Inverse Functions and How to Find Them | 反函数及其求解方法

📚 Inverse Functions and How to Find Them | 反函数及其求解方法

An inverse function reverses the effect of a function. If a function f maps x to y, then its inverse f⁻¹ maps y back to x. In the Edexcel IGCSE syllabus, you need to understand the definition, determine when an inverse exists, and use algebraic steps to find it.

反函数的作用是把一个函数的效果“反转”回来。如果函数 f 把 x 对应到 y，那么它的反函数 f⁻¹ 就把 y 对应回 x。在爱德思 IGCSE 考纲中，你需要理解反函数的定义、判断反函数存在的条件，并掌握用代数步骤求出反函数的方法。

1. What is an Inverse Function? | 什么是反函数

A function is a rule that assigns each input exactly one output. The inverse function performs the opposite rule, recovering the original input from a given output. For example, if f(2) = 5, then f⁻¹(5) = 2.

函数是一种规则，它给每个输入值恰好一个输出值。反函数执行相反的规则，从给定的输出值恢复原来的输入值。例如，如果 f(2) = 5，那么 f⁻¹(5) = 2。

We write the inverse of f as f⁻¹, read “f inverse.” It is important not to confuse f⁻¹(x) with 1 / f(x). They are completely different.

我们把 f 的反函数记作 f⁻¹，读作 “f 的逆”。注意不要把 f⁻¹(x) 与 1 / f(x) 混淆，两者完全不同。

In essence, an inverse function undoes the original function. If you first apply f and then apply f⁻¹, you get back the starting value.

本质上，反函数撤销原函数的操作。如果你先应用 f，再应用 f⁻¹，就会回到初始值。

2. When Does an Inverse Function Exist? | 反函数存在的条件

For a function to have an inverse, it must be one-to-one (injective). This means each output comes from exactly one input. In other words, different inputs cannot produce the same output.

一个函数要有反函数，它必须是一一对应的（单射）。这意味着每个输出值只能来自一个输入值。换句话说，不同的输入不能产生相同的输出。

The horizontal line test is a quick visual check. If any horizontal line intersects the graph of the function at more than one point, then the function is not one-to-one and has no inverse over the entire domain.

水平线检验是一个快速的图形判断方法。如果任意一条水平线与函数图像相交多于一个点，那么这个函数就不是一一对应的，在整个定义域上不存在反函数。

For example, f(x) = x² fails this test because both x = 2 and x = -2 give y = 4. However, we can restrict the domain to x ≥ 0, so the function becomes one-to-one and its inverse is f⁻¹(x) = √x.

例如，f(x) = x² 不满足该检验，因为 x = 2 和 x = -2 都得到 y = 4。但我们可以把定义域限制为 x ≥ 0，这样函数就变成一一对应，它的反函数是 f⁻¹(x) = √x。

In IGCSE questions, if a quadratic has a restricted domain, you must use that restriction when finding the inverse.

在 IGCSE 试题中，如果二次函数有限制定义域，求解反函数时必须使用该限制条件。

3. Steps to Find an Inverse Function | 求反函数的基本步骤

We can find the inverse of a given function using a standard algebraic procedure. Follow these steps carefully.

我们可以按照标准的代数步骤来求一个给定函数的反函数。请仔细按照以下步骤操作。

Step 1: Write y instead of f(x). Replace f(x) with y.

第 1 步：用 y 代替 f(x)。把 f(x) 改写成 y。
Step 2: Swap x and y in the equation. This represents reversing the input and output.

第 2 步：在等式中交换 x 和 y。这一步代表交换输入和输出。
Step 3: Solve the equation for y. This gives y in terms of x.

第 3 步：解方程，求出 y 关于 x 的表达式。
Step 4: Replace y with f⁻¹(x). Also state the domain if necessary.

第 4 步：把 y 替换成 f⁻¹(x)。如有必要，写出定义域。

The key trick is swapping x and y. After that, solving for y gives the inverse rule.

关键技巧是交换 x 和 y。之后解出 y，就得到了反函数的规则。

4. Worked Example: Linear Function | 示例：线性函数

Let us find the inverse of f(x) = 2x + 3.

我们来求 f(x) = 2x + 3 的反函数。

First, replace f(x) with y: y = 2x + 3.

首先，把 f(x) 换成 y：y = 2x + 3。

Then swap x and y: x = 2y + 3.

然后交换 x 和 y：x = 2y + 3。

Solve for y: x – 3 = 2y, so y = (x – 3) / 2.

解出 y：x – 3 = 2y，所以 y = (x – 3) / 2。

f⁻¹(x) = (x – 3) / 2

We can verify this. Compute f(f⁻¹(x)) = 2((x – 3)/2) + 3 = (x – 3) + 3 = x. The inverse works.

我们可以验证一下。计算 f(f⁻¹(x)) = 2((x – 3)/2) + 3 = (x – 3) + 3 = x。反函数成立。

Notice that the slope of the inverse is 1/2, which is the reciprocal of the original slope 2. This happens for all linear functions with slope not equal to zero.

