📚 Example 4.5.3: Stationary Points of a Cubic Function | 例题4.5.3：三次函数的驻点
Welcome to this detailed walkthrough of Example 4.5.3 from the AQA A-Level Mathematics Pure syllabus. This example focuses on one of the most fundamental skills in differentiation: locating the stationary points of a cubic function and classifying them correctly using the second derivative. Understanding this process is essential for curve sketching, optimisation problems and many areas of applied mathematics.
欢迎来到 AQA A-Level 数学纯数部分的例题 4.5.3 详细讲解。本例题聚焦于微分学中最核心的技能之一：找到三次函数的驻点，并利用二阶导数对其进行正确分类。理解这一过程对于曲线作图、最优化问题以及应用数学的众多领域都至关重要。
1. Understanding Stationary Points | 理解驻点
A stationary point is a point on a curve where the gradient is exactly zero. At such a point, the curve is momentarily horizontal: it is neither increasing nor decreasing. Algebraically, this happens when the first derivative, f′(x), is equal to zero. Stationary points can be local maxima, local minima, or stationary points of inflection, and distinguishing between them is a key exam skill.
驻点是指曲线上梯度恰好为零的点。在这一点处，曲线暂时保持水平：既不增加也不减少。从代数角度看，当一阶导数 f′(x) 等于零时，就会出现驻点。驻点可以是局部极大值、局部极小值或驻点型拐点，而区分这三者是考试中的关键技能。
Graphically, imagine a roller coaster: at the top of a hill the gradient is zero, and at the bottom of a valley the gradient is also zero. These correspond to a maximum and a minimum respectively. A stationary point of inflection is rarer; the gradient is zero but the curve continues in the same general direction, just flattening out momentarily.
从图形上想象过山车：在山顶时梯度为零，在山谷底部时梯度也为零。它们分别对应极大值和极小值。驻点型拐点较为少见；此时梯度为零，但曲线整体继续沿相同方向延伸，只是暂时趋于平坦。
2. The Problem Statement | 题目描述
In this example, we investigate the cubic function f(x) = x³ − 6x² + 9x + 1. The task is stated in three parts: first, find the coordinates of the stationary points of this function; second, determine the nature of each stationary point using the second derivative test; and third, briefly confirm that the curve behaves as expected over the entire real line.
本例中，我们考察三次函数 f(x) = x³ − 6x² + 9x + 1。题目分为三个部分：第一，求出该函数驻点的坐标；第二，利用二阶导数判别法判断每个驻点的性质；第三，简要确认曲线在整个实数范围内的行为符合预期。
Before diving into the computation, it is always worth looking at the leading coefficient. Here, the x³ term has a positive coefficient of 1, so we know that as x → +∞, f(x) → +∞, and as x → −∞, f(x) → −∞. This means the curve enters from the bottom left and leaves toward the top right, which will help us interpret our results at the end.
在进行计算之前，观察首项系数总是值得的。这里 x³ 项的系数为 1，为正数，因此我们知道当 x → +∞ 时，f(x) → +∞；当 x → −∞ 时，f(x) → −∞。这意味着曲线从左下方进入，向右上方离开，这将在最后帮助我们解读计算结果。
3. Step 1: Differentiate the Function | 第一步：对函数求导
To locate stationary points, we must first find the derivative of f(x). Applying the power rule 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 the constant 1 is 0 — we obtain the first derivative:
为了确定驻点，我们首先需要求出 f(x) 的导数。运用幂法则逐项求导——x³ 的导数为 3x²，−6x² 的导数为 −12x，9x 的导数为 9，常数项 1 的导数为 0——我们得到一阶导数：
f′(x) = 3x² − 12x + 9
Notice that every term in the derivative has a common factor of 3. Factorising out this common factor simplifies the expression considerably and makes the next step much easier. We can also factorise the quadratic inside the bracket directly.
注意导数中每一项都有公因数 3。提取这个公因数可以大大简化表达式，并使下一步的计算更加容易。我们还可以直接分解括号内的二次式。
Factorising further, we recognise that x² − 4x + 3 = (x − 1)(x − 3). Hence the derivative can be written in fully factorised form, which is ideal for solving the equation f′(x) = 0.
