📚 Example 5.6.2: Stationary Points & Second Derivative Test | 例题 5.6.2：驻点与二阶导数判别法
This worked example examines a classic AQA A-Level question on finding and classifying stationary points using both the first and second derivatives. You will see the full solution method, the reasoning behind each step, and the way to present your answer in an exam.
本例题详解一道 AQA 数学 A-Level 的经典考题：利用一阶导数和二阶导数求函数的驻点，并判断其是极大值点、极小值点还是拐点。你将看到完整的解题过程、每一步的推导逻辑，以及考试中规范的书写方式。
1. Learning Objectives | 学习目标
stationary points; use f″(x) to classify each stationary point; and interpret the results in the context of the curve's graph.
学完本例后，你应该能够：求三次多项式的导数；令 f′(x) = 0 求出驻点；利用 f″(x) 判断每个驻点的类型；并结合函数图像理解所得结果。
- Differentiate polynomials confidently using the power rule.
- 用幂法则熟练地对多项式求导。
- Solve quadratic equations by factorisation.
- 通过因式分解求解二次方程。
- Apply the second derivative test correctly.
- 正确运用二阶导数判别法。
2. The Problem | 问题陈述
the coordinates of all stationary points and determine their nature, showing full working.
设函数 f(x) = x³ − 6x² + 9x + 1。求该函数所有驻点的坐标，并判断其性质，要求写出完整过程。
f(x) = x³ − 6x² + 9x + 1
This question tests two key skills: differentiation and classification. It appears frequently on Paper 1 of the AQA A-Level Mathematics exam.
本题考查两个关键技能：求导与分类判断。这类题目在 AQA 数学 A-Level 的 Paper 1 中出现频率很高。
3. Step 1: Differentiate | 第一步：求导
We begin by applying the power rule to each term. Recall that the derivative of xⁿ is nxⁿ⁻¹, and the derivative of a constant term is zero.
首先对每一项使用幂法则。回忆：xⁿ 的导数是 nxⁿ⁻¹，常数项的导数为零。
f′(x) = 3x² − 12x + 9
Differentiating term by term: the derivative of x³ is 3x²; the derivative of −6x² is −12x; the derivative of 9x is 9; and the derivative of 1 is 0.
逐项求导：x³ 的导数是 3x²；−6x² 的导数是 −12x；9x 的导数是 9；常数 1 的导数是 0。
4. Step 2: Solve f′(x) = 0 | 第二步：解方程 f′(x) = 0
stationary point, the gradient of the curve is zero. So we set f′(x) = 0 and solve for x.
在驻点处，曲线的切线斜率为零。因此令 f′(x) = 0，并解出 x。
3x² − 12x + 9 = 0
Factor out the common factor of 3 to simplify the equation.
先提取公因数 3，以简化方程。
3(x² − 4x + 3) = 0 ⇒ x² − 4x + 3 = 0
Now factorise the quadratic expression x² − 4x + 3. We look for two numbers that multiply to 3 and add to −4. These are −1 and −3.
接着对二次式 x² − 4x + 3 进行因式分解。我们需要找到两个数，它们相乘等于 3，相加等于 −4。这两个数就是 −1 和 −3。
(x − 1)(x − 3) = 0
e solutions are x = 1 and x = 3. These are the x-coordinates of the two stationary points.
因此解得 x = 1 和 x = 3。这就是两个驻点的横坐标。
5. Step 3: Find the Second Derivative | 第三步：求二阶导数
To classify the stationary points, we use the second derivative. The second derivative measures the rate of change of the gradient, telling us how the curve bends at each point.
为了判断驻点的类型，我们使用二阶导数。二阶导数衡量的是导数（梯度）的变化率，反映曲线在每个点的弯曲方向。
f″(x) = d/dx (3x² − 12x + 9) = 6x − 12
gives f″(x) = 6x − 12. This linear expression will be evaluated at each stationary point.
对 f′(x) 逐项求导，得到 f″(x) = 6x − 12。接下来将把每个驻点的横坐标代入这个一次函数。
6. Step 4: Classify Each Point | 第四步：分类判断
We apply the second derivative test: if f″(x) > 0 at a stationary point, the curve is convex and the point is a local minimum; if f″(x) < 0, the curve is concave and the point is a local maximum.
我们应用二阶导数判别法：若某驻点处 f″(x) > 0，曲线在该点附近凹向上（convex），该点为局部极小值点；若 f″(x) < 0，曲线凹向下（concave），该点为局部极大值点。
First evaluate f″(x) at x = 1.
先计算 x = 1 处的二阶导数。
f″(1) = 6(1) − 12 = −6 < 0
Since f″(1) is negative, the point at x = 1 is a local maximum.
