📚 Increasing and Decreasing Functions | 函数的单调性
In A-Level Mathematics, understanding how a function behaves—whether it rises or falls as x increases—is essential for curve sketching and optimisation. The derivative f'(x) provides a powerful tool to determine these intervals of increase and decrease.
在A-Level数学中，理解函数随x增加而上升或下降的变化规律，对于描绘曲线和优化问题至关重要。导数 f'(x) 提供了一个强有力的工具来确定这些递增和递减的区间。
1. Derivative and Slope of Tangent | 导数与切线斜率
The derivative of a function at a point gives the gradient of the tangent line at that point. If the tangent slopes upward (positive gradient), the function is increasing at that instant. If it slopes downward (negative gradient), the function is decreasing. This geometric link forms the foundation for studying monotonicity.
函数在某点的导数给出了该点切线的斜率。如果切线向上倾斜（斜率为正），函数在该点处递增；如果切线向下倾斜（斜率为负），函数在该点处递减。这种几何联系构成了研究单调性的基础。
For a differentiable function f, the sign of f'(x) over an interval tells us the overall trend. A positive derivative means the y-values are growing as x increases, while a negative derivative means they are shrinking.
对于可导函数f，区间上 f'(x) 的符号告诉我们总体趋势。正导数意味着随着x增大y值在增长，而负导数意味着y值在减小。
2. Conditions for Increasing Functions | 递增函数的条件
A function f is said to be strictly increasing on an interval if for any two numbers x₁ and x₂ in that interval, x₁ < x₂ implies f(x₁) < f(x₂). If f is differentiable, this condition is usually equivalent to f'(x) > 0 throughout the interval.
如果对于区间内任意两点x₁和x₂，当x₁ < x₂时有 f(x₁) < f(x₂)，则称f在该区间上严格递增。如果f可导，这一条件通常等价于在整个区间上 f'(x) > 0。
In Edexcel questions, you will typically find 'increasing' intervals by solving f'(x) > 0. Be aware that some functions, like f(x)=x³, have f'(x) ≥ 0 for all x and are increasing everywhere even though f' es not necessarily break the increasing nature; you need to check the sign on either side.
在Edexcel考题中，通常通过求解 f'(x) > 0 来找出“递增”区间。需要注意，某些函数如 f(x)=x³ 满足 f'(x) ≥ 0 对所有x成立，即使 f'(0)=0 但函数整体仍是递增的。导数在孤立点处为零并不一定破坏递增性质；你需要检查两边的符号。
3. Conditions for Decreasing Functions | 递减函数的条件
Similarly, a function f is strictly decreasing on an interval if x₁ < x₂ leads to f(x₁) > f(x₂). With differentiability, this is matched by f'(x) < 0 on that interval. To find where a graph falls, we solve the inequality f'(x) < 0.
类似地，如果在区间上 x₁ < x₂ 导致 f(x₁) > f(x₂)，则函数严格递减。在可导情形下，这对应于该区间上 f'(x) < 0。为了找出图像下降的部分，我们求解不等式 f'(x) < 0。
Just as with increasing functions, isolated points where the derivative equals zero do not automatically make a function non-decreasing; you must investigate the sign chart. A function can be decreasing on two separate intervals with a stationary point in between.
和递增函数一样，导数在孤立点为零不会自动使函数不递减；必须研究符号表。函数可以在两个分开的区间上递减，中间夹着一个驻点。
4. Critical Points and Stationary Points | 临界点与驻点
A critical point occurs where f'(x) = 0 or where f' es not exist (but f is defined). Stationary points are a subset of critical points where f' crease and decrease because the sign of the derivative can only change at critical points.
临界点出现在 f'(x) = 0 或 f'(x) 不存在（但f有定义）的地方。驻点是临界点的一个子集，指 f'(x) = 0 的点。在确定递增和递减区间时，这些点是关键的分界线，因为导数的符号只能在临界点处改变。
Examples of critical points that are not stationary: a cusp where the derivative is undefined, or a vertical tangent. Even if a function is not differentiable at a point, that point still splits the real line and must be considered when building a sign diagram for f'(x).