注意反函数的斜率是 1/2，它是原函数斜率 2 的倒数。所有斜率不为零的线性函数都有这种关系。

5. Worked Example: Quadratic with Restricted Domain | 示例：二次函数（限制定义域）

Consider f(x) = x², x ≥ 0. Because the domain is restricted to non-negative numbers, the function is one-to-one and has an inverse.

考虑 f(x) = x²，x ≥ 0。由于定义域被限制为非负数，该函数是一一对应的，因此存在反函数。

Follow the steps: y = x², then swap x and y: x = y².

按步骤进行：y = x²，然后交换 x 和 y：x = y²。

Solving for y gives y = ±√x. But the original domain was x ≥ 0, so the range of f is y ≥ 0. Therefore the inverse must produce non-negative values. We choose the positive square root:

解出 y，得到 y = ±√x。但原函数的定义域是 x ≥ 0，所以 f 的值域是 y ≥ 0。因此反函数必须产生非负值。我们选择正平方根：

f⁻¹(x) = √x, x ≥ 0

Here the domain of f⁻¹ is x ≥ 0, which is exactly the range of f. This example shows why domain restrictions are essential for quadratics.

这里反函数的定义域是 x ≥ 0，恰好是 f 的值域。这个例子说明了对二次函数进行定义域限制的必要性。

If the original function were f(x) = x², x ≤ 0, the inverse would be f⁻¹(x) = -√x, because the outputs must be non-positive.

如果原函数改为 f(x) = x²，x ≤ 0，那么反函数就是 f⁻¹(x) = -√x，因为输出必须是非正数。

6. Worked Example: Rational Function | 示例：含分式的函数

Let us find the inverse of f(x) = (2x + 3) / (x – 1), where x ≠ 1.

我们来求 f(x) = (2x + 3) / (x – 1) 的反函数，其中 x ≠ 1。

Start by writing y = (2x + 3) / (x – 1). Swap x and y: x = (2y + 3) / (y – 1).

首先写出 y = (2x + 3) / (x – 1)。交换 x 和 y：x = (2y + 3) / (y – 1)。

Now solve for y. Multiply both sides by (y – 1): x(y – 1) = 2y + 3.

现在解出 y。两边同时乘以 (y – 1)：x(y – 1) = 2y + 3。

Expand: xy – x = 2y + 3. Collect y terms on one side: xy – 2y = x + 3.

展开：xy – x = 2y + 3。把含 y 的项移到一边：xy – 2y = x + 3。

Factor out y: y(x – 2) = x + 3. Therefore y = (x + 3) / (x – 2).

提取 y：y(x – 2) = x + 3。因此 y = (x + 3) / (x – 2)。

f⁻¹(x) = (x + 3) / (x – 2), x ≠ 2

The domain of the inverse is x ≠ 2, which matches the range of the original function (since the original has a horizontal asymptote at y = 2).

反函数的定义域是 x ≠ 2，这与原函数的值域一致（因为原函数有水平渐近线 y = 2）。

7. Graphs of Inverse Functions | 反函数的图像

The graph of f⁻¹ is the reflection of the graph of f across the line y = x. This is because swapping x and y is equivalent to reflecting across that line.

反函数的图像是原函数图像关于直线 y = x 的镜像。因为交换 x 和 y 相当于关于这条直线进行反射。

If a point (a, b) lies on the graph of f, then the point (b, a) lies on the graph of f⁻¹.

如果点 (a, b) 在 f 的图像上，那么点 (b, a) 就在 f⁻¹ 的图像上。

For example, f(x) = 2x + 3 contains the point (1, 5). Its inverse f⁻¹(x) = (x – 3)/2 contains the point (5, 1).

例如，f(x) = 2x + 3 经过点 (1, 5)。它的反函数 f⁻¹(x) = (x – 3)/2 经过点 (5, 1)。

When sketching inverse graphs on the IGCSE exam, always draw the line y = x as a dashed guide. Then reflect the given curve across it.

在 IGCSE 考试中画反函数图像时，务必画出直线 y = x 作为虚线参考线，然后将给定曲线关于它反射。

8. Domain and Range Swap | 定义域与值域互换

A crucial property of inverse functions is that the domain and range swap:

反函数的一个重要性质是定义域和值域互相交换：

Domain of f⁻¹ = Range of f, and Range of f⁻¹ = Domain of f

This means if f accepts x-values from set A and produces outputs in set B, then f⁻¹ accepts values from B and produces outputs in A.

这意味着如果 f 接受集合 A 中的 x 值并产生集合 B 中的输出，那么 f⁻¹ 接受集合 B 中的值，并产生集合 A 中的输出。

In practice, when you find an inverse, you must state its domain. Often this domain is exactly the range of the original function.

在实际操作中，求出反函数后，必须写出它的定义域。这个定义域通常正是原函数的值域。

For example, f(x) = x², x ≥ 0 has range y ≥ 0. Therefore f⁻¹(x) = √x has domain x ≥ 0.

