📚 Example 4.2.4: Classifying Stationary Points | 例 4.2.4：驻点判定
This AQA A-level pure mathematics example takes a typical cubic function and shows how to find and classify its stationary points using the second derivative test. The method is a core skill for Paper 1 and appears frequently in differentiation questions.
本 AQA A-level 纯数学例题选取一个典型的三次函数，展示如何利用二阶导数判定法找出并判定其驻点。该方法是卷一的核心技能，经常出现在微分题中。
1. The Worked Example | 例题展示
. We will find all stationary points and determine whether each one is a local maximum or a local minimum.
考虑函数 f(x) = x³ - 3x + 2。我们将找出所有驻点，并判断每个驻点是局部极大值还是局部极小值。
f(x) = x³ - 3x + 2
This example assumes you are confident with differentiating powers of x and solving quadratic equations by factorisation.
本例题假设你已经熟练掌握了 x 的幂函数求导以及用因式分解法解二次方程。
2. Step 1: Differentiate f(x) | 第一步：对 f(x) 求导
Differentiate term by term using the power rule: the derivative of xⁿ is nxⁿ⁻¹. For x³, the derivative is 3x². For -3x, the derivative is -3. The constant 2 has derivative 0.
使用幂函数求导法则逐项求导：xⁿ 的导数为 nxⁿ⁻¹。x³ 的导数是 3x²，-3x 的导数是 -3，常数 2 的导数是 0。
f'(x) = 3x² - 3
Factorise the derivative so that the values of x making f'(x) = 0 can be read quickly. Both terms share a factor of 3.
将导数因式分解，以便快速找出使 f'(x) = 0 的 x 值。两项都有公因数 3。
f'(x) = 3(x² - 1) = 3(x - 1)(x + 1)
Writing the derivative in fully factorised form is a good habit. It reduces sign errors and makes the link to stationary points explicit.
将导数写成完全因式分解的形式是一个好习惯。它可以减少符号错误，并明确展示与驻点的联系。
3. Step 2: Solve f'(x) = 0 | 第二步：解方程 f'(x) = 0
Stationary points occur where the gradient of the curve is zero. Therefore, set the factorised derivative equal to zero and solve for x.
驻点出现在曲线斜率为零的位置。因此，令因式分解后的导数等于零并求解 x。
3(x - 1)(x + 1) = 0
product is zero when either factor is zero. This gives the two solutions x = 1 or x = -1.
因为 3 不可能等于零，所以当任一因式为零时乘积为零。由此得到两个解：x = 1 或 x = -1。
These x-values are the x-coordinates of the stationary points. Do not stop here; you must also find the corresponding y-coordinates.
这些 x 值就是驻点的横坐标。不要到此为止；你还必须求出相应的纵坐标。
4. Step 3: Find the y-coordinates | 第三步：求纵坐标
Substitute each x-value back into the original function f(x), not the derivative. Many students mistakenly substitute into f'(x) and get zero.
将每个 x 值代回原函数 f(x)，而不是导数。许多学生误将 x 值代入 f'(x)，结果得到零。
For x = 1:
当 x = 1 时：
f(1) = (1)³ - 3(1) + 2 = 1 - 3 + 2 = 0
For x = -1:
当 x = -1 时：
f(-1) = (-1)³ - 3(-1) + 2 = -1 + 3 + 2 = 4
The stationary points are therefore (1, 0) and (-1, 4). Always give both coordinates in your final answer unless the question specifies otherwise.
因此驻点为 (1, 0) 和 (-1, 4)。除非题目另有规定，最终答案务必写出两个坐标。
5. Step 4: Compute the Second Derivative | 第四步：计算二阶导数
Differentiate f'(x) = 3x² - 3 to obtain the second derivative, f''(x). Differentiate term by term again.
对 f'(x) = 3x² - 3 求导，得到二阶导数 f''(x)。再次逐项求导。
f''(x) = 6x
The second derivative measures the rate of change of the gradient. Its sign tells us whether the curve is concave up or concave down at a given point.
