A-Level Mathematics: Finding and Applying Critical Values | A-Level 数学：临界值的求法与应用

📚 A-Level Mathematics: Finding and Applying Critical Values | A-Level 数学：临界值的求法与应用

Critical values play a central role in A-Level Mathematics, especially in differentiation, curve sketching and optimisation problems. In this article, we will explain what critical values are, how to find them, how to classify them, and how to apply them to real examination-style questions.

临界值在 A-Level 数学中占据核心地位，尤其是在微分、曲线绘制和最优化问题中。本文将详细讲解什么是临界值、如何求临界值、如何分类它们，以及如何在典型考试题目中应用这些方法。

1. What Are Critical Values? | 什么是临界值？

A critical point on a curve y = f(x) is a point where the derivative f′(x) is equal to zero, or where f′(x) does not exist. The x-coordinate of such a point is often called a critical value, and the corresponding y-coordinate is called the stationary value.

曲线 y = f(x) 上的临界点是指导数 f′(x) 等于零或 f′(x) 不存在的点。该点的横坐标通常称为临界值，对应的纵坐标称为驻值。

In most A-Level exams, the functions you meet are differentiable, so you mainly look for solutions of f′(x) = 0. These points may be maximum points, minimum points, or points of inflection.

在大多数 A-Level 考试中，所遇到的函数都是可微的，因此主要需要解方程 f′(x) = 0。这些点可能是最大值点、最小值点或拐点。

f′(x) = 0

It is important to remember that “critical value” can mean the x-value itself, while “critical point” means the exact point (x, f(x)) on the curve. Always read the question carefully.

需要特别注意，”临界值”可能指 x 值本身，而”临界点”指曲线上具体的点 (x, f(x))。解题时一定要仔细审题。

2. Differentiating to Find Critical Points | 通过求导找临界点

The first step is to find the derivative of the function. For a polynomial function, you apply the power rule term by term. For example, if f(x) = x³ − 3x² + 2, then f′(x) = 3x² − 6x.

第一步是求出函数的导数。对于多项式函数，逐项使用幂法则即可。例如，若 f(x) = x³ − 3x² + 2，则 f′(x) = 3x² − 6x。

f′(x) = 3x² − 6x = 3x(x − 2)

After differentiating, you set f′(x) = 0 and solve for x. This gives the x-coordinates of any stationary points.

求出导数后，令 f′(x) = 0 并解出 x，即可得到驻点的横坐标。

Always simplify the derivative before solving. Factorising is usually quicker than expanding, and it reduces the risk of algebraic errors.

在解方程之前，一定要先化简导数。因式分解通常比展开更快，而且能减少代数错误的风险。

3. Solving f′(x) = 0 | 解 f′(x) = 0

For a quadratic derivative such as 3x² − 6x, factor out the common term: 3x(x − 2) = 0. This gives x = 0 or x = 2. These are the critical values.

对于形如 3x² − 6x 的二次导数，先提取公因式：3x(x − 2) = 0，于是 x = 0 或 x = 2。这两个值就是临界值。

For rational functions, set the numerator of f′(x) equal to zero. For example, if f′(x) = (x² − 4) ⁄ (x − 1)², then x² − 4 = 0 gives x = ±2.

对于有理函数，令 f′(x) 的分子等于零。例如，若 f′(x) = (x² − 4) ⁄ (x − 1)²，则 x² − 4 = 0 得到 x = ±2。

Remember to check the domain. Any x that makes the denominator zero is not in the domain and cannot be a critical value, even if the numerator is zero.

不要忘记检查定义域。任何使分母为零的 x 都不在定义域内，即使分子为零，也不能作为临界值。

When solving, write each step clearly. A single algebraic slip can change the critical values completely.

解方程时，每一步都要写清楚。一个代数错误就可能完全改变临界值。

4. The Second Derivative Test | 二阶导数检验法

Once the critical values are found, you can classify them using the second derivative f″(x). The second derivative measures the rate of change of the gradient.

