📚 Stationary Points: Maxima, Minima and Inflection | 驻点：极大值、极小值与拐点
A stationary point is a point on a curve where the gradient is exactly zero, so the tangent drawn there is horizontal. Finding these points and deciding what type they are is one of the most heavily examined skills in A-Level Mathematics and Further Mathematics: it underpins curve sketching, optimisation and the analysis of real-world models. This guide keeps the original coverage of the topic - the definition, the first and second derivative tests, points of inflection and common pitfalls - and expands it with worked examples, comparison tables and exam technique.
驻点是曲线上梯度恰好为零的点，在该点处曲线的切线是水平的。寻找这些点并判断它们的类型，是 A-Level 数学与进阶数学中考查频率最高的技能之一：它是绘制函数图像、求解优化问题以及分析实际模型的基础。本指南保留原有内容——驻点定义、第一导数检验法、第二导数检验法、拐点以及常见错误——并补充了例题、对比表格与考试技巧。
1. What Is a Stationary Point? | 什么是驻点
se the derivative measures the gradient, a zero derivative means the curve is momentarily flat, and the tangent line at that point is parallel to the x-axis. Stationary points fall into exactly three categories: local maxima, local minima and stationary points of inflection. Every turning point of a differentiable function is a stationary point, but not every stationary point is a turning point.
曲线 y = f(x) 上的驻点，是指满足 f′(x) = 0 的点。由于导数表示梯度，导数为零意味着曲线在此处瞬时"变平"，该点处的切线与 x 轴平行。驻点只分为三类：局部极大值点、局部极小值点以及平稳拐点。可导函数的每一个极值点都是驻点，但并非每一个驻点都是极值点。
The three types are distinguished by what happens to the sign of f′(x) as x increases through the stationary point.
区分这三类点的依据，是当 x 从左到右经过驻点时 f′(x) 的符号如何变化。
| Type | 类型 | Sign of f′(x) before | 左侧 f′(x) 符号 | Sign of f′(x) after | 右侧 f′(x) 符号 | f″(x) | 二阶导数 |
| Local maximum | 局部极大值点 | positive 正 | negative 负 | f″(x) < 0 |
| Local minimum | 局部极小值点 | negative 负 | positive 正 | f″(x) > 0 |
| Stationary point of inflection | 平稳拐点 | positive or negative 正或负 | same sign 同号 | f″(x) = 0 |
Remember that 'local' does not mean 'global'. A local maximum may be lower than another part of the curve further away, so always read the question carefully before claiming a greatest or least value.
注意"局部"并不等于"全局"。一个局部极大值可能比曲线远处某一点还低，因此在断言最大值或最小值之前，一定要看清题目要求。
2. Finding Stationary Points | 寻找驻点的标准步骤
x-value back into the original equation y = f(x) to obtain the corresponding y-coordinate. Marks are almost always awarded for the correct y-values, so never stop at the x-values alone.
寻找驻点是一个三步流程：先对函数求导得到 f′(x)；再令 f′(x) = 0 并解出 x；最后把每个 x 值代回原方程 y = f(x)，求出相应的 y 坐标。评分标准几乎总会给 y 值分配分数，因此千万不要只算出 x 值就停下。
0) = 2 and y(2) = 8 − 12 + 2 = −2. The stationary points are therefore (0, 2) and (2, −2).
例题：对于 y = x³ − 3x² + 2，导数为 y′ = 3x² − 6x = 3x(x − 2)。令 y′ = 0 得 x = 0 或 x = 2。代回原函数：y(0) = 2，y(2) = 8 − 12 + 2 = −2。因此驻点为 (0, 2) 与 (2, −2)。
If the derivative cannot be factorised easily, use the quadratic formula or your calculator's equation solver, but always show the algebraic step in an examination so that method marks are secure.
如果导数不易因式分解，可使用求根公式或计算器的解方程功能，但在考试中必须写出代数步骤，以确保拿到方法分。
3. The First Derivative Test | 第一导数检验法
The first derivative test examines the sign of f′(x) immediately to the left and immediately to the right of the stationary point. Choose a convenient test value slightly smaller than a and slightly larger than a, evaluate f′(x) at each, and record whether the sign is positive or negative.
第一导数检验法考察驻点左右两侧紧邻位置上 f′(x) 的符号。选取略小于 a 和略大于 a 的方便检验值，分别代入 f′(x)，记录符号是正还是负。
ve to negative, so (0, 2) is a local maximum. At x = 2: choose x = 1.9 giving y′ = −0.57 < 0, and x = 2.1 giving y′ = 0.63 > 0. The sign changes from negative to positive, so (2, −2) is a local minimum.
把它用于 y = x³ − 3x² + 2，其中 y′ = 3x² − 6x。在 x = 0 处：取 x = −0.1 得 y′ = 0.63 > 0，取 x = 0.1 得 y′ = −0.57 < 0。符号由正变负，所以 (0, 2) 是局部极大值点。在 x = 2 处：取 x = 1.9 得 y′ = −0.57 < 0，取 x = 2.1 得 y′ = 0.63 > 0。符号由负变正，所以 (2, −2) 是局部极小值点。
You may present the test as a short sign diagram, writing the sign of f′(x) on each interval between and beyond the critical values. A sign diagram is compact, easy to mark and makes the reasoning visible.
