Reciprocal Graphs of Functions | IB数学：函数取倒数的图像变化

📚 Reciprocal Graphs of Functions | IB数学：函数取倒数的图像变化

In IB Mathematics, one of the most elegant and exam-relevant transformations is taking the reciprocal of a function, i.e. transforming y = f(x) into y = 1/f(x). This operation produces a completely new graph that behaves in surprisingly systematic ways near zeros, asymptotes, and infinity.

在IB数学中，一个非常重要且常考的图像变换就是取函数的倒数，即将 y = f(x) 变换为 y = 1/f(x)。这一运算会产生一幅全新的图像，其在零点、渐近线和无穷远处附近的行为具有极强的规律性。

1. Definition of Reciprocal Transformation | 倒数变换的定义

Given a function f(x), its reciprocal is defined as g(x) = 1/f(x), provided that f(x) ≠ 0. The domain of g(x) excludes all values of x where f(x) = 0, because division by zero is undefined.

给定函数 f(x)，其倒数定义为 g(x) = 1/f(x)，前提是 f(x) ≠ 0。g(x) 的定义域排除了所有使 f(x) = 0 的 x 值，因为除以零是没有意义的。

For example, if f(x) = x² − 1, then g(x) = 1/(x² − 1). The domain of g is x ≠ ±1.

例如，若 f(x) = x² − 1，则 g(x) = 1/(x² − 1)。g 的定义域为 x ≠ ±1。

2. Relationship Between y = f(x) and y = 1/f(x) | 原函数与倒数函数的关系

The reciprocal function is not simply a reflected or scaled version of the original graph; it is a complete restructuring. However, there are key correspondences: where f(x) is large, 1/f(x) is small; where f(x) is small (but positive or negative), 1/f(x) is large.

倒数函数并非简单地对原图像进行翻转或缩放，而是对图像的一种彻底重构。不过，它们之间存在关键对应关系：当 f(x) 很大时，1/f(x) 很小；当 f(x) 很小（无论正负）时，1/f(x) 很大。

y = 1/f(x) ⇔ f(x) · y = 1

This product relationship means that the point (x, y) on the original graph and the point (x, 1/y) on the reciprocal graph always multiply to 1. Both graphs share the same x-coordinate for each corresponding point.

这种乘积关系意味着原图像上的点 (x, y) 与倒数图像上的点 (x, 1/y) 的纵坐标之积恒等于 1。两个图像上对应点的横坐标相同。

3. Zeros of f(x) Become Vertical Asymptotes | 零点变为垂直渐近线

If f(a) = 0, then as x approaches a, the value of 1/f(x) tends toward positive or negative infinity. Therefore, every zero of f(x) corresponds to a vertical asymptote in y = 1/f(x).

如果 f(a) = 0，那么当 x 趋近于 a 时，1/f(x) 的值趋向正无穷或负无穷。因此，f(x) 的每一个零点都对应 y = 1/f(x) 中的一条垂直渐近线。

For instance, f(x) = x − 3 has a zero at x = 3. The graph of y = 1/(x − 3) has a vertical asymptote at x = 3. The behaviour on either side depends on the sign of f(x).

例如，f(x) = x − 3 在 x = 3 处有一个零点。y = 1/(x − 3) 的图像在 x = 3 处有一条垂直渐近线。渐近线两侧的行为取决于 f(x) 的符号。

If f(x) → 0⁺, then 1/f(x) → +∞.
若 f(x) → 0⁺，则 1/f(x) → +∞。
If f(x) → 0⁻, then 1/f(x) → −∞.
若 f(x) → 0⁻，则 1/f(x) → −∞。

Note that the vertical asymptotes of 1/f(x) occur at exactly the same x-values as the x-intercepts of f(x).

