Inverse Functions: Solving and Graphical Relationships | 反函数的求解与图像关系

📚 Inverse Functions: Solving and Graphical Relationships | 反函数的求解与图像关系

Inverse functions are a cornerstone of A-Level mathematics. They connect algebraic manipulation, domain and range reasoning, and graph transformations, and they appear in nearly every Pure Mathematics exam paper. Mastering how to find an inverse function and how to visualise it as a reflection across the line y = x will give you a decisive advantage.

反函数是 A-Level 数学的基石。它串联起代数运算、定义域与值域的推理以及图像变换，几乎出现在每一份纯数学试卷中。掌握如何求解反函数，并将其理解为关于直线 y = x 的反射图像，将为你带来决定性的优势。

1. Definition of an Inverse Function | 反函数的定义

A function f maps each input x to exactly one output f(x). The inverse function, denoted f⁻¹, reverses this mapping: it takes each output value and returns the original input. For example, if f(3) = 11, then f⁻¹(11) = 3.

函数 f 将每个输入 x 映射到唯一输出 f(x)。反函数记作 f⁻¹，它逆转这一映射：将每个输出值还原为原始输入。例如，若 f(3) = 11，则 f⁻¹(11) = 3。

The notation f⁻¹(x) is read as “f inverse of x”. It must never be confused with the reciprocal 1/f(x). The superscript −1 here represents the inverse operation, not a power.

记号 f⁻¹(x) 读作”x 的 f 反函数”。绝不可将其与倒数 1/f(x) 混淆。此处的上标 −1 表示逆运算，而非幂次。

2. The One-to-One Condition | 一一对应条件

For an inverse function to exist, the original function f must be one-to-one (injective). This means that every element of its range corresponds to exactly one element of its domain. If two distinct x-values produce the same y-value, then the reverse mapping is ambiguous and cannot be a function.

反函数存在的条件是原函数 f 必须是一一对应的（单射）。这意味着值域中的每个元素恰好对应定义域中的一个元素。若两个不同的 x 值产生同一个 y 值，则逆向映射存在歧义，不能构成函数。

Graphically, we apply the horizontal line test: a function is one-to-one if every horizontal line intersects its graph at most once. For example, a straight line with non-zero gradient passes this test; a parabola does not.

从图像上看，我们使用水平线检验法：若每一条水平直线与函数图像至多相交一次，则该函数是一一对应的。例如，斜率非零的直线通过该检验；而抛物线不通过。

Linear functions with m ≠ 0: one-to-one
Quadratic functions: not one-to-one unless the domain is restricted
eˣ: one-to-one
sin x: not one-to-one unless restricted, e.g. x ∈ [−π/2, π/2]

斜率 m ≠ 0 的线性函数：一一对应
二次函数：除非限制定义域，否则不是一一对应
eˣ：一一对应
sin x：除非限制定义域，例如 x ∈ [−π/2, π/2]，否则不是一一对应

3. Solving for the Inverse Algebraically | 代数求解反函数

To find the inverse of a function y = f(x), follow a four-step procedure. This algebraic method works for any function that is one-to-one on its stated domain.

求函数 y = f(x) 的反函数，遵循四步程序。该代数方法适用于任何在其给定定义域上一一对应的函数。

Step 1: Write y = f(x) → Step 2: Swap x and y → Step 3: Rearrange to make y the subject → Step 4: Write f⁻¹(x)

步骤 1：写出 y = f(x) → 步骤 2：交换 x 与 y → 步骤 3：化简使 y 成为主项 → 步骤 4：写出 f⁻¹(x)

Let us apply this to f(x) = 3x − 5.

我们以 f(x) = 3x − 5 为例。

First, write y = 3x − 5. Then swap the variables to obtain x = 3y − 5. Next, add 5 to both sides: x + 5 = 3y. Finally, divide by 3: y = (x + 5) / 3. Therefore f⁻¹(x) = (x + 5) / 3.

首先，写出 y = 3x − 5。然后交换变量得到 x = 3y − 5。接着，两边加 5：x + 5 = 3y。最后，两边除以 3：y = (x + 5) / 3。因此 f⁻¹(x) = (x + 5) / 3。

4. Domain and Range of the Inverse | 反函数的定义域与值域

A vital relationship connects the domain and range of f to those of f⁻¹. The domain of f⁻¹ is exactly the range of f, and the range of f⁻¹ is exactly the domain of f. This symmetry is the key to solving many exam questions.

f 与 f⁻¹ 的定义域和值域之间存在一种至关重要的对称关系。f⁻¹ 的定义域恰好等于 f 的值域，f⁻¹ 的值域恰好等于 f 的定义域。这一对称性是解答许多考试题目的关键。

Function	Domain	Range
f(x) = x², x ≥ 0	x ≥ 0	y ≥ 0
f⁻¹(x) = √x	x ≥ 0	y ≥ 0
g(x) = 2ˣ	x ∈ ℝ	y > 0
g⁻¹(x) = log₂x	x > 0	y ∈ ℝ

Notice in both examples that the domain and range are swapped. When the question gives you a restricted domain, you must always state the corresponding domain of the inverse.

