📚 IB Mathematics: Finding Stationary Points and Determining Extrema | IB数学：驻点的求解与极值判定
In IB Mathematics, stationary points are among the most frequently tested ideas in calculus. They connect the derivative to the shape of a curve and appear in both Analysis and Approaches (AA) and Applications and Interpretation (AI).
在IB数学中，驻点是微积分中最高频的考点之一。它将导数与曲线形态联系起来，在分析与方法（AA）以及应用与解释（AI）两门课程中都会出现。
1. What Are Stationary Points? | 什么是驻点？
A stationary point of a function f is a point on the graph where the tangent line is horizontal. At such a point, the derivative is equal to zero: f'(x) = 0.
函数 f 的驻点是指图像上切线为水平线的点。在驻点处，导数为零：f'(x) = 0。
The word “stationary” reflects the idea that the function momentarily stops increasing or decreasing. However, this pause alone does not tell us whether the point is a peak, a valley, or simply a flat point.
“驻”这个字体现了函数在该点处暂时停止增大或减小的过程。然而，这种”停顿”本身并不能告诉我们该点是波峰、波谷，还是一个单纯的水平拐点。
There are three common types of stationary points: local maximum, local minimum, and stationary inflection point. You need to be able to find them analytically and classify them correctly.
驻点通常分为三类：局部极大值、局部极小值和驻点拐点。你需要能够通过解析方法求出它们并正确分类。
2. The Condition for a Stationary Point: f'(x) = 0 | 驻点的条件：f'(x) = 0
Let c be a number in the domain of f. If f'(c) = 0, then the point (c, f(c)) is a stationary point of f. This is the defining condition.
设 c 是函数 f 定义域内的一个数。如果 f'(c) = 0，则点 (c, f(c)) 就是 f 的一个驻点。这是驻点的定义性条件。
It is important to distinguish between a stationary point and a critical point. A critical point is any point where f'(x) = 0 or f'(x) is undefined. A stationary point is a special critical point where the derivative is zero.
务必要区分驻点与临界点。临界点是指 f'(x) = 0 或 f'(x) 不存在的点；而驻点是导数为零的那一类特殊临界点。
For most IB questions, f is a differentiable polynomial, trigonometric, exponential, or rational function, so you only need to solve f'(x) = 0.
在大多数IB题目中，f 是可导的多项式函数、三角函数、指数函数或有理函数，因此你只需要解方程 f'(x) = 0 即可。
3. Step-by-Step Procedure for Finding Stationary Points | 求驻点的逐步方法
The procedure for finding stationary points can be summarised in four clear steps.
求驻点的过程可以概括为四个清晰的步骤。
- Step 1: Differentiate f(x) to obtain f'(x).
- 步骤1：对 f(x) 求导，得到 f'(x)。
- Step 2: Solve the equation f'(x) = 0.
- 步骤2：解方程 f'(x) = 0。
- Step 3: Substitute each x value back into f(x) to find the y-coordinate.
- 步骤3：将每个 x 值代回 f(x)，求出对应的 y 坐标。
- Step 4: Classify each stationary point using the first or second derivative test.
- 步骤4：利用一阶或二阶导数判定法对每个驻点进行分类。
For example, if f(x) = x² − 4x + 5, then f'(x) = 2x − 4. Solving 2x − 4 = 0 gives x = 2, and f(2) = 1, so the stationary point is (2, 1).
例如，若 f(x) = x² − 4x + 5，则 f'(x) = 2x − 4。解 2x − 4 = 0 得 x = 2，且 f(2) = 1，因此驻点为 (2, 1)。
4. The First Derivative Test | 一阶导数判定法
The first derivative test looks at how the sign of f'(x) changes as x passes through a stationary point c.
一阶导数判定法观察当 x 经过驻点 c 时，f'(x) 的符号如何变化。
If f'(x) changes from positive to negative at c, then f has a local maximum at c. If f'(x) changes from negative to positive at c, then f has a local minimum at c.
如果在 c 两侧 f'(x) 由正变负，则 f 在 c 处取得局部极大值；如果由负变正，则 f 在 c 处取得局部极小值。
If f'(x) does not change sign, then c is neither a local maximum nor a local minimum; it is a stationary inflection point.
如果 f'(x) 的符号不发生变化，那么 c 既不是局部极大值也不是局部极小值，而是驻点拐点。
Sign change + → − : local maximum | 符号 + → − ：局部极大值
Sign change − → + : local minimum | 符号 − → + ：局部极小值
| Interval | Sign of f'(x) | Conclusion |
| x < c | + | Increasing |
| x = c | 0 | Stationary |
| x > c | − | Decreasing |
In the table above, f changes from increasing to decreasing, so c is a local maximum.