找到临界值后，可以用二阶导数 f″(x) 对它们进行分类。二阶导数衡量的是梯度变化的速率。

If f″(x) < 0 at a critical value, the point is a local maximum.
如果 f″(x) < 0，则该点为局部最大值点。
If f″(x) > 0 at a critical value, the point is a local minimum.
如果 f″(x) > 0，则该点为局部最小值点。
If f″(x) = 0, the test is inconclusive, so use another method.
如果 f″(x) = 0，二阶导数检验无效，需要使用其他方法。

For f(x) = x³ − 3x² + 2, we have f″(x) = 6x − 6. At x = 0, f″(0) = −6 < 0, so (0, 2) is a local maximum. At x = 2, f″(2) = 6 > 0, so (2, −2) is a local minimum.

对于 f(x) = x³ − 3x² + 2，f″(x) = 6x − 6。在 x = 0 处，f″(0) = −6 < 0，所以 (0, 2) 是局部最大值点；在 x = 2 处，f″(2) = 6 > 0，所以 (2, −2) 是局部最小值点。

The second derivative test is fast, but it only works when the second derivative is non-zero. Always state your conclusion clearly in an exam.

二阶导数检验非常快捷，但仅在二阶导数不为零时有效。考试中一定要清楚地写出结论。

5. The First Derivative Test | 一阶导数检验法

The first derivative test involves examining the sign of f′(x) just before and just after the critical value. This is especially useful when f″(x) = 0 or is difficult to calculate.

一阶导数检验法是通过观察临界值前后 f′(x) 的符号变化来判断极值类型。当 f″(x) = 0 或二阶导数难以计算时，这个方法尤其有用。

For x = 0 in the example above, take x = −1 and x = 1. f′(−1) = 3(−1)(−3) = 9 > 0, while f′(1) = 3(1)(−1) = −3 < 0. The gradient changes from positive to negative, so x = 0 is a maximum.

以上面的 x = 0 为例，取 x = −1 和 x = 1。f′(−1) = 3(−1)(−3) = 9 > 0，而 f′(1) = 3(1)(−1) = −3 < 0。梯度从正变负，所以 x = 0 是最大值点。

For x = 2, take x = 1 and x = 3. f′(1) < 0 and f′(3) = 3(3)(1) = 9 > 0. The gradient changes from negative to positive, so x = 2 is a minimum.

对于 x = 2，取 x = 1 和 x = 3。f′(1) < 0，而 f′(3) = 3(3)(1) = 9 > 0。梯度从负变正，所以 x = 2 是最小值点。

If the gradient does not change sign, then the critical point is a point of inflection.

如果梯度未发生变号，则该临界点是一个拐点。

6. Points of Inflection | 拐点

A point of inflection is a point where the curve changes from concave to convex, or vice versa. At such a point, f″(x) = 0, and the sign of f″(x) changes on either side.

拐点是曲线凹凸性发生改变的点。在拐点处，f″(x) = 0，且 f″(x) 在两侧变号。

For f(x) = x³ − 3x² + 2, f″(x) = 6x − 6 = 0 gives x = 1. To confirm, check f″(0) = −6 < 0 and f″(2) = 6 > 0. Since the sign changes, (1, 0) is a point of inflection.

对于 f(x) = x³ − 3x² + 2，f″(x) = 6x − 6 = 0 得到 x = 1。为了确认，检查 f″(0) = −6 < 0 和 f″(2) = 6 > 0。由于符号改变，(1, 0) 是一个拐点。

f″(x) = 0 does not always mean a point of inflection

For example, f(x) = x⁴ has f″(0) = 0, but f″(x) = 12x² ≥ 0 on both sides, so the curve does not change concavity. Therefore x = 0 is not a point of inflection.

例如，f(x) = x⁴ 满足 f″(0) = 0，但 f″(x) = 12x² ≥ 0 在两侧都成立，曲线的凹凸性没有改变，因此 x = 0 不是拐点。

Always test the sign of f″(x) on both sides before classifying a point as an inflection.

