Stationary Points and Extrema: Methods for Finding and Determining | 驻点与极值：求解及判定方法

📚 Stationary Points and Extrema: Methods for Finding and Determining | 驻点与极值：求解及判定方法

In calculus, stationary points are points on a curve where the derivative is zero. They are also known as turning points when the curve changes direction, and they form the foundation for solving optimisation problems in mathematics, physics and economics.

在微积分中，驻点是曲线上导数为零的点。当曲线改变方向时，它们也称为转向点，并且它们是解决数学、物理和经济学中优化问题的基础。

For a differentiable function f(x), a stationary point occurs at x = a if f'(a) = 0. However, not every stationary point is an extremum; some are points of inflection with a horizontal tangent. This article explains how to locate stationary points and how to determine whether they correspond to a local maximum, a local minimum or neither.

对于可微函数 f(x)，当 x = a 时若 f'(a) = 0，则称 x = a 为驻点。然而，并非每个驻点都是极值点；有些是具有水平切线的拐点。本文将解释如何找到驻点，以及如何判断它们对应局部极大值、局部极小值还是两者都不是。

1. What Are Stationary Points? | 什么是驻点？

A stationary point is a point on the graph of a function where the derivative is exactly zero. Graphically, the tangent line at that point is horizontal.

驻点是函数图像上导数恰好为零的点。从图像上看，该点处的切线是水平的。

For example, for f(x) = x², the derivative is f'(x) = 2x. Setting f'(x) = 0 gives x = 0, so the point (0, 0) is a stationary point. For f(x) = x³, f'(x) = 3x², so x = 0 is also a stationary point, but it is not a local extremum because the function is increasing on both sides of x = 0.

例如，对于 f(x) = x²，导数为 f'(x) = 2x。令 f'(x) = 0 得到 x = 0，因此点 (0, 0) 是驻点。对于 f(x) = x³，f'(x) = 3x²，所以 x = 0 也是驻点，但它不是局部极值，因为函数在 x = 0 两侧都递增。

The name “stationary” means the function is momentarily at rest; it is neither increasing nor decreasing at that instant.

“驻”这个名字意味着函数在该瞬间处于静止状态；它既不增加也不减少。

2. Types of Turning Points | 极值点的类型

Stationary points can be classified into three main types:

驻点可分为三种主要类型：

Local maximum: the function changes from increasing to decreasing as x passes through the point.

局部极大值：当 x 通过该点时，函数从递增变为递减。
Local minimum: the function changes from decreasing to increasing as x passes through the point.

局部极小值：当 x 通过该点时，函数从递减变为递增。
Horizontal point of inflection: the function continues increasing or continues decreasing on both sides, while the tangent is horizontal.

水平拐点：函数在两侧继续递增或继续递减，而切线却是水平的。

For example, f(x) = -x² has a local maximum at x = 0; f(x) = x² has a local minimum at x = 0; and f(x) = x³ has a horizontal point of inflection at x = 0.

例如，f(x) = -x² 在 x = 0 处有局部极大值；f(x) = x² 在 x = 0 处有局部极小值；而 f(x) = x³ 在 x = 0 处有水平拐点。

3. Finding Stationary Points | 求解驻点

To find stationary points of a function f(x), follow these steps:

要求函数 f(x) 的驻点，请遵循以下步骤：

Compute the derivative f'(x).

计算导数 f'(x)。
Solve the equation f'(x) = 0.

解方程 f'(x) = 0。
Substitute the x-values back into f(x) to obtain the coordinates of the stationary points.

将 x 值代回 f(x)，得到驻点的坐标。

Consider f(x) = x³ – 3x² + 2. Then f'(x) = 3x² – 6x = 3x(x – 2). Setting f'(x) = 0 gives x = 0 or x = 2. The corresponding y-values are f(0) = 2 and f(2) = 8 – 12 + 2 = -2. Therefore, the stationary points are (0, 2) and (2, -2).

考虑 f(x) = x³ – 3x² + 2。则 f'(x) = 3x² – 6x = 3x(x – 2)。令 f'(x) = 0 得到 x = 0 或 x = 2。对应的 y 值为 f(0) = 2 和 f(2) = 8 – 12 + 2 = -2。因此，驻点为 (0, 2) 和 (2, -2)。

In general, a polynomial of degree n has at most n – 1 stationary points, but some may be repeated or complex.

