A-Level Maths Question 13: Stationary Points and Curve Sketching | A-level 数学第十三题：驻点与曲线绘制

📚 A-Level Maths Question 13: Stationary Points and Curve Sketching | A-level 数学第十三题：驻点与曲线绘制

This article works through a typical AQA A-Level Mathematics question on finding stationary points and determining their nature. The function we will analyse is f(x)=x²e^-x, a classic example that tests your understanding of the product rule, solving equations, and the second derivative test.

本文将为 AQA A-Level 数学考试中的一道典型题目进行详细解析。我们将分析函数 f(x)=x²e^-x，它考查了乘积法则、解方程以及二阶导数判别法。

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1. Introduction to the Question | 问题介绍

The question asks: "The function f is defined by f(x)=x²e^-x for x ∈ ℝ. Find the stationary points of y=f(x) and determine their nature." This is a standard pure mathematics problem that appears in Paper 1 of the AQA A-Level specification.

题目要求：“已知函数 f(x)=x²e^-x，其中 x ∈ ℝ。求曲线 y=f(x) 的驻点，并判断其性质。”这是 AQA A-Level 数学卷一中的经典纯数问题。

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2. The Product Rule for Differentiation | 乘积法则求导

Since f(x) is a product of x² and e^-x, we apply the product rule: if u=x² and v=e^-x, then u'=2x and v'=-e^-x. Therefore:

由于 f(x) 是 x² 与 e^-x 的乘积，我们用乘积法则：设 u=x²，v=e^-x，则 u'=2x，v'=-e^-x。因此：

f'(x)=2x e^-x - x²e^-x

Factorising gives f'(x)=e^-x(2x-x²).

提取公因式得 f'(x)=e^-x(2x-x²)。

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3. Setting the Derivative to Zero | 令导数等于零

Stationary points occur when f'(x)=0. Since e^-x is always positive (never zero), we only need to solve 2x-x²=0.

驻点出现在 f'(x)=0 时。由于 e^-x 恒为正（不为零），我们只需解 2x-x²=0。

2x - x² = 0

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4. Solving for x | 解出 x

Factorise the equation: x(2-x)=0. Hence x=0 or x=2. These are the x-coordinates of the stationary points.

对方程因式分解：x(2-x)=0。因此 x=0 或 x=2，这两个就是驻点的横坐标。

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5. Finding the y-Coordinates | 求 y 坐标

Substitute the x-values back into f(x)=x²e^-x:

将 x 值代回 f(x)=x²e^-x：

x	y=f(x)
0	02e0=0
2	22e-2=4e-2

Thus the stationary points are (0,0) and (2,4e^-2). Note that 4e^-2≈ 0.541.

因此驻点为 (0,0) 和 (2,4e^-2)。注意 4e^-2≈ 0.541。

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6. The Second Derivative | 二阶导数

To determine whether each point is a local minimum, maximum, or point of inflection, we compute f''(x). Starting from f'(x)=e^-x(2x-x²), we differentiate again using the product rule:

为了判断每个驻点是局部极小值、极大值还是拐点，我们计算 f''(x)。从 f'(x)=e^-x(2x-x²) 出发，再次用乘积法则求导：

f''(x)=e^-x(2-2x) - e^-x(2x-x²)

Simplify:

化简：

f''(x)=e^-x(2-2x-2x+x²)=e^-x(x²-4x+2)

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7. Applying the Second Derivative Test | 应用二阶导数测试

Evaluate f''(x) at x=0 and x=2:

分别在 x=0 与 x=2 处计算 f''(x)：

• At x=0: f''(0)=e⁰(0-0+2)=2>0, so (0,0) is a local minimum.
• 在 x=0：f''(0)=e⁰(0-0+2)=2>0，所以 (0,0) 是局部极小值点。
• At x=2: f''(2)=e^-2(4-8+2)=e^-2(-2)=-2e^-2<0, so (2,4e^-2) is a local maximum.
• 在 x=2：f''(2)=e^-2(4-8+2)=e^-2(-2)=-2e^-2<0，所以 (2,4e^-2) 是局部极大值点。

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8. Curve Sketching | 曲线绘制

To sketch the curve, note that as x → -∞, x² → ∞ and e^-x → ∞, so f(x) → ∞. As x → +∞, x² grows but e^-x → 0 much faster; in fact f(x) → 0 from above. The curve passes through (0,0), rises to a maximum at (2,4e^-2), then decays towards the x-axis. The x-axis is a horizontal asymptote as x → ∞.

画草图时，注意当 x → -∞ 时，x² → ∞，且 e^-x → ∞，所以 f(x) → ∞。当 x → +∞ 时，x² 增大但 e^-x → 0 的速度快得多；事实上 f(x) → 0（从上侧）。曲线经过 (0,0)，上升到最高点 (2,4e^-2)，然后逐渐衰减逼近 x 轴。x 轴是曲线在 x → ∞ 时的水平渐近线。

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9. The First Derivative Test (Alternative) | 一阶导数测试（替代方法）

If you are unsure about the second derivative, you can use the first derivative test. For x<0, choose x=-1: f'(-1)=e¹(-2-1)<0. For \(00\). For x>2, choose x=3: f'(3)=e^-3(6-9)<0. The sign changes from negative to positive at x=0 (minimum), and from positive to negative at x=2 (maximum).

如果对二阶导数不确定，可以用一阶导数测试。例如取 x=-1：f'(-1)=e¹(-2-1)<0；取 x=1：f'(1)=e^-1(2-1)>0；取 x=3：f'(3)=e^-3(6-9)<0。在 x=0 处导数由负变正（极小值），在 x=2 处由正变负（极大值）。

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10. Common Exam Mistakes | 常见考试错误

• Forgetting that e^-x is never zero, so the equation 2x-x²=0 is the only necessary condition.
• 忘记 e^-x 永远不为零，因此只需解 2x-x²=0。
• Misapplying the product rule, especially the derivative of e^-x (some write +e^-x instead of -e^-x).
• 乘积法则运用错误，尤其是 e^-x 的导数（有人写成 +e^-x 而不是 -e^-x）。
• When evaluating f''(x), failing to simplify and then making sign errors.
• 计算 f''(x) 时没有化简，导致符号判断错误。
• Not giving the y-coordinates as exact values (e.g., writing 0.54 instead of 4e^-2).
• 没有给出 y 坐标的精确值（例如写 0.54 而不是 4e^-2）。

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11. Practice Question | 练习题

Try this similar question: "Find the stationary points of y=x³e^-x and classify them." The derivative will be e^-x(3x²-x³). Set it to zero, solve x²(3-x)=0, giving x=0 and x=3. Check the nature using the second derivative. (Answer: (0,0) is a horizontal inflection, (3,27e^-3) is a maximum.)

尝试一道类似的题目：“求 y=x³e^-x 的驻点并分类。”导数为 e^-x(3x²-x³)，令其为零，解得 x=0 与 x=3。用二阶导数判断性质。（答案：(0,0) 是水平拐点，(3,27e^-3) 是极大值点。）

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12. Summary | 总结

In this question we used the product rule to find the derivative, set it to zero, solved for x, found the y-coordinates, and classified the stationary points using the second derivative. Always remember to check that your x-values are valid by confirming the exponential factor never vanishes. This method works for any function of the form P(x)e^kx.

本题中我们运用乘积法则求导，令导数为零，解出 x，求出 y 坐标，再用二阶导数判断驻点性质。务必记住检查 x 值是否有效，因为指数因子永远不会为零。这个方法适用于任何 P(x)e^kx 形式的函数。

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