在判定一个点为拐点之前，一定要检验 f″(x) 在两侧的符号。

7. Critical Values in Optimisation | 最优化问题中的临界值

One of the most common applications of critical values is in optimisation problems, where you have to find the maximum or minimum value of a quantity under given constraints.

临界值最常见的应用之一是最优化问题，即在一定约束条件下求某个量的最大值或最小值。

For example, suppose a rectangular field has a perimeter of 100 metres. Let the length be x metres and the width be 50 − x metres. The area is A = x(50 − x) = 50x − x².

例如，假设一块矩形土地的周长为 100 米。设长为 x 米，则宽为 50 − x 米。面积为 A = x(50 − x) = 50x − x²。

A = 50x − x², so A′(x) = 50 − 2x

Setting A′(x) = 0 gives x = 25. Since A″(x) = −2 < 0, this is a maximum. The maximum area is A(25) = 25 × 25 = 625 m².

令 A′(x) = 0，得到 x = 25。由于 A″(x) = −2 < 0，该点为最大值点。最大面积为 A(25) = 25 × 25 = 625 平方米。

In optimisation questions, always consider the domain. Here 0 < x < 50, and x = 25 lies inside the domain, so the answer is valid.

在最优化问题中，一定要考虑定义域。此处 0 < x < 50，x = 25 在定义域内，因此答案有效。

8. Applications to Sketching Curves | 曲线绘制应用

Critical values are essential when sketching curves. They tell you where turning points occur and help you draw an accurate graph.

绘制曲线时，临界值至关重要。它们能告诉你转向点出现的位置，帮助你画出准确的图像。

To sketch a curve, follow these steps:

绘制曲线时，可以遵循以下步骤：

Find the y-intercept by setting x = 0.
求 y 截距，令 x = 0。
Find the x-intercepts by solving f(x) = 0.
求 x 截距，解方程 f(x) = 0。
Find critical points by solving f′(x) = 0.
求临界点，解方程 f′(x) = 0。
Classify each critical point as maximum, minimum or inflection.
将每个临界点分类为最大值点、最小值点或拐点。
Consider the behaviour as x → ±∞.
考虑 x → ±∞ 时曲线的趋势。

For f(x) = x³ − 3x² + 2, the critical points are (0, 2) and (2, −2), and the inflection is at (1, 0). The curve rises to the maximum, falls through the inflection, reaches the minimum, then rises again.

对于 f(x) = x³ − 3x² + 2，临界点为 (0, 2) 和 (2, −2)，拐点为 (1, 0)。曲线先上升到最大值，然后下降经过拐点，到达最小值，再继续上升。

Always label critical points on your sketch. This shows the examiner that you understand the structure of the curve.

在草图上一律要标出临界点。这可以向考官展示你理解曲线的结构。

9. Sign Diagrams and Tables | 符号图与表格

A sign diagram is a quick way to classify critical values. You draw a number line, mark the critical values, and test the sign of f′(x) in each interval.

符号图是一种快速分类临界值的方法。画一条数轴，标出临界值，然后在每个区间内检验 f′(x) 的符号。

For f(x) = x³ − 3x² + 2, the critical values are 0 and 2. The sign of f′(x) = 3x(x − 2) can be summarised in a table.

对于 f(x) = x³ − 3x² + 2，临界值为 0 和 2。f′(x) = 3x(x − 2) 的符号可以用下表总结。

Interval	x < 0	0 < x < 2	x > 2
Sign of f′(x)	3(−)(−) = +	3(+)(−) = −	3(+)(+) = +
Nature	Increasing	Decreasing	Increasing

This table shows clearly that x = 0 is a maximum because the gradient changes from + to −, and x = 2 is a minimum because the gradient changes from − to +.

该表清晰地表明 x = 0 是最大值点，因为梯度由 + 变为 −；x = 2 是最小值点，因为梯度由 − 变为 +。

When drawing a sign diagram, always include the critical values on the number line and test one point in each interval.

