function, always consider the domain and range. Composite functions require that the range of the inner function lies within the domain of the outer function. This can affect the possible forms of the unknown function.
在求函数时，务必考虑定义域和值域。复合函数要求内函数的值域包含在外函数的定义域内。这会影响未知函数的可能形式。
For example, if f(x) = x² with domain x ≥ 0, then the inverse is f⁻¹(x) = √x, not −√x. Similarly, when solving for g(x) in fg(x) = h(x), you must ensure that the resulting expression satisfies any given domain restrictions.
例如，若 f(x) = x² 的定义域为 x ≥ 0，则其反函数为 f⁻¹(x) = √x，而不是 −√x。类似地，在 fg(x) = h(x) 中求解 g(x) 时，必须确保所得表达式满足给定的定义域限制。
9. Common Mistakes and Special Cases | 常见错误与特殊情况
One common mistake is confusing fg(x) with f(x) × g(x). Remember that fg(x) always means f(g(x)) unless brackets indicate multiplication. Another mistake is forgetting to consider the domain when taking square roots or logarithms.
一个常见错误是混淆 fg(x) 与 f(x) × g(x)。记住 fg(x) 始终表示 f(g(x))，除非括号明确表示乘法。另一个错误是在取平方根或对数时忘记考虑定义域。
Special cases include self-inverse functions, where f(f(x)) = x, such as f(x) = 1/x or f(x) = −x. In these cases, f = f⁻¹, which can simplify the process of finding the function.
特殊情况包括自反函数，即满足 f(f(x)) = x 的函数，例如 f(x) = 1/x 或 f(x) = −x。在这些情况下，f = f⁻¹，这可以简化求函数的过程。
10. Summary and Practice Advice | 总结与练习建议
To find a function, identify whether the given information involves composition, inverses, or functional equations. Then choose an appropriate method: direct substitution, change of variable, or rearrangement. Always verify your answer by substituting back into the original condition.
要求一个函数，首先要判断给定信息涉及的是复合、反函数还是函数方程。然后选择合适的方法：直接代换、换元或整理方程。最后务必通过将答案代回原条件进行验证。
Practice with a variety of questions, especially those where the unknown function is not immediately obvious. Pay close attention to domains and ranges, as these often determine whether a solution is valid. With systematic algebraic manipulation, you can master this topic.
要多练习各种题型，尤其是未知函数不太明显的题目。密切关注定义域和值域，因为它们往往决定解是否有效。只要系统地进行代数变形，你一定能够掌握这个知识点。
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