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

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。

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

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。

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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

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

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。

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。

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。

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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。

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

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。

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。

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。

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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。

ap 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。

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)。

-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| 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。

ck 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。

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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 n((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。

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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。

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9. Self-Inverse Functions | 自反函数

nput: 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

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。

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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: 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。

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11. Summary | 总结

omain-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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