绘制符号图时，一定要在数轴上标出临界值，并在每个区间内测试一个点。

10. Common Pitfalls and Exam Tips | 常见误区与考试技巧

Many students lose marks on critical value questions because of small but avoidable mistakes. Here are the most common pitfalls and how to avoid them.

很多学生在临界值题目中失分，往往是因为一些可避免的小错误。以下是最常见的误区及其避免方法。

Confusing f′(x) = 0 with f(x) = 0. The first finds stationary points, the second finds intercepts.
混淆 f′(x) = 0 与 f(x) = 0。前者求驻点，后者求交点。
Forgetting to solve f′(x) = 0 fully. Factorised equations should each be set to zero.
忘记完整求解 f′(x) = 0。因式分解后的每个因子都应分别令其为零。
Using the second derivative test when f″(x) = 0 and stating a conclusion without checking signs.
当 f″(x) = 0 时，仍然使用二阶导数检验，并且不检查符号就下结论。
Ignoring the domain of the function, especially in optimisation questions.
忽略函数的定义域，尤其是在最优化问题中。
Not showing the y-coordinate of a critical point. The question usually asks for the point, not just the x-value.
没有写出临界点的 y 坐标。题目通常要求点，而不仅仅是 x 值。

In the exam, write down the derivative, solve the equation, classify the point, and state the coordinates clearly. This structure earns full marks.

在考试中，写清导数、解出方程、分类点并明确写出坐标。这种答题结构可以帮助你获得满分。

11. Worked Example | 典型例题

Let us work through a complete example from start to finish.

让我们从头到尾完整地解一道典型例题。

Question: Find and classify the critical points of y = 2x³ − 9x² + 12x − 3.

题目：求 y = 2x³ − 9x² + 12x − 3 的临界点并分类。

Step 1: Differentiate the function.

步骤一：对函数求导。

y′ = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2)

Step 2: Set y′ = 0.

步骤二：令 y′ = 0。

6(x − 1)(x − 2) = 0, so x = 1 or x = 2

Step 3: Find the y-coordinates.

步骤三：求 y 坐标。

For x = 1, y = 2 − 9 + 12 − 3 = 2. For x = 2, y = 16 − 36 + 24 − 3 = 1.

当 x = 1 时，y = 2 − 9 + 12 − 3 = 2。当 x = 2 时，y = 16 − 36 + 24 − 3 = 1。

Step 4: Use the second derivative to classify.

步骤四：使用二阶导数分类。

y″ = 12x − 18

At x = 1, y″ = 12 − 18 = −6 < 0, so (1, 2) is a local maximum. At x = 2, y″ = 24 − 18 = 6 > 0, so (2, 1) is a local minimum.

在 x = 1 处，y″ = 12 − 18 = −6 < 0，所以 (1, 2) 是局部最大值点。在 x = 2 处，y″ = 24 − 18 = 6 > 0，所以 (2, 1) 是局部最小值点。

Step 5: Write the final answer clearly.

步骤五：清楚写出最终答案。

The critical points are a local maximum at (1, 2) and a local minimum at (2, 1).

临界点为局部最大值 (1, 2) 和局部最小值 (2, 1)。

12. Summary | 总结

Critical values are found by solving f′(x) = 0. They can be classified as maxima, minima or points of inflection using the second derivative test or the first derivative test.

临界值通过解 f′(x) = 0 来求得。可以利用二阶导数检验法或一阶导数检验法将它们分类为最大值点、最小值点或拐点。

In optimisation problems, critical values give the locations of maximum or minimum quantities. In curve sketching, they show the turning points and the overall shape of the graph.

在最优化问题中，临界值给出了最大值或最小值的位置。在曲线绘制中，它们显示了转向点和图形的整体形状。

Always check the domain, verify the nature of each critical point, and present your working in a logical order. With practice, critical value questions become straightforward and highly rewarding in exams.

始终检查定义域，验证每个临界点的性质，并按逻辑顺序展示你的解题过程。通过练习，临界值题目会变得简单直接，并在考试中带来丰厚的回报。

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A-Level Mathematics: Finding and Applying Critical Values | A-Level 数学：临界值的求法与应用