注意在上述两个例子中，定义域与值域正好互换。当题目给出受限定义域时，你必须始终标明反函数对应的定义域。

5. Graphical Relationship: Reflection in y = x | 图像关系：关于 y = x 对称

The graph of f⁻¹ is the reflection of the graph of f in the line y = x. This is the most important graphical fact about inverse functions. If (a, b) lies on y = f(x), then (b, a) lies on y = f⁻¹(x).

f⁻¹ 的图像是 f 的图像关于直线 y = x 的反射。这是关于反函数最重要的图像事实。若点 (a, b) 在 y = f(x) 上，则点 (b, a) 在 y = f⁻¹(x) 上。

This reflection property has practical uses. If you are asked to sketch f⁻¹ on a coordinate grid, you can simply reflect the graph of f across the 45° line y = x. The intersection points of f and f⁻¹, if any, must lie on the line y = x because only those points are unchanged by reflection.

这一反射性质具有实用价值。如果题目要求你在坐标网格上画出 f⁻¹，你只需将 f 的图像沿 45° 直线 y = x 反射即可。f 与 f⁻¹ 的交点（若存在）必然位于 y = x 上，因为只有这些点在反射后保持不变。

The graph of y = x acts as a mirror. A function that crosses y = x at a 90° angle will have an inverse that also crosses at 90°, preserving the symmetry of the two curves.

直线 y = x 充当一面镜子。若一条函数图像以 90° 穿过 y = x，其反函数也同样以 90° 穿过 y = x，保持两条曲线的对称性。

6. Worked Example: Linear and Quadratic Functions | 例题：线性与二次函数

Example 1: Find f⁻¹(x) for f(x) = 4x − 7, and verify the graphical relationship.

例题 1：求 f(x) = 4x − 7 的 f⁻¹(x)，并验证图像关系。

Write y = 4x − 7. Swap x and y: x = 4y − 7. Solve: x + 7 = 4y, so y = (x + 7) / 4. Hence f⁻¹(x) = (x + 7) / 4.

写出 y = 4x − 7。交换 x 与 y：x = 4y − 7。求解：x + 7 = 4y，所以 y = (x + 7) / 4。因此 f⁻¹(x) = (x + 7) / 4。

Check: choose x = 3. f(3) = 12 − 7 = 5. Then f⁻¹(5) = (5 + 7) / 4 = 3. The round trip returns to the original input, confirming the inverse property.

验证：取 x = 3。f(3) = 12 − 7 = 5。然后 f⁻¹(5) = (5 + 7) / 4 = 3。往返运算回到原始输入，验证了反函数性质。

Example 2: Given g(x) = x² − 4x with domain x ≥ 2, find g⁻¹(x).

例题 2：已知 g(x) = x² − 4x，定义域 x ≥ 2，求 g⁻¹(x)。

Start with y = x² − 4x. Swap: x = y² − 4y. Complete the square on the right-hand side: x = (y − 2)² − 4. Add 4 to both sides: x + 4 = (y − 2)². Take the square root: √(x + 4) = |y − 2|. Since the domain of g is x ≥ 2, the range of g⁻¹ must be y ≥ 2, so y − 2 ≥ 0. Therefore y = 2 + √(x + 4), and g⁻¹(x) = 2 + √(x + 4).

先写 y = x² − 4x。交换：x = y² − 4y。在右侧配方：x = (y − 2)² − 4。两边加 4：x + 4 = (y − 2)²。取平方根：√(x + 4) = |y − 2|。由于 g 的定义域是 x ≥ 2，g⁻¹ 的值域必须是 y ≥ 2，所以 y − 2 ≥ 0。因此 y = 2 + √(x + 4)，即 g⁻¹(x) = 2 + √(x + 4)。

The ± sign is resolved by referring back to the original domain. This is exactly why the one-to-one condition and domain restrictions matter.

± 号的选择依赖于原函数的定义域。这正是一一对应条件和定义域限制至关重要的原因。

7. Worked Example: Exponential and Logarithmic Functions | 例题：指数与对数函数

Exponential and logarithmic functions are the natural inverse pair in Pure Mathematics. For any base a > 0, a ≠ 1:

指数函数与对数函数是纯数学中天然的反函数对。对于任意底数 a > 0，a ≠ 1：

y = aˣ ⇌ x = logₐy

因此，若 f(x) = aˣ 则 f⁻¹(x) = logₐx

Example: f(x) = 2e^(3x) + 1. Find f⁻¹(x).

例题：f(x) = 2e^(3x) + 1。求 f⁻¹(x)。

Write y = 2e^(3x) + 1. Swap x and y: x = 2e^(3y) + 1. Subtract 1: x − 1 = 2e^(3y). Divide by 2: (x − 1) / 2 = e^(3y). Take the natural logarithm of both sides: ln((x − 1) / 2) = 3y. Divide by 3: y = ⅓ ln((x − 1) / 2). Hence f⁻¹(x) = ⅓ ln((x − 1) / 2).

