📚 Mastering IB Math Exercise 10D.2: Differentiation Applications | 掌握IB数学练习10D.2：微分应用
IB Mathematics Exercise 10D.2 focuses on applying differentiation to curves – finding tangents, normals, stationary points, and intervals of increase or decrease. This article walks you through a typical exam-style problem, explains each step with clear reasoning, and provides tips to avoid common pitfalls.
IB数学练习10D.2的核心内容是微分在曲线中的应用——求切线、法线、驻点以及增减区间。本文将带你逐步解析一道典型考题，清晰阐述每一步的推理过程，并提供避开常见错误的实用建议。
1. Overview of the Exercise | 练习概览
the function is increasing or decreasing, locate stationary points, and sketch the curve.
练习10D.2通常会给出一个多项式函数，要求你探究其性态。一道典型题目可能是：对于曲线 f(x) = x³ − 3x + 2，求在 x = 2 处的切线方程，确定函数的递增和递减区间，找出驻点，并绘制曲线草图。
2. The Power Rule for Derivatives | 导数的幂函数法则
Before tackling the problem, recall the power rule: if f(x) = xⁿ, then f' tive of x³ is 3x², the derivative of −3x is −3, and the derivative of the constant 2 is 0.
在解题之前，回顾幂函数法则：若 f(x) = xⁿ，则 f'(x) = n·xⁿ⁻¹。对于我们的函数 f(x) = x³ − 3x + 2，逐项求导：x³ 的导数是 3x²，−3x 的导数是 −3，常数 2 的导数为 0。
Thus the first derivative is f'(x) = 3x² − 3. This expression gives the gradient of the curve at any point x.
因此一阶导数为 f'(x) = 3x² − 3。该表达式给出了曲线上任意一点 x 处的切斜率。
3. Evaluating the Gradient at a Specific Point | 计算指定点的梯度
ngent equation at x = 2, we first need the gradient at that point. Substitute x = 2 into f'(x): f'(2) = 3·(2)² − 3 = 3·4 − 3 = 9. So the slope of the tangent is 9.
要写出 x = 2 处的切线方程，我们首先需要该点的梯度。将 x = 2 代入 f'(x)：f'(2) = 3·(2)² − 3 = 3·4 − 3 = 9。因此切线斜率为 9。
point on the curve: f(2) = 2³ − 3·2 + 2 = 8 − 6 + 2 = 4. The point of tangency is (2, 4).
我们还需要曲线上该点的 y 坐标：f(2) = 2³ − 3·2 + 2 = 8 − 6 + 2 = 4。切点为 (2, 4)。
4. Equation of the Tangent Line | 切线方程
y − y₁ = m(x − x₁). Substituting gives y − 4 = 9(x − 2), which simplifies to y = 9x − 14.
有了斜率 m = 9 以及点 (2, 4)，使用点斜式：y − y₁ = m(x − x₁)。代入得 y − 4 = 9(x − 2)，化简后为 y = 9x − 14。
ten y = mx + c. Double-check by substituting x = 2: 9·2 − 14 = 4, which matches the curve.
始终按照题目要求的格式给出最终答案，通常写成 y = mx + c 形式。代入 x = 2 进行验证：9·2 − 14 = 4，与曲线上的点相符。
5. Equation of the Normal Line | 法线方程
ne is perpendicular to the tangent. Its gradient is the negative reciprocal of the tangent's slope. Here m_tangent = 9, so mnormal = −1/9.
法线垂直于切线，其斜率为切线斜率的负倒数。此处 m_切线 = 9，因此 m_法线 = −1/9。
Multiply through by 9 to clear the fraction: 9y − 36 = −x + 2, or rearranged: x + 9y = 38.
利用同一点 (2, 4)，法线方程为 y − 4 = (−1/9)(x − 2)。两边乘以 9 以消去分数：9y − 36 = −x + 2，整理得 x + 9y = 38。
6. Sign of the First Derivative and Monotonicity | 一阶导数的符号与单调性
To find where the function is increasing or decreasing, examine the sign of f'(x) = 3x² − 3. Set f'(x) = 0 to find critical values: 3x² − 3 = 0 ⇒ x² = 1 ⇒ x = ±1.
为了确定函数在何处递增或递减，考察 f'(x) = 3x² − 3 的符号。令 f'(x) = 0 求临界值：3x² − 3 = 0 ⇒ x² = 1 ⇒ x = ±1。
These two x-values divide the real line into three intervals: x < −1, −1 < x < 1, and x > 1. Test a point from each interval in f'(x).
这两个 x 值将实数轴划分为三个区间：x < −1、−1 < x < 1 以及 x > 1。在每个区间内选取一个测试点代入 f'(x)。
7. Intervals of Increase and Decrease | 递增与递减区间
Choose x = −2 (for x < −1): f'(−2) = 3·4 − 3 = 9 > 0 ⇒ increasing. Choose x = 0 (for −1 < x < 1): f'(0) = −3 < 0 ⇒ decreasing. Choose x = 2 (for x > 1): f'(2) = 9 > 0 ⇒ increasing.