由于 f″(1) 为负数，x = 1 处为局部极大值点。
Next evaluate f″(x) at x = 3.
再计算 x = 3 处的二阶导数。
f″(3) = 6(3) − 12 = 6 > 0
Since f″(3) is positive, the point at x = 3 is a local minimum.
由于 f″(3) 为正数，x = 3 处为局部极小值点。
7. Step 5: Find the y-Coordinates | 第五步：求纵坐标
For a complete answer, we must substitute each x-value back into the original function f(x), not into the derivative. A common error is to substitute into f′(x) instead.
为了给出完整答案，我们必须把每个 x 值代回原函数 f(x)，而不是代回导数。常见的错误是把 x 代入了 f′(x)。
At x = 1:
当 x = 1 时：
f(1) = 1³ − 6(1)² + 9(1) + 1 = 1 − 6 + 9 + 1 = 5
So the local maximum has coordinates (1, 5).
因此局部极大值点的坐标为 (1, 5)。
At x = 3:
当 x = 3 时：
f(3) = 3³ − 6(3)² + 9(3) + 1 = 27 − 54 + 27 + 1 = 1
So the local minimum has coordinates (3, 1).
因此局部极小值点的坐标为 (3, 1)。
| Point | 点 | x | 横坐标 | y | 纵坐标 | f″(x) | Nature | 性质 |
|---|---|---|---|---|
| A | 1 | 5 | −6 | Local maximum | 局部极大值 |
| B | 3 | 1 | 6 | Local minimum | 局部极小值 |
8. Sign-Diagram Verification | 符号表检验
ith a sign diagram of f′(x). This also helps if f″(x) = 0, where the test is inconclusive.
由于二阶导数判别法只反映局部性质，在考试中养成用 f′(x) 的符号表进行验证的好习惯会很有帮助。当 f″(x) = 0 时该方法失效，符号表就显得尤为重要。
the intervals x < 1, 1 < x < 3, and x > 3. We may pick x = 0, x = 2 and x = 4.
在区间 x < 1、1 < x < 3 和 x > 3 内选取测试点，例如 x = 0、x = 2 和 x = 4。
f′(0) = 3(0)² − 12(0) + 9 = 9 > 0
f′(2) = 3(2)² − 12(2) + 9 = 12 − 24 + 9 = −3 < 0
f′(4) = 3(4)² − 12(4) + 9 = 48 − 48 + 9 = 9 > 0
confirms a maximum; moving from negative to positive gradient at x = 3 confirms a minimum.
从左到右的符号模式是：正、零、负、零、正。在 x = 1 处梯度由正变负，证实为极大值；在 x = 3 处梯度由负变正，证实为极小值。
9. Common Mistakes | 常见错误
Below are the most frequently seen errors in AQA mark schemes for this type of question, together with advice on avoiding each one.
以下是在 AQA 评分标准中此类题最常见的错误类型，以及相应的避免建议。
- Substituting x-values into f′(x) instead of f(x) when finding the y-coordinates.
- 求纵坐标时，把 x 值代入了 f′(x) 而不是原函数 f(x)。
- Forgetting to factor out the 3 before solving the quadratic, leading to arithmetic slips.
- 解二次方程前忘记提取公因数 3，导致计算失误。
- Misinterpreting the inequality: remember f″ < 0 gives a maximum, f″ > 0 gives a minimum.
- 搞错不等式方向：记住 f″ < 0 对应极大值，f″ > 0 对应极小值。
- Writing "x = 1 is a maximum" instead of "the point (1, 5) is a maximum" — always give coordinates.
- 只写"x = 1 是极大值点"，而没有给出完整坐标——务必写出坐标 (1, 5)。
10. Practice Question and Summary | 练习与总结
g your answer: Find the stationary points of g(x) = 2x³ − 9x² + 12x + 4 and classify them.
请独立完成以下同类练习后再核对答案：求 g(x) = 2x³ − 9x² + 12x + 4 的驻点并判断其性质。
e (1, 9), a maximum since g″(1) = −6 < 0, and (2, 8), a minimum since g″(2) = 6 > 0.
提示：g′(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2)。驻点为 (1, 9)，g″(1) = −6 < 0，为极大值点；(2, 8)，g″(2) = 6 > 0，为极小值点。
In summary, the method has five steps: differentiate, set the derivative equal to zero, solve for x, compute the second derivative, and classify each stationary point. Then substitute back to give full coordinates. This systematic approach guarantees full marks on any AQA stationary-point question.
总结：解题方法分为五步——求导、令导数为零、解出 x、计算二阶导数、分类判断每个驻点；最后代入原函数得到完整坐标。这种系统化的方法能够帮助你在任何 AQA 驻点类题目中获得满分。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图