不是驻点的临界点例子：尖点处导数无定义，或垂直切线情形。即使函数在某点不可导，该点仍会分割实数轴，在制作 f'(x) 的符号图时必须予以考虑。
5. Step-by-Step: Finding Intervals of Increase/Decrease | 逐步法：求函数的增减区间
To systematically determine where a function is increasing or decreasing, follow these steps:
- Find the domain of f, noting any breaks or excluded values.
- Compute f'(x) and simplify.
- Solve f'(x) = 0 and identify where f'(x) is undefined.
- List these critical points in increasing order and mark them on a number line, splitting the domain into intervals.
- Choose a test point in each interval and evaluate the sign of f'(x).
- Conclude: intervals where f'(x) > 0 are increasing, where f'(x) < 0 are decreasing.
系统地确定函数增减区间的步骤如下：
- 找出f的定义域，注意间断点或排除的值。
- 计算 f'(x) 并进行化简。
- 解方程 f'(x) = 0 并找出 f'(x) 无定义的点。
- 将这些临界点按从小到大的顺序排列，标注在数轴上，将定义域分割成若干区间。
- 在每个区间内选择一个测试点，确定 f'(x) 的符号。
- 结论：f'(x) > 0 的区间为递增区间，f'(x) < 0 的区间为递减区间。
6. Worked Example 1: Polynomial Function | 例题1：多项式函数
Consider f(x) = x³ − 6x² + 9x. Its derivative is f'(x) = 3x² − 12x + 9. Factorising gives f'(x) = 3(x² − 4x + 3) = 3(x − 1)(x − 3). Set f'(x) = 0:
3(x − 1)(x − 3) = 0 → x = 1 or x = 3
考虑函数 f(x) = x³ − 6x² + 9x。其导数为 f'(x) = 3x² − 12x + 9。因式分解得 f'(x) = 3(x² − 4x + 3) = 3(x − 1)(x − 3)。令 f'(x) = 0：
3(x − 1)(x − 3) = 0 → x = 1 或 x = 3
No points where f' Create a sign table with intervals (−∞, 1), (1, 3) and (3, ∞). Test values: x = 0 gives f'(0) = 9 > 0; x = 2 gives f'(2) = 3(2−1)(2−3) = −3 < 0; x = 4 gives f'(4) = 3(3)(1) = 9 > 0.
由于是多项式，f'(x) 没有无定义的点。创建符号表，区间为 (−∞, 1), (1, 3) 和 (3, ∞)。测试值：x = 0 得 f'(0) = 9 > 0；x = 2 得 f'(2) = 3(2−1)(2−3) = −3 < 0；x = 4 得 f'(4) = 3(3)(1) = 9 > 0。
Therefore, f is increasing on (−∞, 1) ∪ ationary points at x = 1 (local maximum) and x = 3 (local minimum) match this sign change.
因此，f 在 (−∞, 1) ∪ (3, ∞) 上递增，在 (1, 3) 上递减。x = 1 处的驻点是局部极大值，x = 3 处的驻点是局部极小值，与符号变化一致。
7. Worked Example 2: Rational Function | 例题2：有理函数
Let f(x) = x + 1/x, defined for x ≠ 0. First, find f'(x) = 1 − 1/x² = (x² − 1)/x². f'(x) = 0 when x² − 1 = 0, i.e. x = ±1. Also, f' is undefined there too, we do not include 0 in intervals but it still splits the domain).
设 f(x) = x + 1/x，定义域为 x ≠ 0。首先求导：f'(x) = 1 − 1/x² = (x² − 1)/x²。当 x² − 1 = 0 时 f'(x) = 0，即 x = ±1。同时，f'(x) 在 x = 0 处无定义，该点是临界点（尽管f在此处也无定义，我们不将0纳入区间，但它仍分割定义域）。
s into intervals: (−∞, −1), (−1, 0), (0, 1) and (1, ∞). Choose test points: x = −2 gives f'(−2) = (4 − 1)/4 = 0.75 > 0; x = −0.5 gives f'(−0.5) = (0.25 − 1)/0.25 = −3 < 0; x = 0.5 gives f'(0.5) = −3 < 0; x = 2 gives f'(2) = 0.75 > 0.