写出 y = 2e^(3x) + 1。交换 x 和 y：x = 2e^(3y) + 1。两边减 1：x − 1 = 2e^(3y)。两边除以 2：(x − 1) / 2 = e^(3y)。两边取自然对数：ln((x − 1) / 2) = 3y。两边除以 3：y = ⅓ ln((x − 1) / 2)。因此 f⁻¹(x) = ⅓ ln((x − 1) / 2)。

The domain of f⁻¹ is x > 1 because the natural logarithm requires a positive argument. This matches the range of the original function: since 2e^(3x) > 0 for all x, f(x) > 1.

f⁻¹ 的定义域为 x > 1，因为自然对数要求自变量为正。这与原函数的值域一致：由于对所有 x 都有 2e^(3x) > 0，所以 f(x) > 1。

8. Composite Functions and Inverses | 复合函数与反函数

One of the most powerful properties of inverse functions is their cancellation property. When a function and its inverse are composed in either order, they cancel each other out:

反函数最强大的性质之一是相消性质。当函数与其反函数以任意顺序复合时，它们相互抵消：

f⁻¹(f(x)) = x 且 f(f⁻¹(x)) = x

The two equations above only hold for x-values within the appropriate domain. They provide a quick way to check whether two functions are indeed inverses.

以上两个等式仅对恰当定义域内的 x 值成立。它们提供了一种快速检验两个函数是否互为反函数的方法。

When dealing with composite functions, note that the inverse of a composition reverses the order of operations:

处理复合函数时，注意复合函数的反函数会颠倒运算顺序：

(g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹

This is similar to the rule for the inverse of matrix products. If you apply f first and then g, you must undo g first and then f.

这类似于矩阵乘积的逆矩阵法则。如果你先应用 f 再应用 g，那么必须先撤销 g 再撤销 f。

9. Self-Inverse Functions | 自反函数

A self-inverse function is a function that is its own inverse. For such a function, applying it twice returns the original input: f(f(x)) = x. The graph of a self-inverse function is symmetric about the line y = x.

自反函数是自身反函数的函数。对于这种函数，连续应用两次将回到原始输入：f(f(x)) = x。自反函数的图像关于直线 y = x 对称。

Classic examples include:

经典例子包括：

f(x) = x, the identity function
f(x) = −x, the reflection through the origin
f(x) = 1/x, x ≠ 0, the reciprocal function
f(x) = (ax + b) / (cx − a) with appropriate conditions

f(x) = x，恒等函数
f(x) = −x，关于原点的反射
f(x) = 1/x，x ≠ 0，倒数函数
f(x) = (ax + b) / (cx − a)，在适当条件下

Exam questions sometimes ask you to find the value of a or b such that a given function is self-inverse. Set f(f(x)) = x, simplify, and equate coefficients.

考试题目有时会要求你求某个参数 a 或 b 的值，使给定函数成为自反函数。令 f(f(x)) = x，化简后比较系数即可。

10. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Students frequently lose marks on inverse function questions for avoidable reasons. The most common mistakes are listed below.

学生常在反函数题目中因可避免的原因丢分。最常见的错误列举如下。

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

错误 1：将 f⁻¹(x) 与 1/f(x) 混淆。记住上标 −1 表示反函数，不是倒数。

Pitfall 2: Forgetting to restrict the domain of a quadratic before finding its inverse. When the domain is not one-to-one, the inverse is not a function.

错误 2：在求二次函数的反函数前忘记限制定义域。当定义域不是一一对应时，反函数就不是函数。

Pitfall 3: Swapping x and y but failing to also swap the domain and range. Always state the domain of the inverse.

错误 3：交换了 x 和 y，但忘记同时交换定义域和值域。务必标明反函数的定义域。

Exam tip: Always verify your answer by checking that f(f⁻¹(x)) simplifies to x. This two-line check catches most algebraic errors.

考试技巧：始终通过验证 f(f⁻¹(x)) 是否化简为 x 来检验答案。这短短两行的检查能发现大多数代数错误。

Exam tip: When sketching, draw the line y = x first as a dotted line, then reflect key points (intercepts and turning points) across it.

考试技巧：画图时，先用虚线画出 y = x，再将关键点（截距和顶点）逐一反射过去。

11. Summary | 总结

In summary, inverse functions reverse the action of a function. In order to find an inverse, the function must be one-to-one; if not, restrict its domain. Algebraically, swap x and y and solve for y. Graphically, reflect the curve in the line y = x. Remember the domain-and-range swap, use the cancellation property to check your work, and keep self-inverse functions in mind.

总的来说，反函数逆转了函数的作用。要求反函数，函数必须是一一对应的；如果不是，则需要限制定义域。代数上，交换 x 和 y 并解出 y。图像上，将曲线沿 y = x 反射。牢记定义域和值域的互换，使用相消性质检验结果，并留意自反函数。

As you practise, draw every function alongside its inverse. The visual habit will deepen your intuition and make exam questions on inverse functions faster and more accurate.

在练习时，请将每个函数与它的反函数同时画出来。这个绘图习惯会加深你的直觉，使考试中的反函数题目解答更快、更准确。

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Inverse Functions: Solving and Graphical Relationships | 反函数的求解与图像关系