Example 6.2.1: Stationary Points and Their Nature | 示例6.2.1：驻点及其性质

📚 Example 6.2.1: Stationary Points and Their Nature | 示例6.2.1：驻点及其性质

This article works through Example 6.2.1 from the AQA A-Level Mathematics pure mathematics syllabus. The problem asks us to find the stationary points of a cubic function and to determine whether each one is a local maximum or a local minimum. This type of question appears regularly in the AS and A-Level exams, so mastering the method is very important.

本文选自AQA A-Level数学纯数部分中的示例6.2.1。题目要求我们求出一个三次函数的驻点，并判断每个驻点是局部极大值还是局部极小值。这类问题在AS和A-Level考试中频繁出现，因此掌握其方法是十分重要的。

1. Problem Statement | 问题陈述

The curve \(y = x^3 – 6x^2 + 9x + 1\) is given in the question. Find the stationary points on the curve and determine their nature.

题目给出了曲线 \(y = x^3 – 6x^2 + 9x + 1\)。求曲线上的驻点并判断其性质。

y = x³ – 6x² + 9x + 1

2. Step 1: Differentiate the Function | 第一步：对函数求导

To locate stationary points, we first differentiate the function with respect to x. This gives the gradient function, also known as the first derivative.

为了找到驻点，我们首先对函数关于x求导。这样就得到了斜率函数，也就是一阶导数。

dy/dx = 3x² – 12x + 9

Differentiating term by term:

逐项求导得：

The derivative of x³ is 3x². x³的导数为3x²。
The derivative of -6x² is -12x. -6x²的导数为-12x。
The derivative of 9x is 9. 9x的导数为9。
The derivative of the constant 1 is 0. 常数1的导数为0。

3. Step 2: Solve dy/dx = 0 | 第二步：解 dy/dx = 0

Stationary points occur where the gradient is zero. Therefore we set the first derivative equal to zero and solve for x.

驻点出现在梯度为零的地方。因此我们令一阶导数等于零，并解出x。

3x² – 12x + 9 = 0

Divide both sides by 3 to simplify the equation.

两边同时除以3，以简化方程。

x² – 4x + 3 = 0

Factorise the quadratic expression.

对二次表达式进行因式分解。

(x – 1)(x – 3) = 0

Therefore the solutions are:

因此解为：

x = 1
x = 3

4. Step 3: Find the Corresponding y-coordinates | 第三步：求对应的y值

Substitute x = 1 and x = 3 back into the original equation to obtain the y-values of the stationary points.

将x = 1和x = 3分别代入原方程，求出驻点对应的y值。

For x = 1:

当x = 1时：

y = 1³ – 6(1)² + 9(1) + 1 = 1 – 6 + 9 + 1 = 5

The first stationary point is (1, 5).

第一个驻点为 (1, 5)。

For x = 3:

当x = 3时：

y = 3³ – 6(3)² + 9(3) + 1 = 27 – 54 + 27 + 1 = 1

The second stationary point is (3, 1).

第二个驻点为 (3, 1)。

5. Step 4: Determine the Nature Using the Second Derivative | 第四步：用二阶导数判断性质

The second derivative tells us about the curvature of the curve. It is obtained by differentiating dy/dx again with respect to x.

二阶导数描述曲线的凹凸性。我们对dy/dx再次关于x求导，得到二阶导数。

d²y/dx² = 6x – 12

Now evaluate the second derivative at each stationary point.

现在在每个驻点处计算二阶导数的值。

At x = 1:

当x = 1时：

d²y/dx² = 6(1) – 12 = -6

Since the second derivative is negative, the curve is concave down, so (1, 5) is a local maximum.

因为二阶导数为负，曲线是向下凹的，所以(1, 5)是局部极大值。

At x = 3:

当x = 3时：

d²y/dx² = 6(3) – 12 = 6

Since the second derivative is positive, the curve is concave up, so (3, 1) is a local minimum.

因为二阶导数为正，曲线是向上凹的，所以(3, 1)是局部极小值。

6. Step 5: Interpret the Results | 第五步：解释结果

The curve has a local maximum at (1, 5) and a local minimum at (3, 1). Since the leading coefficient of the cubic is positive, the graph rises from the left, reaches a maximum, falls down through a minimum, and then rises again to the right.

曲线在(1, 5)处有局部极大值，在(3, 1)处有局部极小值。由于三次函数的首项系数为正，图像从左侧上升，到达极大值后下降，经过极小值，然后向右继续上升。

We can also find the y-intercept by setting x = 0 in the original equation.