一般来说，n 次多项式至多有 n – 1 个驻点，但某些可能是重根或复数根。

4. First Derivative Test | 一阶导数判定法

The first derivative test examines the sign of f'(x) immediately to the left and right of a stationary point x = a.

一阶导数判定法通过检查驻点 x = a 左右两侧 f'(x) 的符号来判断极值类型。

Create a sign table for f'(x) around the stationary point:

为驻点附近的 f'(x) 构造一个符号表：

Interval	x < a	x = a	x > a
Sign of f'(x)	+ / −	0	− / +

If f'(x) changes from positive to negative, then f has a local maximum at x = a. If f'(x) changes from negative to positive, then f has a local minimum at x = a. If the sign does not change, the stationary point is a horizontal point of inflection.

如果 f'(x) 从正变为负，则 f 在 x = a 处有局部极大值。如果 f'(x) 从负变为正，则 f 在 x = a 处有局部极小值。如果符号没有变化，则该驻点为水平拐点。

For f(x) = x³ – 3x² + 2, we have f'(x) = 3x(x – 2). The critical points split the real line into three intervals: x < 0, 0 < x < 2, and x > 2. Testing values: f'(-1) = 3(-1)(-3) = 9 > 0; f'(1) = 3(1)(-1) = -3 < 0; f'(3) = 3(3)(1) = 9 > 0. Hence x = 0 is a local maximum and x = 2 is a local minimum.

对于 f(x) = x³ – 3x² + 2，f'(x) = 3x(x – 2)。临界点将实轴分成三个区间：x < 0，0 < x < 2，以及 x > 2。测试值：f'(-1) = 3(-1)(-3) = 9 > 0；f'(1) = 3(1)(-1) = -3 < 0；f'(3) = 3(3)(1) = 9 > 0。因此 x = 0 是局部极大值，x = 2 是局部极小值。

5. Second Derivative Test | 二阶导数判定法

The second derivative test is often quicker. It uses the value of f”(x) at the stationary point.

二阶导数判定法通常更快。它利用驻点处的 f”(x) 值来判断。

If x = a is a stationary point (f'(a) = 0), then:

如果 x = a 是驻点（f'(a) = 0），则：

If f”(a) > 0, then f has a local minimum at x = a, because the curve is concave up.

如果 f”(a) > 0，则 f 在 x = a 处有局部极小值，因为曲线是凹向上的。
If f”(a) < 0, then f has a local maximum at x = a, because the curve is concave down.

如果 f”(a) < 0，则 f 在 x = a 处有局部极大值，因为曲线是凹向下的。
If f”(a) = 0, the test is inconclusive, and you must use the first derivative test.

如果 f”(a) = 0，则判定法失效，必须使用一阶导数判定法。

For f(x) = x³ – 3x² + 2, we find f”(x) = 6x – 6. At x = 0, f”(0) = -6 < 0, so a local maximum. At x = 2, f''(2) = 6 > 0, so a local minimum.

对于 f(x) = x³ – 3x² + 2，f”(x) = 6x – 6。在 x = 0 处，f”(0) = -6 < 0，因此是局部极大值。在 x = 2 处，f''(2) = 6 > 0，因此是局部极小值。

6. When the Second Derivative Is Zero | 当二阶导数为零时

If f”(a) = 0 at a stationary point, the second derivative test gives no information. This situation often occurs at a horizontal point of inflection, but it can also occur at a local extremum if the function is not smooth enough.

如果在驻点处 f”(a) = 0，二阶导数判定法无法提供信息。这种情况通常出现在水平拐点处，但如果函数光滑性不足，也可能出现在局部极值处。

For example, f(x) = x⁴ has f'(x) = 4x³ and f”(x) = 12x². At x = 0, f'(0) = 0 and f”(0) = 0, yet x = 0 is clearly a local minimum because f(x) ≥ 0 for all x. In such cases, examine the sign of f'(x) on both sides or use higher-order derivatives.

例如，f(x) = x⁴ 有 f'(x) = 4x³ 和 f”(x) = 12x²。在 x = 0 处，f'(0) = 0 且 f”(0) = 0，但 x = 0 显然是局部极小值，因为对所有 x，f(x) ≥ 0。在这种情况下，检查 f'(x) 在两侧的符号或使用更高阶导数。

Another example is f(x) = x³, where f”(0) = 0 and the stationary point is a horizontal point of inflection. The first derivative test is reliable: if the sign of f'(x) does not change, it is not an extremum.