注意，1/f(x) 的垂直渐近线恰好出现在 f(x) 的 x 轴截距处。

4. Horizontal Asymptotes and End Behaviour | 水平渐近线与末端行为

The horizontal asymptote of y = f(x) directly determines the horizontal asymptote of y = 1/f(x). If f(x) → L as x → ±∞, where L is a finite non-zero constant, then 1/f(x) → 1/L.

y = f(x) 的水平渐近线直接决定了 y = 1/f(x) 的水平渐近线。如果当 x → ±∞ 时 f(x) → L，其中 L 是非零有限常数，那么 1/f(x) → 1/L。

Special cases:

特殊情况：

If f(x) → 0 at infinity, then 1/f(x) grows unbounded — there is no horizontal asymptote, but possibly a slant or curve asymptote.
若 f(x) 在无穷远处趋于 0，则 1/f(x) 无界增长——没有水平渐近线，但可能存在斜渐近线或曲线渐近线。
If f(x) → ±∞, then 1/f(x) → 0. Thus y = 0 becomes the horizontal asymptote.
若 f(x) → ±∞，则 1/f(x) → 0。此时 y = 0 成为水平渐近线。

For example, if f(x) = 2x/(x + 1), then f(x) → 2 as x → ∞, so 1/f(x) → 1/2. The reciprocal graph has y = 1/2 as its horizontal asymptote.

例如，若 f(x) = 2x/(x + 1)，则当 x → ∞ 时 f(x) → 2，因此 1/f(x) → 1/2。倒数图像的水平渐近线为 y = 1/2。

5. Invariant Points: Where f(x) = ±1 | 不变点：f(x) = ±1 处

Points where f(x) = 1 or f(x) = −1 remain fixed under the reciprocal transformation, because 1/1 = 1 and 1/(−1) = −1. These are called invariant points, and they lie on the lines y = 1 and y = −1.

当 f(x) = 1 或 f(x) = −1 时，该点在倒数变换下保持不变，因为 1/1 = 1，1/(−1) = −1。这些点称为不变点，它们位于直线 y = 1 和 y = −1 上。

Solving f(x) = 1 and f(x) = −1 gives the x-coordinates of the invariant points. These are the only points where the original graph and the reciprocal graph intersect.

解方程 f(x) = 1 和 f(x) = −1 即可得到不变点的横坐标。这是原图像与倒数图像仅有的交点。

For example, if f(x) = x², then f(x) = 1 gives x = ±1. The reciprocal graph y = 1/x² passes through (1, 1) and (−1, 1).

例如，若 f(x) = x²，则 f(x) = 1 解得 x = ±1。倒数图像 y = 1/x² 经过点 (1, 1) 和 (−1, 1)。

6. Sign Analysis: Same Sign as f(x) | 符号分析：与原函数同号

Since 1/f(x) has the same sign as f(x) for every x in the domain (positive divided by positive is positive; negative divided by negative is positive result with negative sign), the reciprocal graph never crosses the x-axis and never changes sign except at vertical asymptotes.

由于 1/f(x) 在其定义域内每一点的符号都与 f(x) 相同（正数除以正数为正，负数除以负数为正），倒数图像永远不会跨越 x 轴，除了在垂直渐近线处，其符号不会发生改变。

This means that if f(x) > 0 on an interval, then 1/f(x) > 0 on the same interval. If f(x) < 0, then 1/f(x) < 0. The reciprocal graph stays entirely above the x-axis where f(x) is positive and entirely below where f(x) is negative.

这意味着如果 f(x) 在某区间内大于 0，则 1/f(x) 在同一区间内也大于 0。如果 f(x) 小于 0，则 1/f(x) 也小于 0。倒数图像在 f(x) 为正的区间完全位于 x 轴上方，在 f(x) 为负的区间完全位于 x 轴下方。

sign(1/f(x)) = sign(f(x)) for all x ∈ Domain

This property is extremely useful when sketching reciprocal graphs: you can first mark the sign regions of f(x) and then sketch accordingly.

这一性质在绘制倒数图像时非常有用：你可以先标出 f(x) 的符号区间，再据此绘制图像。

7. Local Maxima and Minima Interchange | 极值点互换

If f(x) has a local maximum at x = a with value M, then 1/f(x) has a local minimum at x = a with value 1/M, provided M ≠ 0. Similarly, a local minimum of f(x) becomes a local maximum of 1/f(x).