选取 x = −2（区间 x < −1）：f'(−2) = 3·4 − 3 = 9 > 0 ⇒ 递增。选取 x = 0（区间 −1 < x < 1）：f'(0) = −3 < 0 ⇒ 递减。选取 x = 2（区间 x > 1）：f'(2) = 9 > 0 ⇒ 递增。
−1) and (1, ∞), and decreasing on (−1, 1). Always state intervals using correct notation.
因此，f(x) 在 (−∞, −1) 和 (1, ∞) 上递增，在 (−1, 1) 上递减。务必使用正确的区间记号表述。
8. Locating and Classifying Stationary Points | 定位并分类驻点
Stationary points occur where f' 2 = −1 + 3 + 2 = 4. f(1) = 1³ − 3·1 + 2 = 0. So stationary points are (−1, 4) and (1, 0).
驻点出现在 f'(x) = 0 处，我们已经在 x = −1 和 x = 1 处求得。计算 y 坐标：f(−1) = (−1)³ − 3(−1) + 2 = −1 + 3 + 2 = 4。f(1) = 1³ − 3·1 + 2 = 0。因此驻点为 (−1, 4) 和 (1, 0)。
To classify them, use the sign change of f'(x) or the second derivative. From the intervals above: at x = −1, f' changes from positive to negative ⇒ local maximum. At x = 1, f' changes from negative to positive ⇒ local minimum.
通过 f'(x) 的符号变化或二阶导数进行分类。从上面的区间分析可知：在 x = −1，f' 由正变负 ⇒ 局部极大值；在 x = 1，f' 由负变正 ⇒ 局部极小值。
9. Second Derivative Test | 二阶导数检验
As a cross-check, find the second derivative: f''(x) = 6x. At x = −1, f''(−1) = −6 < 0, confirming a concave-down shape and a local maximum. At x = 1, f''(1) = 6 > 0, confirming a concave-up shape and a local minimum.
作为验证，求出二阶导数：f''(x) = 6x。在 x = −1 处，f''(−1) = −6 < 0，确认曲线下凹，为局部极大值；在 x = 1 处，f''(1) = 6 > 0，确认曲线上凹，为局部极小值。
When the second derivative is zero, the test is inconclusive, but here it gives a clear conclusion that matches the first-derivative sign analysis.
当二阶导数为零时，该检验方法无法得出结论，但此处所得结论明确，且与一阶导数符号分析一致。
10. Sketching the Curve | 绘制曲线草图
With the key features identified, we can sketch the cubic. Plot the stationary points (−1, 4) and (1, 0); the y-intercept is at (0, 2); mark the tangent point (2, 4). The curve rises from the left, turns at the local maximum, falls to the local minimum, then rises again.
明确了关键特征后，我们可以绘制三次曲线草图。标出驻点 (−1, 4) 和 (1, 0)；y 轴截距在 (0, 2)；标出切点 (2, 4)。曲线从左端起上升，在局部极大值处转向，下降至局部极小值，然后再次上升。
to illustrate how the line just touches the curve. The sketch should clearly show the increasing and decreasing intervals.
在 (2, 4) 处添加切线 y = 9x − 14，以显示直线如何恰好与曲线相切。草图中应清晰标出递增和递减区间。
11. Common Mistakes in Exercise 10D.2 | 练习10D.2中的常见错误
A frequent error is forgetting to find the y-coordinate before writing the tangent equation; using only the derivative value gives an incorrect line that may not even touch the curve. Always compute f(x₀).
常见的错误是在写切线方程之前忘记求出 y 坐标；仅使用导数值会导致直线错误，甚至可能根本不触及曲线。务必计算 f(x₀)。
nt, provided the tangent gradient is not zero. Also be careful with algebraic signs when expanding brackets.
另一个错误是错误计算法线斜率。请记住，只要切线斜率不为零，法线斜率 = −1 / 切线斜率。在展开括号时也要注意代数符号。
Students often write intervals of increase/decrease without properly testing values or confuse the sign of the derivative. Always use a sign chart or test points between critical values.
学生常常在未恰当测试数值的情况下直接写出增减区间，或混淆导数的符号。请务必在临界值之间使用符号表或代入测试点。
12. Exam-Style Tips and Summary | 考试技巧与总结
In IB exams, show all steps clearly: derivative calculation, setting f' ifying stationary points using either the first or second derivative test. Both methods are accepted, but the sign-change method is often faster for polynomials.
在 IB 考试中，要清晰展示所有步骤：导数计算、令 f'(x) = 0、进行区间测试，以及使用一阶或二阶导数检验对驻点进行分类。两种方法均可接受，但对于多项式，符号变化法通常更快捷。
Practice Exercise 10D.2 with different functions – quadratics, cubics, and polynomials of higher degree – to become fluent in identifying tangents and curve behaviour. This skill is fundamental for optimisation and modelling problems later in the course.
用不同的函数练习10D.2——二次函数、三次函数以及更高次的多项式——以熟练地识别切线和曲线性态。这项技能是课程后续优化和建模问题的基础。
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