另一个例子是 f(x) = x³，其中 f”(0) = 0 且该驻点是水平拐点。一阶导数判定法可靠：如果 f'(x) 的符号不变，则不是极值。

7. Extrema on Closed Intervals | 闭区间上的极值

When the domain of f is a closed interval [a, b], the global maximum and minimum may occur at stationary points inside the interval or at the endpoints. The derivative is not required to be zero at endpoints.

当 f 的定义域是闭区间 [a, b] 时，全局最大值和最小值可能出现在区间内部的驻点处，也可能出现在端点处。在端点处导数不要求为零。

To find the absolute extrema on [a, b]:

要找到在 [a, b] 上的绝对极值：

Find all stationary points inside the interval.

找出区间内的所有驻点。
Evaluate f at every stationary point and at both endpoints a and b.

计算每个驻点以及两个端点 a 和 b 处的 f 值。
The largest value is the global maximum; the smallest value is the global minimum.

最大值即为全局最大值；最小值即为全局最小值。

For example, let f(x) = x³ – 3x² + 2 on [-1, 3]. The stationary points are (0, 2) and (2, -2). Evaluate f(-1) = -1 – 3 + 2 = -2, f(3) = 27 – 27 + 2 = 2. Therefore the global maximum is 2 (attained at x = 0 and x = 3), and the global minimum is -2 (attained at x = 2 and x = -1).

例如，令 f(x) = x³ – 3x² + 2，定义域为 [-1, 3]。驻点为 (0, 2) 和 (2, -2)。计算 f(-1) = -1 – 3 + 2 = -2，f(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 和 x = -1 处取得）。

8. Worked Example 1: Quadratic and Cubic | 例题1：二次与三次函数

Find and classify the stationary points of f(x) = x² – 4x + 5.

求 f(x) = x² – 4x + 5 的驻点并分类。

Step 1: f'(x) = 2x – 4. Set f'(x) = 0, so 2x – 4 = 0 ⇒ x = 2.

第一步：f'(x) = 2x – 4。令 f'(x) = 0，得 2x – 4 = 0 ⇒ x = 2。

Step 2: f(2) = 4 – 8 + 5 = 1. So (2, 1) is the only stationary point.

第二步：f(2) = 4 – 8 + 5 = 1。因此 (2, 1) 是唯一的驻点。

Step 3: f”(x) = 2 > 0, so it is a local minimum. Since it is a quadratic with a positive leading coefficient, it is also the global minimum.

第三步：f”(x) = 2 > 0，因此是局部极小值。由于这是首项系数为正的二次函数，它也是全局极小值。

Now consider f(x) = 2x³ – 9x² + 12x – 3.

现在考虑 f(x) = 2x³ – 9x² + 12x – 3。

f'(x) = 6x² – 18x + 12 = 6(x² – 3x + 2) = 6(x – 1)(x – 2). Setting f'(x) = 0 gives x = 1 and x = 2.

f'(x) = 6x² – 18x + 12 = 6(x² – 3x + 2) = 6(x – 1)(x – 2)。令 f'(x) = 0 得 x = 1 和 x = 2。

Evaluate f(1) = 2 – 9 + 12 – 3 = 2, f(2) = 16 – 36 + 24 – 3 = 1. Now f”(x) = 12x – 18, so f”(1) = -6 < 0 (local maximum at (1, 2)) and f''(2) = 6 > 0 (local minimum at (2, 1)).

计算 f(1) = 2 – 9 + 12 – 3 = 2，f(2) = 16 – 36 + 24 – 3 = 1。现在 f”(x) = 12x – 18，因此 f”(1) = -6 < 0（在 (1, 2) 处为局部极大值），f''(2) = 6 > 0（在 (2, 1) 处为局部极小值）。

9. Worked Example 2: Exponential and Trigonometric | 例题2：指数与三角函数

Find the stationary points of f(x) = eˣ – x.

求 f(x) = eˣ – x 的驻点。

f'(x) = eˣ – 1. Set eˣ – 1 = 0 ⇒ eˣ = 1 ⇒ x = 0. Then f(0) = 1 – 0 = 1, so (0, 1) is the only stationary point.

f'(x) = eˣ – 1。令 eˣ – 1 = 0 ⇒ eˣ = 1 ⇒ x = 0。则 f(0) = 1 – 0 = 1，因此 (0, 1) 是唯一的驻点。

f”(x) = eˣ, and f”(0) = 1 > 0, so it is a local minimum. In fact, this is the global minimum because eˣ > 0 for all real x and the function tends to infinity as x → ±∞.

f”(x) = eˣ，且 f”(0) = 1 > 0，因此是局部极小值。实际上，这是全局极小值，因为 eˣ > 0 对所有实数 x 成立，且当 x → ±∞ 时函数趋于无穷。

Now consider f(x) = sin²x on the interval [0, 2π].