二阶导数衡量斜率的变化率。它的符号告诉我们曲线在给定点处是凹向上还是凹向下。
In this example f''(x) is linear, so it is easy to evaluate at any point. This makes the second derivative test very efficient.
在本例中 f''(x) 是一个线性函数，因此在任意点求值都很容易。这使得二阶导数判定法非常高效。
6. Step 5: Apply the Second Derivative Test | 第五步：应用二阶导数判定法
Evaluate f''(x) at each x-coordinate of the stationary points. If f''(a) > 0, the point is a local minimum. If f''(a) < 0, the point is a local maximum.
在每个驻点的 x 坐标处计算 f''(x)。如果 f''(a) > 0，该点为局部极小值；如果 f''(a) < 0，该点为局部极大值。
At x = 1, f'' 6 × 1 = 6. This is positive, so the curve is concave up at x = 1, giving a local minimum.
在 x = 1 处，f''(1) = 6 × 1 = 6。该值为正，所以曲线在 x = 1 处凹向上，形成局部极小值。
f''(1) = 6 > 0 ⇒ (1, 0) is a local minimum
f''(1) = 6 > 0 ⇒ (1, 0) 为局部极小值
At x = -1, f'' 1) = -6. This is negative, so the curve is concave down at x = -1, giving a local maximum.
在 x = -1 处，f''(-1) = 6 × (-1) = -6。该值为负，所以曲线在 x = -1 处凹向下，形成局部极大值。
f''(-1) = -6 < 0 ⇒ (-1, 4) is a local maximum
f''(-1) = -6 < 0 ⇒ (-1, 4) 为局部极大值
Always quote the sign of f''(x) and link it clearly to the nature of the stationary point. This is what examiners look for.
务必写出 f''(x) 的符号，并将其与驻点的性质清晰地联系起来。这正是阅卷人所关注的。
7. Interpreting the Results | 结果解读
um at (1, 0). The leading term x³ means that as x → -∞, f(x) → -∞, and as x → ∞, f(x) → ∞.
f(x) = x³ - 3x + 2 的图像在 (-1, 4) 处有局部极大值，在 (1, 0) 处有局部极小值。首项 x³ 表明当 x → -∞ 时 f(x) → -∞，当 x → ∞ 时 f(x) → ∞。
The table below summarises the working. Keeping a summary table can help you check your answers quickly.
下表总结了计算过程。保留一个总结表可以帮助你快速检查答案。
| x | Stationary point | f''(x) | Nature |
|---|---|---|---|
| -1 | (-1, 4) | -6 | Local maximum |
| 1 | (1, 0) | 6 | Local minimum |
Remember that 'local' means the point is a maximum or minimum only within a small neighbourhood, not necessarily the highest or lowest point on the whole curve.
请记住，“局部”意味着该点仅在一个小邻域内是极大值或极小值，不一定是整条曲线上的最高点或最低点。
8. Alternative: First Derivative Sign Test | 替代方法：一阶导数符号法
If f'' d to evaluate, you can use the first derivative sign test instead. Choose test values on either side of each stationary point and check the sign of f'(x).
如果 f''(a) = 0 或二阶导数难以求值，你可以改用一阶导数符号法。在每个驻点两侧选取测试值，并检查 f'(x) 的符号。
For x < -1, say x = -2, f'(-2) = 3(-2)² - 3 = 12 - 3 = 9 > 0. For -1 < x < 1, say x = 0, f' fore the gradient changes from positive to negative at x = -1, confirming a local maximum.
当 x < -1 时，例如 x = -2，f'(-2) = 3(-2)² - 3 = 12 - 3 = 9 > 0。当 -1 < x < 1 时，例如 x = 0，f'(0) = -3 < 0。因此斜率在 x = -1 处由正变负，确认为局部极大值。
Similarly, for x > 1, say x = 2, f' > 0. The gradient changes from negative to positive at x = 1, confirming a local minimum.
类似地，当 x > 1 时，例如 x = 2，f'(2) = 12 - 3 = 9 > 0。斜率在 x = 1 处由负变正，确认为局部极小值。
The first derivative test is often longer, but it is more general. It also works when the second derivative is zero.
一阶导数符号法通常步骤更多
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