你可以用简短的符号表来呈现检验过程，在临界值分隔出的每个区间上写出 f′(x) 的正负号。符号表简洁、便于阅卷，也能让推理过程一目了然。
4. The Second Derivative Test | 第二导数检验法
imum; if f″(a) > 0 the point is a local minimum; if f″(a) = 0 the test is inconclusive and you must fall back on the first derivative test or examine higher derivatives.
第二导数检验法通常更快。先求出驻点 a，再计算 f″(a)，然后套用规则：若 f″(a) < 0，该点为局部极大值点；若 f″(a) > 0，该点为局部极小值点；若 f″(a) = 0，此检验法失效，必须改用第一导数检验法或考察更高阶导数。
Why does it work? The second derivative measures the rate at which the gradient is changing, which is the concavity of the curve. A negative second derivative means the curve is concave down, like the top of a hill, so a flat point there must be a peak. A positive second derivative means the curve is concave up, like the bottom of a valley, so a flat point there must be a trough.
为什么它成立？二阶导数衡量的是梯度本身的变化率，也就是曲线的凹凸性。二阶导数为负说明曲线向下凹，形如山顶，此处若梯度为零必然是峰点；二阶导数为正说明曲线向上凹，形如谷底，此处若梯度为零必然是谷点。
0, confirming a local minimum. Both tests agree, which is a useful self-check in the exam.
继续上例 y′ = 3x² − 6x，得 y″ = 6x − 6。当 x = 0 时，y″ = −6 < 0，确认为局部极大值；当 x = 2 时，y″ = 6 > 0，确认为局部极小值。两种检验法结论一致，这在考试中是很好的自我检查方式。
5. Points of Inflection | 拐点与平稳拐点
twice differentiable, f″(x) = 0 is a necessary condition, but it is not sufficient: the second derivative must actually change sign.
拐点是曲线凹凸性发生改变的点，由向上凹变为向下凹，或反之。在拐点处，若函数二阶可导，则 f″(x) = 0 是必要条件，但并不充分：二阶导数必须真正发生符号变化。
int of inflection. For y = x⁴, y′ = 4x³ and y″ = 12x². Here y″(0) = 0 but y″ ≥ 0 everywhere, so the concavity never changes and (0, 0) is a local minimum, not an inflection point.
对于 y = x³，y′ = 3x²，y″ = 6x。当 x = 0 时一阶与二阶导数同时为零，且 y″ 由负变正，所以 (0, 0) 是平稳拐点。对于 y = x⁴，y′ = 4x³，y″ = 12x²。此处 y″(0) = 0，但恒有 y″ ≥ 0，凹凸性从未改变，因此 (0, 0) 是局部极小值点，而非拐点。
nt (2, 3). At that point y′ = 3(2)² − 12(2) + 9 = −3, so the curve is still falling steeply; the inflection point is not a stationary point.
拐点不必是驻点。对于 y = x³ − 6x² + 9x + 1，二阶导数为 y″ = 6x − 12，在 x = 2 处为零，得拐点 (2, 3)。在该点 y′ = 3(2)² − 12(2) + 9 = −3，曲线仍在下行，因此该拐点不是驻点。
6. Optimisation Problems | 优化问题
a single variable, state the domain allowed by the physical situation, differentiate, solve f′(x) = 0, and justify that the answer is a maximum or a minimum using the second derivative.
优化问题要求在真实情境中求某个量的最大值或最小值。方法固定：建立只含一个变量的表达式，说明实际情境允许的定义域，求导，解 f′(x) = 0，并用二阶导数论证所得答案是最大值还是最小值。
A′ = 20 − 2x, so A′ = 0 gives x = 10. Since A″ = −2 < 0, this is a maximum, and the greatest area is A = 10 × 10 = 100 cm².
经典例题：矩形周长为 40 cm，设一边为 x cm，另一边为 (20 − x) cm。面积为 A = x(20 − x) = 20x − x²，定义域 0 < x < 20。求导 A′ = 20 − 2x，令 A′ = 0 得 x = 10。由于 A″ = −2 < 0，这是最大值，最大面积为 A = 10 × 10 = 100 cm²。
x. Then S′ = 2x − 2000 ÷ x² = 0 gives x³ = 1000, so x = 10 and h = 5. Since S″ = 2 + 4000 ÷ x³ > 0, this is a minimum, and S = 100 + 200 = 300 cm².
体积例题：一个无盖长方体的底面为边长 x cm 的正方形，体积为 500 cm³。高为 h = 500 ÷ x²，表面积为 S = x² + 4xh = x² + 2000 ÷ x。令 S′ = 2x − 2000 ÷ x² = 0 得 x³ = 1000，故 x = 10，h = 5。由于 S″ = 2 + 4000 ÷ x³ > 0，这是最小值，且 S = 100 + 200 = 300 cm²。
Always finish with a sentence that answers the question in context, for example 'the minimum surface area is 300 cm²'. Examiners award a specific mark for that concluding statement.
最后一定要用一句话结合情境回答问题，例如"最小表面积为 300 cm²"。阅卷人对此结论句设有专门的分值。
7. Stationary Points of Polynomial Functions | 多项式函数的驻点
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