📚 Stationary Points | 驻点
In IGCSE Mathematics, understanding stationary points is crucial for analysing curves and solving optimisation problems. A stationary point occurs where the gradient of a function is zero — that is, where the tangent to the curve is horizontal. These points can be local maxima, local minima, or points of inflection. Mastering how to find and classify them using derivatives is a key skill assessed frequently in Edexcel IGCSE exams.
在 IGCSE 数学中，理解驻点对分析曲线和解决优化问题至关重要。驻点出现在函数梯度为零的位置，即曲线切线为水平线。这些点可以是局部极大值、局部极小值或拐点。掌握如何利用导数求出并分类驻点是 Edexcel IGCSE 考试中经常考查的核心技能。
1. What Are Stationary Points? | 什么是驻点？
A stationary point on a curve y = f(x) is any point where the first derivative f' alled turning points, but strictly, a turning point is where the curve changes from increasing to decreasing (or vice versa). A stationary point can also be a point of inflection where the gradient is zero but the curve does not turn — it continues rising or falling after flattening.
曲线 y = f(x) 上的驻点是指一阶导数 f'(x) 等于零的点，即切线斜率为零。在这些点处，曲线瞬间变平。驻点有时也被称为转向点，但严格来说，转向点是指曲线由递增变为递减（或相反）的位置。驻点也可能是拐点，此时梯度为零但曲线并未转向——变平后继续上升或下降。
differentiable everywhere over their domain. The stationary points are found by solving f'(x) = 0, and then you classify them using either the first or second derivative test.
在 Edexcel IGCSE 中，主要处理的都是在其定义域内处处可导的函数。驻点通过求解 f'(x) = 0 得到，然后使用一阶导数检验法或二阶导数检验法进行归类。
2. Finding Stationary Points: Setting f'(x) = 0 | 求驻点：令 f'(x) = 0
ionary points for y = f(x), follow these steps: (i) Differentiate the function to obtain f' onding y-coordinate. (iv) Write the stationary points as coordinates (x, y). Remember that a cubic can give two stationary points, a quadratic usually gives one, and higher-degree polynomials can give more.
求 y = f(x) 的驻点，遵循以下步骤：(i) 对函数求导得到 f'(x) 或 dy/dx。(ii) 令导数为零并解出 x 值。(iii) 将每个 x 值代回原函数求出相应的 y 坐标。(iv) 将驻点以坐标 (x, y) 的形式写出。需要注意的是，三次函数可能有两个驻点，二次函数通常只有一个，更高次多项式可能有更多。
Sometimes the equation f' real; in that case, the function has no stationary points. Also, always express coordinates in exact form unless the question asks for decimal approximations.
有时方程 f'(x) = 0 解出的 x 值不是实数，此时函数没有驻点。此外，除非题目要求近似值，否则坐标务必保留准确形式。
3. Classifying Using the First Derivative Test | 用一阶导数检验法分类
The first derivative test examines how the sign of f'(x) changes as we cross the stationary point. Choose a value just to the left and just to the right of the stationary point's x-coordinate, and calculate the sign of f'(x) at these points. If f'(x) changes from positive to negative, the stationary point is a local maximum. If f'(x) changes from negative to positive, it is a local minimum. If the sign does not change (stays positive or stays negative), the point is a stationary point of inflection.
一阶导数检验法观察经过驻点时 f'(x) 的符号如何变化。选取驻点横坐标左侧和右侧非常接近的数值，计算 f'(x) 的符号。若 f'(x) 由正变负，则该驻点为局部极大值；若由负变正，则为局部极小值；若符号不变（始终为正或始终为负），该点是驻点拐点。
nts at x = −1 and x = 1. For x = −1, pick test points x = −2 (left) and x = 0 (right). You'll find f'(−2) > 0 and f'(0) < 0, so the gradient changes from positive to negative — d is completely reliable and especially useful when the second derivative test gives zero.
例如，考虑函数 f(x) = x³ − 3x，它在 x = −1 和 x = 1 处有驻点。对于 x = −1，选取测试点 x = −2（左侧）和 x = 0（右侧）。计算得 f'(−2) > 0 且 f'(0) < 0，梯度由正变负——故为局部极大值。这种方法完全可靠，在二阶导数检验得零时尤其有用。
4. Classifying Using the Second Derivative Test | 用二阶导数检验法分类
The second derivative test uses the second derivative f''(x) or d²y/dx². After finding a stationary point at x = a, evaluate f''(a). If f''(a) > 0, the curve is concave up (∪-shaped), so the stationary point is a local minimum. If f''(a) < 0, the curve is concave down (∩-shaped), so it is a local maximum. If f''(a) = 0, the test is inconclusive — be a minimum, maximum, or inflection, and you must use the first derivative test instead.
二阶导数检验法利用二阶导数 f''(x) 或 d²y/dx²。求出 x = a 处的驻点后，计算 f''(a)。若 f''(a) > 0，曲线呈凹向上（∪ 形），驻点为局部极小值；若 f''(a) < 0，曲线呈凹向下（∩ 形），则为局部极大值。若 f''(a) = 0，无法判定——此点可能是极小值、极大值或拐点，必须改用一阶导数检验法。
ther method as long as the reasoning is clear. You should write a short conclusion such as "f''(2) = 6 > 0, so (2, −1) is a minimum point."
当二阶导数容易计算且不为零时，该方法通常更快。不过，很多 IGCSE 评分标准只要推理清晰，两种方法都接受。建议写出简短结论，例如 "f''(2) = 6 > 0，故 (2, −1) 为极小值点"。
5. Worked Example 1: Quadratic Function | 例题 1：二次函数
Find the coordinates and nature of the stationary point on y = x² − 4x + 3.
求 y = x² − 4x + 3 的驻点坐标并判断其性质。
Step 1: Differentiate. dy/dx = 2x − 4.
步骤 1：求导。dy/dx = 2x − 4.
Step 2: Set dy/dx = 0. 2x − 4 = 0 ⇒ x = 2.
步骤 2：令 dy/dx = 0。2x − 4 = 0 ⇒ x = 2.
Step 3: Find y. y = (2)² − 4(2) + 3 = 4 − 8 + 3 = −1. So the point is (2, −1).
步骤 3：求 y。y = (2)² − 4(2) + 3 = 4 − 8 + 3 = −1。因此点为 (2, −1)。
e. d²y/dx² = 2. At x = 2, d²y/dx² = 2 > 0, so the point is a local minimum. The curve is a ∪-shaped parabola, confirming this.
步骤 4：二阶导数。d²y/dx² = 2。在 x = 2 处，d²y/dx² = 2 > 0，故该点为局部极小值。曲线为 ∪ 形抛物线，印证了这一点。
6. Worked Example 2: Cubic Function | 例题 2：三次函数
Determine the stationary points of y = x³ − 3x and classify them.
求 y = x³ − 3x 的驻点并进行分类。
Differentiate: dy/dx = 3x² − 3. Set the derivative equal to zero: 3x² − 3 = 0 ⇒ x² = 1 ⇒ x = 1 or x = −1.
求导：dy/dx = 3x² − 3。令导数为零：3x² − 3 = 0 ⇒ x² = 1 ⇒ x = 1 或 x = −1.
s: For x = 1, y = (1)³ − 3(1) = −2 → (1, −2). For x = −1, y = (−1)³ − 3(−1) = 2 → (−1, 2).
求 y 坐标：当 x = 1，y = (1)³ − 3(1) = −2 → (1, −2)。当 x = −1，y = (−1)³ −
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