定义域分割为区间：(−∞, −1), (−1, 0), (0, 1) 和 (1, ∞)。选择测试点：x = −2 得 f'(−2) = (4 − 1)/4 = 0.75 > 0；x = −0.5 得 f'(−0.5) = (0.25 − 1)/0.25 = −3 < 0；x = 0.5 得 f'(0.5) = −3 < 0；x = 2 得 f'(2) = 0.75 > 0。
Thus, f is increasing on (−∞, −1) ∪ (1, ∞) and decreasing on (−1, 0) ∪ (0, 1). Note that x = 0 is a vertical asymptote, so behaviour there is separate.
因此，f 在 (−∞, −1) ∪ (1, ∞) 上递增，在 (−1, 0) ∪ (0, 1) 上递减。注意 x = 0 是垂直渐近线，该点附近的行为需单独考虑。
8. Using the First Derivative Test | 使用一阶导数检验
aviour around critical points. After finding critical points, observe the sign change of f'(x): if f' e sign does not change, it is an inflection point with horizontal tangent (e.g., f(x)=x³).
一阶导数检验不仅有助于确定局部极值点，还能确认临界点附近的增减行为。找出临界点后，观察 f'(x) 的符号变化：如果 f' 在临界点处由正变负，函数有局部极大值；由负变正则对应局部极小值。如果符号不变，则是具有水平切线的拐点（如 f(x)=x³）。
This test is particularly useful when the second derivative is cumbersome to compute. Simply build the sign chart for f'(x) and read off the intervals of increase and decrease, which automatically reveals the nature of each stationary point.
当二阶导数计算麻烦时，这一检验特别有用。只需构建 f'(x) 的符号表，读出递增和递减区间，即可自动揭示每个驻点的性质。
9. Common Mistakes and Exam Tips | 常见错误与考试技巧
Mistake 1: Forgetting to consider the domain. If a function has restrictions (denominators, square roots, logs), the sign chart must respect those breaks. A point where f is undefined cannot belong to an increasing or decreasing interval.
错误一：忘记考虑定义域。如果函数有限制（分母、平方根、对数），符号表必须尊重这些间断点。函数无定义的点不能包含在递增或递减区间内。
Mistake 2: Assuming the sign changes at every critical point. Some stationary points are points of inflection where f'(x) does not change sign. Always plug in test values rather than guessing alternate signs.
错误二：假定符号在每个临界点都改变。有些驻点是拐点，f'(x) 的符号并不改变。一定要代入测试值，而不是靠猜测符号交替。
g (e.g., f(x) = x with x negative). Always refer to the derivative, not the original function's output sign.
错误三：混淆递增/递减与 f(x) 的符号。函数值本身可以为负但函数却在递增（例如 f(x)=x 在x为负时）。务必以导数为准，而非原函数的输出符号。
signs) to earn method marks. Clearly state final intervals using interval notation, e.g., 'Increasing for x ∈ (−∞, 1) ∪ (3, ∞)'.
考试建议：绘制整洁的符号图（数轴附测试值和符号），以便获得方法分。用区间记号清晰陈述最终区间，如 “递增区间为 x ∈ (−∞, 1) ∪ (3, ∞)”。
10. Graphical Interpretation and Real-World Applications | 图像解释与实际应用
On a graph, an increasing function moves upwards as we go from left to right; a decreasing function moves downwards. By studying f'(x), we can sketch the shape of the curve without plotting every point. This is the core of curve sketching: you first locate turning points, then decide where the graph rises and falls.
在图上，从左到右看，递增函数向上攀升，递减函数向下滑落。通过研究 f'(x)，我们无需描出每个点也能勾画曲线的形状。这是曲线描绘的核心：首先定位转折点，然后确定图像何处上升何处下降。
and decreasing functions describe growth and decay. For example, if s(t) is displacement, then s' itive direction (increasing displacement). In business, a profit function P(q) increasing for q < 100 and decreasing afterwards suggests an optimal production level at the stationary point.
在实际应用中，递增和递减函数描述增长与衰减。例如，若 s(t) 是位移，则 s'(t) = v(t) > 0 表示向正方向运动（位移递增）。在商业中，利润函数 P(q) 在 q < 100 时递增、之后递减，暗示着在驻点处存在最优生产水平。
By mastering the sign of the derivative, you gain the ability to analyse dynamic systems, optimise outcomes, and understand the behaviour of mathematical models found in physics, economics and biology.
通过掌握导数的符号，你能够分析动态系统、优化结果，并理解物理、经济学和生物学中数学模型的行为。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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