如果 f(x) 在 x = a 处有局部最大值 M，那么 1/f(x) 在 x = a 处有局部最小值 1/M，前提是 M ≠ 0。类似地，f(x) 的局部最小值变为 1/f(x) 的局部最大值。

This happens because the reciprocal function is strictly decreasing on intervals where f(x) > 0 and strictly decreasing on intervals where f(x) < 0. The x-coordinate of the extremum remains the same.

这是因为倒数函数在 f(x) > 0 的区间上严格递减，在 f(x) < 0 的区间上也严格递减。极值点的横坐标保持不变。

For example, f(x) = 4 − x² has a maximum of 4 at x = 0. Then y = 1/(4 − x²) has a minimum of 1/4 at x = 0.

例如，f(x) = 4 − x² 在 x = 0 处有最大值 4。则 y = 1/(4 − x²) 在 x = 0 处有最小值 1/4。

8. Sketching Steps for y = 1/f(x) | y = 1/f(x) 的作图步骤

A systematic approach to sketching reciprocal graphs is essential for exam success. Follow these steps:

系统地绘制倒数图像的方法对考试取得好成绩至关重要。请遵循以下步骤：

Sketch or analyse y = f(x) first, marking all x-intercepts and key points.
先绘制或分析 y = f(x)，标出所有 x 轴截距和关键点。
Draw vertical asymptotes at every x-intercept of f(x).
在 f(x) 的每个 x 轴截距处画出垂直渐近线。
Find the horizontal asymptote: if f(x) → L, then 1/f(x) → 1/L.
求水平渐近线：若 f(x) → L，则 1/f(x) → 1/L。
Locate invariant points by solving f(x) = 1 and f(x) = −1.
通过解 f(x) = 1 和 f(x) = −1 来确定不变点。
Determine the sign of f(x) in each region and sketch 1/f(x) accordingly.
确定每个区间内 f(x) 的符号，据此绘制 1/f(x)。
Check extrema: local maxima of f(x) become local minima of 1/f(x) and vice versa.
检查极值：f(x) 的局部最大值变为 1/f(x) 的局部最小值，反之亦然。

9. Worked Example 1: Quadratic Function | 例题一：二次函数

Consider f(x) = x² − 4. Sketch y = 1/f(x) = 1/(x² − 4).

考虑 f(x) = x² − 4。绘制 y = 1/f(x) = 1/(x² − 4) 的图像。

Step 1: Zeros: x² − 4 = 0 gives x = ±2. Vertical asymptotes at x = −2 and x = 2.

步骤一：零点：x² − 4 = 0 解得 x = ±2。垂直渐近线在 x = −2 和 x = 2 处。

Step 2: As x → ±∞, f(x) → +∞, so 1/f(x) → 0. Horizontal asymptote: y = 0.

步骤二：当 x → ±∞ 时，f(x) → +∞，因此 1/f(x) → 0。水平渐近线为 y = 0。

Step 3: Invariant points: x² − 4 = 1 gives x = ±√5; x² − 4 = −1 gives x = ±√3.

步骤三：不变点：x² − 4 = 1 解得 x = ±√5；x² − 4 = −1 解得 x = ±√3。

Step 4: Sign analysis: f(x) > 0 for x < −2 or x > 2; f(x) < 0 for −2 < x < 2. Therefore, 1/(x² − 4) is positive outside (−2, 2) and negative inside.

步骤四：符号分析：当 x < −2 或 x > 2 时 f(x) > 0；当 −2 < x < 2 时 f(x) < 0。因此 1/(x² − 4) 在 (−2, 2) 之外为正，在区间内为负。

Step 5: f(x) has a minimum of −4 at x = 0, so 1/f(x) has a maximum of −1/4 at x = 0.

步骤五：f(x) 在 x = 0 处有最小值 −4，因此 1/f(x) 在 x = 0 处有最大值 −1/4。

10. Worked Example 2: Linear Function | 例题二：线性函数

Consider f(x) = 2x + 1. Sketch y = 1/(2x + 1).

