Example 4.2.4: Optimizing a Cubic Function | 例题4.2.4：三次函数的最优化

📚 Example 4.2.4: Optimizing a Cubic Function | 例题4.2.4：三次函数的最优化

This worked example, taken from an AQA A-Level Mathematics practice set, demonstrates how to find the maximum and minimum values of a cubic function on a closed interval using differentiation. In particular, it highlights the importance of checking both critical points and endpoints.

这个例题选自 AQA A-Level 数学练习集，演示如何通过微分求三次函数在闭区间上的最大值和最小值。它特别强调同时检查临界点和区间端点的重要性。

1. Problem Statement | 问题陈述

Consider the function f(x) = 2x³ − 9x² + 12x + 5 on the closed interval [0, 3]. Determine the maximum and minimum values of f(x) on this interval.

考虑函数 f(x) = 2x³ − 9x² + 12x + 5 在闭区间 [0, 3] 上。求 f(x) 在该区间上的最大值和最小值。

2. Find the First Derivative | 求一阶导数

To locate stationary points, we first compute the first derivative of f(x). Differentiating term by term:

为了找到驻点，我们首先计算 f(x) 的一阶导数。逐项求导：

f ′(x) = 6x² − 18x + 12

This derivative will be used to find where the gradient of the curve is zero.

此导数将用于找到曲线斜率为零的位置。

3. Find Critical Points | 求临界点

Set the derivative equal to zero and solve for x:

令导数等于零并解出 x：

6x² − 18x + 12 = 0

Divide both sides by 6 to simplify:

两边同时除以 6 以化简：

x² − 3x + 2 = 0

Factorise the quadratic:

对二次式进行因式分解：

(x − 1)(x − 2) = 0

Thus the critical points are x = 1 and x = 2. Both lie inside the interval [0, 3].

因此临界点为 x = 1 和 x = 2。两者都在区间 [0, 3] 内。

4. Second Derivative Test | 二阶导数检验

To classify the critical points, we compute the second derivative:

为了判断临界点的类型，我们计算二阶导数：

f ′′(x) = 12x − 18

Evaluate f ′′(x) at each critical point:

在每个临界点处求 f ′′(x) 的值：

At x = 1: f ′′(1) = 12(1) − 18 = −6 < 0, so x = 1 is a local maximum.

在 x = 1 处：f ′′(1) = 12(1) − 18 = −6 < 0，因此 x = 1 是局部极大值点。
At x = 2: f ′′(2) = 12(2) − 18 = 6 > 0, so x = 2 is a local minimum.

在 x = 2 处：f ′′(2) = 12(2) − 18 = 6 > 0，因此 x = 2 是局部极小值点。

5. Evaluate Function at Critical Points | 计算临界点处的函数值

Now we substitute x = 1 and x = 2 into the original function:

现在将 x = 1 和 x = 2 代入原函数：

f(1) = 2(1)³ − 9(1)² + 12(1) + 5 = 2 − 9 + 12 + 5 = 10

f(2) = 2(2)³ − 9(2)² + 12(2) + 5 = 16 − 36 + 24 + 5 = 9

So the local maximum value is 10 and the local minimum value is 9. But we must also consider the endpoints.

因此局部极大值为 10，局部极小值为 9。但我们必须同时考虑区间端点。

6. Evaluate Function at Endpoints | 计算端点处的函数值

For a closed interval, the absolute maximum and minimum can occur at the endpoints. Evaluate f at x = 0 and x = 3:

对于闭区间，绝对最大值和最小值可能出现在端点处。计算 f 在 x = 0 和 x = 3 的值：

f(0) = 2(0)³ − 9(0)² + 12(0) + 5 = 5

f(3) = 2(3)³ − 9(3)² + 12(3) + 5 = 54 − 81 + 36 + 5 = 14

Thus the function values at the endpoints are 5 and 14.

因此端点处的函数值为 5 和 14。

7. Determine Absolute Maximum and Minimum | 确定绝对最大值和最小值

We now compare all candidate values: f(0) = 5, f(1) = 10, f(2) = 9, and f(3) = 14.

现在比较所有候选值：f(0) = 5，f(1) = 10，f(2) = 9，f(3) = 14。

x	0	1	2	3
f(x)	5	10	9	14

The absolute maximum is 14, which occurs at x = 3. The absolute minimum is 5, which occurs at x = 0.

绝对最大值为 14，出现在 x = 3 处；绝对最小值为 5，出现在 x = 0 处。

8. Graphical Interpretation | 图像解释

A graph of f(x) on [0, 3] shows a cubic curve rising from (0, 5), reaching a local maximum at (1, 10), dipping to a local minimum at (2, 9), and then rising sharply to (3, 14). The endpoint values dominate the local extrema in this case.

在 [0, 3] 上绘制 f(x) 的图像，可以看到一条三次曲线：从 (0, 5) 上升，在 (1, 10) 达到局部极大值，下降到 (2, 9) 的局部极小值，然后急剧上升到 (3, 14)。本例中端点值支配了局部极值。

9. Common Mistakes | 常见错误

Forgetting to evaluate the endpoints when finding absolute extrema on a closed interval.

在闭区间上求绝对极值时忘记计算端点值。
Making a sign error when differentiating the cubic term, for example writing 6x instead of 6x².

对三次项求导时出现符号错误，例如写成 6x 而不是 6x²。
Assuming that a local maximum or minimum found via f ′′(x) is automatically the absolute maximum or minimum.

认为通过 f ′′(x) 找到的局部极大值或极小值自动就是绝对最大值或最小值。
Not checking whether critical points lie inside the given interval.

没有检查临界点是否位于给定区间内。

10. Summary of Steps | 步骤总结

For any cubic function on a closed interval, follow these steps:

对于闭区间上的任何三次函数，请遵循以下步骤：

Find f ′′(x) and solve f ′(x) = 0 to locate critical points.

求 f ′(x) 并解 f ′(x) = 0 以确定临界点。
Check that the critical points lie in the interval.

检查临界点是否在区间内。
Evaluate f(x) at each valid critical point.

在每个有效临界点处计算 f(x)。
Evaluate f(x) at both endpoints of the interval.

在区间的两个端点处计算 f(x)。
Compare all obtained values; the largest is the absolute maximum, the smallest is the absolute minimum.

比较所有得到值；最大值为绝对最大值，最小值为绝对最小值。

11. Practice Exercise | 练习

Try the same technique for g(x) = x³ − 3x² + 2 on [−1, 3]. Find the absolute maximum and minimum values.

请用相同方法计算 g(x) = x³ − 3x² + 2 在 [−1, 3] 上的绝对最大值和最小值。

Hint: The derivative is g ′(x) = 3x² − 6x, giving critical points at x = 0 and x = 2.

提示：导数为 g ′(x) = 3x² − 6x，临界点为 x = 0 和 x = 2。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Example 4.2.4: Optimizing a Cubic Function | 例题4.2.4：三次函数的最优化