现在考虑 f(x) = sin²x 在区间 [0, 2π] 上。

Using the product rule or the identity sin²x = ½(1 – cos 2x), we get f'(x) = 2 sin x cos x = sin 2x. Setting sin 2x = 0 gives 2x = 0, π, 2π, 3π, 4π, so x = 0, π/2, π, 3π/2, 2π.

使用乘积法则或恒等式 sin²x = ½(1 – cos 2x)，得到 f'(x) = 2 sin x cos x = sin 2x。令 sin 2x = 0 得 2x = 0, π, 2π, 3π, 4π，所以 x = 0, π/2, π, 3π/2, 2π。

Now f”(x) = 2 cos 2x. At x = 0, f”(0) = 2 > 0 (minimum); x = π/2, f”(π/2) = 2 cos π = -2 < 0 (maximum); x = π, f''(π) = 2 > 0 (minimum); x = 3π/2, f”(3π/2) = -2 < 0 (maximum); x = 2π, f''(2π) = 2 > 0 (minimum). Since the interval is closed, these are also the extrema, with the global maximum equal to 1 and global minimum equal to 0.

现在 f”(x) = 2 cos 2x。在 x = 0 处，f”(0) = 2 > 0（极小值）；x = π/2 处，f”(π/2) = 2 cos π = -2 < 0（极大值）；x = π 处，f''(π) = 2 > 0（极小值）；x = 3π/2 处，f”(3π/2) = -2 < 0（极大值）；x = 2π 处，f''(2π) = 2 > 0（极小值）。由于区间是闭区间，这些也是极值，全局最大值为 1，全局最小值为 0。

10. Common Mistakes and Tips | 常见错误与技巧

Students often confuse stationary points with points where the derivative does not exist. Remember that a stationary point requires f'(x) = 0; a point where f'(x) is undefined may still be a local extremum (for example, the corner of f(x) = |x| at x = 0), but it is not a stationary point.

学生经常将驻点与导数不存在的点混淆。请记住，驻点要求 f'(x) = 0；而 f'(x) 未定义的点仍可能是局部极值（例如 f(x) = |x| 在 x = 0 处的尖点），但它不是驻点。

Another common error is relying solely on the second derivative test without checking that f'(a) = 0 first. The second derivative test is only valid at stationary points.

另一个常见错误是只依赖二阶导数判定法，而没有先检查 f'(a) = 0。二阶导数判定法仅在驻点处有效。

When solving f'(x) = 0, always factor the expression fully and check for all solutions, including those outside the domain of interest. Also remember that trigonometric equations often have infinitely many solutions; restrict the search to the given interval.

在解 f'(x) = 0 时，务必完全因式分解并检查所有解，包括那些在目标定义域之外的解。还要记住，三角方程通常有无穷多个解；应在给定区间内搜索。

Finally, draw a rough graph or use a sign table to verify your conclusions, especially when f”(x) = 0 or when the function has discontinuities.

最后，画一个草图或使用符号表来验证你的结论，特别是当 f”(x) = 0 或函数具有不连续性时。

11. Summary | 总结

Stationary points are found by solving f'(x) = 0 and then substituting back to get the coordinates. To classify them, use the first derivative test (sign changes of f’) or the second derivative test (sign of f”). The second derivative test fails when f”(a) = 0; in that case, fall back on the first derivative test.

驻点通过解 f'(x) = 0 并代回得到坐标来找到。要分类它们，可以使用一阶导数判定法（f’ 的符号变化）或二阶导数判定法（f” 的符号）。当 f”(a) = 0 时，二阶导数判定法失效；此时应改用一阶导数判定法。

For global extrema on a closed interval, always compare values at stationary points and endpoints. With practice, locating and classifying extrema becomes a routine but powerful tool in calculus.

对于闭区间上的全局极值，始终比较驻点和端点处的值。通过练习，找到并分类极值将成为微积分中常规但强大的工具。

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Stationary Points and Extrema: Methods for Finding and Determining | 驻点与极值：求解及判定方法