考虑 f(x) = 2x + 1。绘制 y = 1/(2x + 1) 的图像。

The zero of f is at x = −1/2, giving a vertical asymptote there. As x → ±∞, f(x) → ±∞, so 1/f(x) → 0; hence y = 0 is the horizontal asymptote.

f 的零点在 x = −1/2 处，因此垂直渐近线在此处。当 x → ±∞ 时，f(x) → ±∞，所以 1/f(x) → 0；因此水平渐近线为 y = 0。

The invariant points satisfy 2x + 1 = 1 ⇔ x = 0, giving the point (0, 1), and 2x + 1 = −1 ⇔ x = −1, giving the point (−1, −1).

不变点满足 2x + 1 = 1 ⇔ x = 0，得到点 (0, 1)；以及 2x + 1 = −1 ⇔ x = −1，得到点 (−1, −1)。

For x > −1/2, f(x) > 0, so 1/f(x) > 0, decreasing from +∞ (near the asymptote) toward 0. For x < −1/2, f(x) < 0, so 1/f(x) < 0, decreasing from 0 to −∞.

当 x > −1/2 时，f(x) > 0，所以 1/f(x) > 0，从渐近线附近的 +∞ 递减趋向 0。当 x < −1/2 时，f(x) < 0，所以 1/f(x) < 0，从 0 递减到 −∞。

11. Common Mistakes and Exam Tips | 常见错误与应试技巧

Students frequently make the following errors when dealing with reciprocal graphs:

学生在处理倒数图像时经常犯以下错误：

Mistake 1: Forgetting that x-intercepts of f(x) become vertical asymptotes of 1/f(x).
错误一：忘记 f(x) 的 x 轴截距变为 1/f(x) 的垂直渐近线。
Mistake 2: Drawing 1/f(x) crossing the x-axis. The reciprocal graph never crosses y = 0.
错误二：绘制 1/f(x) 跨越 x 轴。倒数图像永远不会穿过 y = 0。
Mistake 3: Confusing invariant points (f(x) = ±1) with x-intercepts.
错误三：将不变点（f(x) = ±1）与 x 轴截距混淆。
Mistake 4: Forgetting to exclude zeros of f(x) from the domain of the reciprocal function.
错误四：忘记从倒数函数的定义域中排除 f(x) 的零点。

Exam Tip: Always state the domain restriction explicitly: Domain of 1/f(x) = {x ∈ ℝ : f(x) ≠ 0}. This demonstrates full understanding and earns method marks.

应试技巧：务必明确写出定义域限制：1/f(x) 的定义域 = {x ∈ ℝ : f(x) ≠ 0}。这能体现你的完整理解并帮助你获得方法分。

12. Multiple Transformations: f(x) → af(bx + c) + d | 多重变换：f(x) → af(bx + c) + d

Exam questions often combine reciprocal transformations with other transformations. For example, if y = f(x) is given, you may be asked to sketch y = 1/(f(x) − 2).

考试题常将倒数变换与其他变换结合。例如，已知 y = f(x)，要求绘制 y = 1/(f(x) − 2) 的图像。

In this case, the vertical asymptotes occur where f(x) − 2 = 0, i.e. where f(x) = 2. The horizontal asymptote shifts accordingly: if f(x) → L, then 1/(f(x) − 2) → 1/(L − 2).

在这种情况下，垂直渐近线出现在 f(x) − 2 = 0 处，即 f(x) = 2 处。水平渐近线也相应移动：若 f(x) → L，则 1/(f(x) − 2) → 1/(L − 2)。

The invariant points now satisfy f(x) − 2 = 1 ⇔ f(x) = 3 and f(x) − 2 = −1 ⇔ f(x) = 1.

此时不变点满足 f(x) − 2 = 1 ⇔ f(x) = 3 以及 f(x) − 2 = −1 ⇔ f(x) = 1。

Always analyse such composite transformations step by step: first locate the zeros of the new denominator, then find the asymptotes, then the invariant points.

处理这类复合变换时一定要逐步分析：先定位新分母的零点，再找渐近线，然后确定不变点。

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Reciprocal Graphs of Functions | IB数学：函数取倒数的图像变化