在上表中，f 从递增变为递减，因此 c 是局部极大值。
5. The Second Derivative Test | 二阶导数判定法
The second derivative test is usually faster than the first derivative test. You evaluate f”(x) at the stationary point c.
二阶导数判定法通常比一阶导数判定法更快捷。你只需要在驻点 c 处计算 f”(c)。
If f”(c) < 0, then c is a local maximum.
If f”(c) > 0, then c is a local minimum.
If f”(c) = 0, the test is inconclusive.
When f”(c) = 0, the second derivative does not tell us whether the point is a maximum, a minimum, or a stationary inflection. In that case, return to the first derivative test.
当 f”(c) = 0 时，二阶导数无法判断该点是极大值、极小值还是驻点拐点。此时应回到一阶导数判定法。
A common misconception is that f”(c) = 0 always means a point of inflection. This is not true. You must check that the concavity actually changes on either side of c.
一个常见误解是：f”(c) = 0 就一定代表拐点。这是不正确的。你必须检验在 c 的两侧凹凸性是否真的发生了改变。
6. Local Maximum, Local Minimum, and Stationary Inflection | 极大值、极小值与驻点拐点
A local maximum is a point where f(x) is greater than or equal to all nearby values. A local minimum is a point where f(x) is less than or equal to all nearby values.
局部极大值是指 f(x) 不小于附近所有函数值的点；局部极小值是指 f(x) 不大于附近所有函数值的点。
A stationary inflection point occurs where f'(x) = 0 and the concavity of the graph changes. At such a point, the tangent is horizontal, and the curve crosses the tangent line.
驻点拐点出现在 f'(x) = 0 且图像凹凸性改变的位置。在该点处切线水平，并且曲线穿过这条切线。
For example, f(x) = x³ has f'(x) = 3x² and f'(0) = 0. But f”(0) = 0, and the sign of f'(x) is positive on both sides of 0. Therefore (0, 0) is a stationary inflection point, not a local maximum or minimum.
例如，f(x) = x³ 的导数为 f'(x) = 3x²，且 f'(0) = 0。但 f”(0) = 0，并且 f'(x) 在 0 两侧都为正。因此 (0, 0) 是驻点拐点，而不是局部极大值或极小值。
Visually, a stationary inflection point has a horizontal tangent but the curve continues in the same direction after passing through the point.
从图像上看，驻点拐点的切线水平，但曲线经过该点后继续沿同一方向延伸。
7. Worked Example: A Cubic Function | 实例：三次函数
Consider the function f(x) = x³ − 3x² + 2. Find its stationary points and classify them.
考虑函数 f(x) = x³ − 3x² + 2，求其驻点并分类。
First, differentiate f(x).
首先，对 f(x) 求导。
f'(x) = 3x² − 6x = 3x(x − 2)
Set f'(x) = 0 to find stationary points.
令 f'(x) = 0，求出驻点。
3x(x − 2) = 0 ⇒ x = 0 or x = 2
Now calculate the y-coordinates. f(0) = 2 and f(2) = 8 − 12 + 2 = −2, so the stationary points are (0, 2) and (2, −2).
接着计算 y 坐标。f(0) = 2，f(2) = 8 − 12 + 2 = −2，因此驻点为 (0, 2) 和 (2, −2)。
Next, find the second derivative: 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.
接着求二阶导数：f”(x) = 6x − 6。在 x = 0 处，f”(0) = −6 < 0，因此 (0, 2) 是局部极大值；在 x = 2 处，f”(2) = 6 > 0，因此 (2, −2) 是局部极小值。
You can confirm this with a quick sign table around x = 0 and x = 2.
你可以用 x = 0 和 x = 2 附近的符号表来验证这一结果。
8. Worked Example: A Trigonometric Function | 实例：三角函数
Consider g(x) = sin x + cos x on the interval 0 ≤ x ≤ 2π. Find and classify all stationary points.
考虑区间 0 ≤ x ≤ 2π 上的函数 g(x) = sin x + cos x，求所有驻点并分类。
Differentiating gives g'(x) = cos x − sin x. Set g'(x) = 0.
求导得 g'(x) = cos x − sin x。令 g'(x) = 0。
cos x − sin x = 0 ⇒ tan x = 1
In the interval 0 ≤ x ≤ 2π, the solutions are x = π/4 and x = 5π/4.
在区间 0 ≤ x ≤ 2π 内，解为 x = π/4 和 x = 5π/4。
Now g”(x) = −sin x − cos x. At x = π/4, g”(π/4) = −√2 < 0, so this point is a local maximum. At x = 5π/4, g”(5π/4) = √2 > 0, so this point is a local minimum.
现在求 g”(x) = −sin x − cos x。在 x = π/4 处，g”(π/4) = −√2 < 0，因此该点为局部极大值；在 x = 5π/4 处，g”(5π/4) = √2 > 0，因此该点为局部极小值。
The corresponding y-values are g(π/4) = √2 and g(5π/4) = −√2.
