📚 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 \(f(x)=x^{2}e^{-x}\), 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 \(f(x)=x^{2}e^{-x}\)，它考查了乘积法则、解方程以及二阶导数判别法。
1. Introduction to the Question | 问题介绍
The question asks: “The function \(f\) is defined by \(f(x)=x^{2}e^{-x}\) for \(x \in \mathbb{R}\). 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 \(f(x)=x^{2}e^{-x}\)，其中 \(x \in \mathbb{R}\)。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。
2. The Product Rule for Differentiation | 乘积法则求导
Since \(f(x)\) is a product of \(x^{2}\) and \(e^{-x}\), we apply the product rule: if \(u=x^{2}\) and \(v=e^{-x}\), then \(u’=2x\) and \(v’=-e^{-x}\). Therefore:
由于 \(f(x)\) 是 \(x^{2}\) 与 \(e^{-x}\) 的乘积，我们用乘积法则：设 \(u=x^{2}\)，\(v=e^{-x}\)，则 \(u’=2x\)，\(v’=-e^{-x}\)。因此：
\(f'(x)=2x e^{-x} – x^{2}e^{-x}\)
Factorising gives \(f'(x)=e^{-x}(2x-x^{2})\).
提取公因式得 \(f'(x)=e^{-x}(2x-x^{2})\)。
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 \(2x-x^{2}=0\).
驻点出现在 \(f'(x)=0\) 时。由于 \(e^{-x}\) 恒为正（不为零），我们只需解 \(2x-x^{2}=0\)。
\(2x – x^{2} = 0\)
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。
5. Finding the y-Coordinates | 求 y 坐标
Substitute the x-values back into \(f(x)=x^{2}e^{-x}\):
将 x 值代回 \(f(x)=x^{2}e^{-x}\)：
| \(x\) | \(y=f(x)\) |
| 0 | \(0^{2}e^{0}=0\) |
| 2 | \(2^{2}e^{-2}=4e^{-2}\) |
Thus the stationary points are \((0,0)\) and \((2,4e^{-2})\). Note that \(4e^{-2}\approx 0.541\).
因此驻点为 \((0,0)\) 和 \((2,4e^{-2})\)。注意 \(4e^{-2}\approx 0.541\)。
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 \(f'(x)=e^{-x}(2x-x^{2})\), 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 \(f'(x)=e^{-x}(2x-x^{2})\) 出发，再次用乘积法则求导：
\(f”(x)=e^{-x}(2-2x) – e^{-x}(2x-x^{2})\)
Simplify:
化简：
\(f”(x)=e^{-x}(2-2x-2x+x^{2})=e^{-x}(x^{2}-4x+2)\)
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 \(x=0\): \(f”(0)=e^{0}(0-0+2)=2>0\), so \((0,0)\) is a local minimum.
- 在 \(x=0\)：\(f”(0)=e^{0}(0-0+2)=2>0\)，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。
8. Curve Sketching | 曲线绘制
To sketch the curve, note that as \(x \to -\infty\), \(x^{2} \to \infty\) and \(e^{-x} \to \infty\), so \(f(x) \to \infty\). As \(x \to +\infty\), \(x^{2}\) grows but \(e^{-x} \to 0\) much faster; in fact \(f(x) \to 0\) 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 \(x \to \infty\).
画草图时，注意当 \(x \to -\infty\) 时，\(x^{2} \to \infty\)，且 \(e^{-x} \to \infty\)，所以 \(f(x) \to \infty\)。当 \(x \to +\infty\) 时，\(x^{2}\) 增大但 \(e^{-x} \to 0\) 的速度快得多；事实上 \(f(x) \to 0\)（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 \(x \to \infty\) 时的水平渐近线。
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 \(x=-1\): \(f'(-1)=e^{1}(-2-1)<0\). 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 \(x=-1\)：\(f'(-1)=e^{1}(-2-1)<0\)；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）。
10. Common Exam Mistakes | 常见考试错误
- Forgetting that \(e^{-x}\) is never zero, so the equation \(2x-x^{2}=0\) is the only necessary condition.
- 忘记 \(e^{-x}\) 永远不为零，因此只需解 \(2x-x^{2}=0\)。
- 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 \(4e^{-2}\)）。
11. Practice Question | 练习题
Try this similar question: “Find the stationary points of \(y=x^{3}e^{-x}\) and classify them.” The derivative will be \(e^{-x}(3x^{2}-x^{3})\). Set it to zero, solve \(x^{2}(3-x)=0\), 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 \(y=x^{3}e^{-x}\) 的驻点并分类。”导数为 \(e^{-x}(3x^{2}-x^{3})\)，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。）
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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