继续分解，我们可以看出 x² − 4x + 3 = (x − 1)(x − 3)。因此导数可以写成完全因式分解的形式，这对于求解方程 f′(x) = 0 最为理想。
f′(x) = 3(x − 1)(x − 3)
4. Step 2: Solve f′(x) = 0 | 第二步：解方程 f′(x) = 0
By definition, stationary points occur where the first derivative equals zero. We therefore set the factorised derivative equal to zero and solve for x:
根据定义，驻点出现在一阶导数等于零的地方。因此，我们令分解后的导数等于零，并求解 x：
3(x − 1)(x − 3) = 0
Since 3 is a nonzero constant, it does not affect the solution; at least one of the two brackets must be zero. This gives us two linear equations: x − 1 = 0 or x − 3 = 0. Solving each gives the x-coordinates of the stationary points:
由于 3 是非零常数，它不影响解；两个括号中至少有一个必须为零。这给出了两个一次方程：x − 1 = 0 或 x − 3 = 0。分别求解得到驻点的 x 坐标：
x = 1 or x = 3
It is important to remember that a quadratic derivative always yields at most two stationary points. Here, because the discriminant of the derivative is positive, we obtain two distinct real values of x, meaning the curve has two separate stationary points.
请务必记住，二次导数最多只能产生两个驻点。此处，由于导数的判别式为正，我们得到两个不同的实数 x 值，这意味着曲线有两个独立的驻点。
5. Step 3: Calculate the y-Coordinates | 第三步：计算 y 坐标
A stationary point is a coordinate pair, not just an x-value. We must substitute each x-value back into the original function f(x). Substituting x = 1:
驻点是一个坐标对，而不仅仅是一个 x 值。我们必须将每个 x 值代回原函数 f(x)。代入 x = 1：
f(1) = 1³ − 6(1)² + 9(1) + 1 = 1 − 6 + 9 + 1 = 5
Similarly, substituting x = 3:
同样地，代入 x = 3：
f(3) = 3³ − 6(3)² + 9(3) + 1 = 27 − 54 + 27 + 1 = 1
Thus the two stationary points have coordinates (1, 5) and (3, 1). A very common mistake in the exam is to stop here and give only the x-values. Always remember to go back to the original function — never substitute into the derivative.
因此，两个驻点的坐标为 (1, 5) 和 (3, 1)。考试中一个非常常见的错误是停在第一步，只给出 x 值。务必记得代回原函数——绝不要代入导数。
6. Step 4: The Second Derivative Test | 第四步：二阶导数判别法
To determine whether each stationary point is a maximum or a minimum, we use the second derivative. For the function f(x) = x³ − 6x² + 9x + 1, we differentiate f′(x) = 3x² − 12x + 9 term by term:
为了判断每个驻点是极大值还是极小值，我们使用二阶导数。对于函数 f(x) = x³ − 6x² + 9x + 1，我们对 f′(x) = 3x² − 12x + 9 逐项求导：
f″(x) = 6x − 12
The rule is straightforward: if f″(x) < 0 at a stationary point, the curve bends downward, giving a local maximum; if f″(x) > 0, the curve bends upward, giving a local minimum. We evaluate the second derivative at each stationary x-value in turn.
判别规则很简单：如果在驻点处 f″(x) < 0，曲线向下弯曲，得到局部极大值；如果 f″(x) > 0，曲线向上弯曲，得到局部极小值。我们依次在每个驻点的 x 值处计算二阶导数。
At x = 1:
在 x = 1 处：
f″(1) = 6(1) − 12 = −6 < 0
Since the second derivative is negative, the point (1, 5) is a local maximum. At x = 3:
由于二阶导数为负，点 (1, 5) 是局部极大值。在 x = 3 处：
f″(3) = 6(3) − 12 = 6 > 0
Since the second derivative is positive, the point (3, 1) is a local minimum. The second derivative test is quick and reliable when f″(x) is nonzero, which is the case here.
由于二阶导数为正，点 (3, 1) 是局部极小值。当 f″(x) 非零时，二阶导数判别法快速可靠，本例正是如此。
7. Summary Table and Interpretation | 汇总表格与结果解读
It is often helpful to organise the results in a table. This makes your reasoning clear to the examiner and reduces the risk of misreading your own work. Below is the standard summary for this example.
将结果整理成表格往往很有帮助。这能让考官清楚地看到你的推理过程，并降低误读自己作答的风险。以下是本例题的标准汇总表。
| Stationary Point | x | f″(x) | Sign of f″(x) | Conclusion |
| (1, 5) | 1 | −6 | Negative (< 0) | Local Maximum |
| (3, 1) | 3 | 6 | Positive (> 0) | Local Minimum |
Since the leading coefficient of the cubic is positive, the curve rises from the bottom left, reaches a maximum at (1, 5), descends to a minimum at (3, 1), and then rises again toward the top right. This is consistent with the classical cubic shape and confirms that our classification is correct.