我们还可以在原方程中令x = 0来求y轴截距。

y = 0³ – 6(0)² + 9(0) + 1 = 1

So the curve crosses the y-axis at (0, 1).

因此曲线与y轴的交点为(0, 1)。

7. Graphical Interpretation | 图像意义

On the graph, the local maximum appears as a ‘hill’ and the local minimum appears as a ‘valley’. At each stationary point, the tangent is horizontal, meaning that the gradient is zero.

在图像上，局部极大值表现为“山峰”，局部极小值表现为“山谷”。在每个驻点处，切线是水平的，也就是说梯度为零。

For this particular cubic, the graph starts from negative infinity on the left, passes through (0, 1), rises to the peak at (1, 5), then descends to the trough at (3, 1), and finally climbs toward positive infinity on the right.

对于这个特定的三次函数，图像从左侧的负无穷开始，经过(0, 1)，上升到顶点(1, 5)，然后下降到谷底(3, 1)，最后向右上方趋向正无穷。

This shape is characteristic of a cubic function with three distinct real roots or one repeated root; here there is only one real root because the minimum is above the x-axis.

这种形状是三次函数的典型特征。注意此曲线的最小值高于x轴，因此该函数只有一个实数根。

8. Common Mistakes to Avoid | 容易犯的错误

Students often lose marks due to small errors in algebra or notation. Here are some common pitfalls to watch out for.

学生在代数或记号上的小错误经常导致失分。以下是一些常见的陷阱，需要格外注意。

Forgetting to set the derivative equal to zero. 忘记令导数等于零。
Failing to substitute x-values back into the original equation. 忘记将x值代回原方程求y值。
Mixing up the sign of the second derivative test. 混淆二阶导数检验的正负号。
Making a mistake when factorising the quadratic. 因式分解二次式时出错。
Using dy/dx incorrectly or omitting the dx. 错误使用dy/dx或漏掉dx。

9. Exam Tips for AQA | AQA考试提示

AQA examiners expect clear setting out. Always state the derivative, solve the equation explicitly, and clearly show the second derivative test. Use the correct notation dy/dx and d²y/dx², and write all equations on a new line where possible.

AQA考官希望考生步骤清晰。务必写出导数，明确地解方程，并清楚展示二阶导数检验。使用正确的记号dy/dx和d²y/dx²，并且尽可能将方程另起一行书写。

Furthermore, check whether the question asks for “stationary points” or “turning points”. Stationary points include points of inflection where the gradient is zero, though in this example both are turning points.

此外，注意题目要求的是“驻点”还是“转向点”。驻点包括梯度为零的拐点，不过在这个例子中两个点都是转向点。

In the AQA exam, these skills are often assessed in the Pure Mathematics papers (Paper 1 and Paper 2). You may be asked to apply this method to different types of functions, such as quadratics, exponentials, or trigonometric functions.

在AQA考试中，这些技能通常在纯数试卷（卷1和卷2）中考查。你可能会被要求对不同类型的函数运用这一方法，例如二次函数、指数函数或三角函数。

10. Practice Question | 练习题

Try the following question independently, using the same method as Example 6.2.1.

请独立尝试以下问题，使用与示例6.2.1相同的方法。

Find the stationary points of y = x³ – 3x + 2 and determine their nature.

The answer is: local maximum at (-1, 4) and local minimum at (1, 0).

答案是：局部极大值在(-1, 4)，局部极小值在(1, 0)。

dy/dx = 3x² – 3 = 0 ⇒ x = -1 or x = 1

d²y/dx² = 6x, so at x = -1, d²y/dx² = -6 (maximum); at x = 1, d²y/dx² = 6 (minimum).

11. Summary | 总结

The method for finding and classifying stationary points can be summarised in four steps:

求驻点并判断性质的方法可以总结为四个步骤：

Differentiate the function to find dy/dx. 对函数求导，得到dy/dx。
Solve dy/dx = 0 to find the x-coordinates of the stationary points. 解dy/dx = 0，求出驻点的x坐标。
Substitute these x-values into the original equation to find the y-coordinates. 将这些x值代入原方程，求出y坐标。
Use the second derivative d²y/dx² to determine whether each stationary point is a maximum, minimum, or point of inflection. 利用二阶导数d²y/dx²判断每个驻点是极大值、极小值还是拐点。

Mastering this procedure is essential for AQA A-Level Mathematics. It links together differentiation, algebra, and graph sketching in a single unified skill.

掌握这一过程对AQA A-Level数学至关重要。它将微分、代数和图像绘制统一成一个综合技能。

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Example 6.2.1: Stationary Points and Their Nature | 示例6.2.1：驻点及其性质