对应的 y 值为 g(π/4) = √2，g(5π/4) = −√2。
9. Global vs Local Extrema on a Closed Interval | 闭区间上的全局与局部极值
A local extremum is only compared with nearby values. A global extremum is the largest or smallest value of the function on its entire domain or on a given interval.
局部极值只与附近的值比较；全局极值则是函数在整个定义域或给定区间上的最大值或最小值。
For a continuous function on a closed interval [a, b], the global maximum and minimum occur either at stationary points or at the endpoints a and b.
对于闭区间 [a, b] 上的连续函数，全局最大值和最小值只会出现在驻点处或端点 a、b 处。
To find the global extrema, follow this method.
求全局极值时，按以下方法操作。
- Find all stationary points inside (a, b).
- 找到 (a, b) 内所有驻点。
- Evaluate f at every stationary point.
- 计算 f 在所有驻点处的值。
- Evaluate f(a) and f(b).
- 计算 f(a) 和 f(b)。
- Compare all values. The largest is the global maximum, and the smallest is the global minimum.
- 比较所有值。最大者即为全局最大值，最小者即为全局最小值。
In IB exam questions, always check the endpoints even if the question only asks about stationary points.
在IB考试中，即使题目只问驻点，也要记得检查端点。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
One of the most common mistakes is solving f'(x) = 0 but forgetting to substitute the x-values back into f(x) to write the full coordinate pair.
最常见的错误之一是解出 f'(x) = 0 后，忘记把 x 代回 f(x)，从而没有写出完整的坐标点。
Another common error is using the second derivative test when f”(c) = 0 and concluding “inflection point” without checking the change in concavity.
另一个常见错误是当 f”(c) = 0 时仍使用二阶导数判定法，并且不检查凹凸性变化就直接得出”拐点”的结论。
You should also remember that stationary points are different from turning points. A stationary inflection point is stationary but not a turning point.
你还应记住，驻点与转向点不同。驻点拐点是驻点，但不是转向点。
When solving trigonometric derivative equations, always restrict your solutions to the given domain and use the unit circle carefully.
解三角函数导数方程时，务必把解限制在给定区间内，并谨慎使用单位圆。
Finally, show all working clearly. In IB mark schemes, the derivative, the solution of f'(x) = 0, and the classification each receive separate marks.
最后，要清晰展示全部过程。在IB评分标准中，导数、解 f'(x) = 0 和分类分别独立给分。
11. Practice Questions | 练习
Question 1. Find and classify the stationary points of f(x) = 2x³ − 9x² + 12x − 3.
题目1. 求函数 f(x) = 2x³ − 9x² + 12x − 3 的驻点并分类。
Solution. f'(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2). The stationary points are x = 1 and x = 2. Since f”(x) = 12x − 18, we have f”(1) = −6 < 0, so (1, 2) is a local maximum. Also f”(2) = 6 > 0, so (2, 1) is a local minimum.
解答. f'(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2)，得到驻点 x = 1 和 x = 2。由于 f”(x) = 12x − 18，f”(1) = −6 < 0，因此 (1, 2) 为局部极大值；f”(2) = 6 > 0，因此 (2, 1) 为局部极小值。
Question 2. Find the global maximum and minimum of h(x) = x³ − 3x² + 2 on the interval [0, 3].
题目2. 求函数 h(x) = x³ − 3x² + 2 在区间 [0, 3] 上的全局最大值与最小值。
Solution. From the earlier example, the stationary points in (0, 3) are x = 0? Note x = 0 is an endpoint here? Actually interval [0,3], x=0 is endpoint and also stationary. The stationary point x=2 is inside. h(0)=2, h(2)=−2, h(3)=27−27+2=2. The global maximum is 2, achieved at x=0 and x=3; the global minimum is −2 at x=2.
解答. 由前面的例子，在 (0, 3) 内的驻点主要是 x = 2，而 x = 0 是端点。h(0)=2，h(2)=−2，h(3)=27−27+2=2。因此全局最大值为 2，在 x=0 和 x=3 处取得；全局最小值为 −2，在 x=2 处取得。
Question 3. Show that f(x) = x⁴ has a local minimum at x = 0 even though f”(0) = 0.
题目3. 证明 f(x) = x⁴ 在 x = 0 处有局部极小值，尽管 f”(0) = 0。
Solution. f'(x) = 4x³, so f'(0) = 0. For x < 0, f'(x) < 0; for x > 0, f'(x) > 0. Since f’ changes from negative to positive, x = 0 is a local minimum. This shows why the first derivative test is more reliable when the second derivative is zero.
解答. f'(x) = 4x³，所以 f'(0) = 0。当 x < 0 时，f'(x) < 0；当 x > 0 时，f'(x)
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