由于三次函数的首项系数为正，曲线从左下方上升，在 (1, 5) 处达到极大值，下降到 (3, 1) 处的极小值，然后再次向右上方上升。这与经典三次函数形状一致，也证实了我们的分类是正确的。
8. Alternative Method: The First Derivative Test | 备用方法：一阶导数符号判别法
Although the second derivative test works beautifully here, there are occasions when f″(x) = 0 at a stationary point. In such cases, the test is inconclusive, and we must fall back on the first derivative test. This involves checking the sign of f′(x) on either side of each stationary point.
虽然二阶导数判别法在本例中非常有效，但在某些情况下，驻点处可能满足 f″(x) = 0。此时二阶导数判别法失效，我们必须改用一阶导数符号判别法。这种方法是在每个驻点的两侧检查 f′(x) 的符号。
Let us apply this to x = 1 as a verification. Choose a value just below 1, say x = 0, and a value just above 1, say x = 2. Using the factorised derivative f′(x) = 3(x − 1)(x − 3): at x = 0, we have f′(0) = 3(−1)(−3) = 9 > 0; at x = 2, we have f′(2) = 3(1)(−1) = −3 < 0. The gradient changes from positive to negative, confirming that (1, 5) is a local maximum.
让我们以 x = 1 为例进行验证。选择略小于 1 的值，如 x = 0，以及略大于 1 的值，如 x = 2。利用因式分解后的导数 f′(x) = 3(x − 1)(x − 3)：在 x = 0 处，f′(0) = 3(−1)(−3) = 9 > 0；在 x = 2 处，f′(2) = 3(1)(−1) = −3 < 0。梯度从正变为负，确认 (1, 5) 是局部极大值。
For x = 3, test x = 2 and x = 4: f′(2) = −3 < 0 and f′(4) = 3(3)(1) = 9 > 0. The gradient changes from negative to positive, confirming a local minimum at (3, 1). This sign-change method is slower but always works, even when the second derivative is zero.
对于 x = 3，取 x = 2 和 x = 4 进行测试：f′(2) = −3 < 0，f′(4) = 3(3)(1) = 9 > 0。梯度从负变为正，确认 (3, 1) 是局部极小值。这种符号变化法虽然较慢，但始终有效，即使二阶导数为零时也能使用。
9. Common Mistakes to Avoid | 需要避免的常见错误
Students frequently lose marks on stationary-point questions for avoidable reasons. Here are the most common pitfalls, along with guidance on how to avoid each one. The first mistake is forgetting to factorise the derivative before solving, which makes the algebra unnecessarily difficult and invites errors.
学生经常因一些可以避免的原因在驻点题目上丢分。以下是最常见的陷阱以及相应的规避建议。第一个错误是忘记在求解前对导数进行因式分解，这会使代数运算不必要地复杂化，并容易引入错误。
The second mistake is substituting the x-values back into the derivative instead of the original function when finding y-coordinates. The derivative gives the gradient, not the height of the curve. The third mistake is confusing the sign rule: negative second derivative means maximum, positive means minimum — many students write this backwards.
第二个错误是在求 y 坐标时，将 x 值代回导数而不是原函数。导数给出的是梯度，而不是曲线的高度。第三个错误是混淆符号规则：二阶导数为负意味着极大值，为正意味着极小值——很多学生会写反。
- Always solve f′(x) = 0, not f(x) = 0. | 始终解 f′(x) = 0，而不是 f(x) = 0。
- Substitute back into f(x) to find y-coordinates. | 代回 f(x) 求 y 坐标。
- Write the final conclusion using full sentences: “local maximum at (1, 5)”. | 用完整句子写出最终结论，例如 “局部极大值点为 (1, 5)”。
- If f″(x) = 0 at a stationary point, use the first derivative test. | 如果驻点处 f″(x) = 0，请使用一阶导数符号判别法。
Finally, do not forget to state the coordinates of both stationary points in your final answer. An answer that gives only the x-values or only the classification earns only partial credit. A clear, structured layout is rewarded by examiners.
最后，别忘了在最终答案中写出两个驻点的完整
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