例如，f(x) = x²，x ≥ 0 的值域是 y ≥ 0。因此 f⁻¹(x) = √x 的定义域是 x ≥ 0。

9. Composite Functions and Inverse Relationship | 复合函数与反函数的关系

For a function f and its inverse f⁻¹, the following identities hold for all x in the appropriate domains:

对于函数 f 和它的反函数 f⁻¹，在适当的定义域内，以下恒等式成立：

f(f⁻¹(x)) = x and f⁻¹(f(x)) = x

These two equations provide a powerful verification tool. If you compute both compositions and simplify to x, then the inverse you found is correct.

这两个等式提供了强大的验证工具。如果你计算两个复合函数，都化简为 x，那么你求出的反函数就是正确的。

Also, the inverse of an inverse is the original function: (f⁻¹)⁻¹ = f.

另外，反函数的反函数就是原函数：(f⁻¹)⁻¹ = f。

In exam problems, you may be asked to show that two functions are inverses of each other by composing them. Use the identities above.

在考试题中，你可能会被要求通过复合来证明两个函数互为反函数。使用上述恒等式即可。

10. Common Mistakes and Tips | 常见错误与技巧

Students often make similar errors when working with inverse functions. Here are some pitfalls to avoid.

学生在处理反函数时常犯一些相似的错误。以下是需要避免的常见陷阱。

Mistake: Confusing f⁻¹(x) with 1/f(x). Remember that the superscript -1 means inverse, not reciprocal.

错误：把 f⁻¹(x) 与 1/f(x) 混淆。记住上标 -1 表示反函数，不是倒数。
Mistake: Forgetting to swap x and y. Without swapping, you simply solve for x, which does not give the inverse.

错误：忘记交换 x 和 y。如果不交换，你只是在解 x，并不能得到反函数。
Mistake: Ignoring domain restrictions for quadratics. Always check whether the original function is one-to-one on its stated domain.

错误：忽略二次函数的定义域限制。始终检查原函数在其给定定义域上是否一一对应。
Mistake: Not stating the domain of the inverse. The domain of f⁻¹ is the range of f; make sure you write it down.

错误：没有写出反函数的定义域。f⁻¹ 的定义域是 f 的值域；一定要写清楚。

Tip: After finding f⁻¹, quickly check that f(f⁻¹(x)) simplifies to x. If it does not, review your algebra.

技巧：求出 f⁻¹ 后，快速检查 f(f⁻¹(x)) 是否化简为 x。如果不是，请检查你的代数运算。

11. Exam-Style Practice Questions | 考试风格练习题

Let us look at a typical Edexcel IGCSE question structure. You are often given a function and asked to find its inverse, state a domain, or sketch both graphs.

我们来看一道典型的爱德思 IGCSE 题目结构。题目通常会给出一个函数，要求你求它的反函数、写出定义域，或者画出两个函数的图像。

Question: Given f(x) = 3x – 5 for x ∈ ℝ, find f⁻¹(x) and verify your answer.

题目：已知 f(x) = 3x – 5，x ∈ ℝ，求 f⁻¹(x) 并验证答案。

Solution: y = 3x – 5. Swap x and y: x = 3y – 5. Solve: x + 5 = 3y, so y = (x + 5) / 3. Therefore f⁻¹(x) = (x + 5) / 3.

解答：y = 3x – 5。交换 x 和 y：x = 3y – 5。解得：x + 5 = 3y，所以 y = (x + 5) / 3。因此 f⁻¹(x) = (x + 5) / 3。

Verification: f(f⁻¹(x)) = 3((x + 5)/3) – 5 = (x + 5) – 5 = x. Correct.

验证：f(f⁻¹(x)) = 3((x + 5)/3) – 5 = (x + 5) – 5 = x。正确。

Another question style gives a graph and asks you to sketch the inverse. Remember the reflection across y = x.

另一种题型是给一个图像，要求你画出反函数的图像。记住关于 y = x 做反射。

12. Summary | 总结

An inverse function reverses the action of a function. It exists only when the original function is one-to-one. To find it, replace f(x) with y, swap x and y, then solve for y. The resulting expression is f⁻¹(x).

反函数撤销一个函数的操作。它只在一一对应的函数中存在。求法步骤是：把 f(x) 替换为 y，交换 x 和 y，然后解出 y。得到的表达式就是 f⁻¹(x)。

Always remember that the domain and range swap between a function and its inverse. The graph of f⁻¹ is the reflection of the graph of f in the line y = x. Use the composition identities f(f⁻¹(x)) = x and f⁻¹(f(x)) = x to check your work.

始终记住函数与其反函数的定义域和值域互换。f⁻¹ 的图像是 f 的图像关于直线 y = x 的镜像。使用复合恒等式 f(f⁻¹(x)) = x 和 f⁻¹(f(x)) = x 来检查你的答案。

With these steps and tips, you are ready to solve inverse function problems confidently on the IGCSE exam.

掌握了这些步骤和技巧，你就能自信地解决 IGCSE 考试中的反函数问题。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Inverse Functions and How to Find Them | 反函数及其求解方法