# 中值定理（MVT）

> AP 微积分 AB · 微分的分析应用
> 来源: https://www.owlsprep.com/zh/study/ap-calculus-ab-u5-mean-value-theorem/

本指南讲解AP 微积分AB中值定理（MVT）的正式表述、前提假设和应用，包括罗尔定理、寻找保证存在的点，以及FRQ证明练习。

**先修:** 判断闭区间上的连续性; 判断开区间上的可导性; 计算常用函数的导数

## 学习目标

- 陈述中值定理（MVT）的正式表述和前提假设
- 验证中值定理何时可应用于区间上的函数
- 找出中值定理保证存在的所有c值
- 应用中值定理证明函数界和实际问题结论
- 认识到罗尔定理是中值定理的特殊情况

## 什么是中值定理（MVT）？

中值定理（MVT）是微分微积分的核心理论结论，在AP微积分AB考试中经常考察，占总分的约4-7%，同时出现在选择题和自由问答题部分，通常和其他知识点（如函数行为的证明）结合考察。

**中值定理（MVT）** — 若函数$f$满足两个前提假设：(1) $f$在闭区间$[a,b]$上连续，且(2) $f$在开区间$(a,b)$上可导，则在$(a,b)$中至少存在一个数$c$，使得$c$点的瞬时变化率等于整个区间上的平均变化率。

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

> **tip**
>
> 在AP自由问答题中，你必须明确陈述并验证中值定理的两个前提假设才能获得完整的证明分数。

## 中值定理的前提假设与罗尔定理（特殊情况）

MVT是条件定理：只有当两个前提假设都满足时，才能保证点$c$存在。如果任意一个假设不成立，就不能保证这样的$c$存在。第一个要求是*闭*区间$[a,b]$上连续（包括端点在内处处无间断点），第二个要求是*开*区间$(a,b)$上可导（不要求端点处可导，因为区间端点无法定义双侧导数）。

**罗尔定理** — 中值定理的一个特殊情况，额外增加条件$f(a) = f(b)$。若中值定理的所有前提假设都成立且$f(a)=f(b)$，则至少存在一个$c \in (a,b)$使得$f'(c) = 0$，即区间内部某点存在水平切线。

**例题:** 中值定理是否适用于区间$[-1, 3]$上的函数$f(x) = \frac{x^2}{x-2}$？证明你的结论。

1. 首先检查中值定理的第一个前提假设：闭区间$[-1, 3]$上的连续性。
2. $f(x)$是有理函数，因此在分母不为零的所有$x$处连续。分母在$x=2$处为零，而$x=2$严格位于$[-1, 3]$内部，因此$f$在$x=2$处有无穷间断点。
3. 由于连续性假设不成立，我们不需要检查可导性假设：只要任意一个假设不成立，定理就不适用。
4. 结论：中值定理不适用于$[-1, 3]$上的$f$。

> **exam_tip**
>
> 如果你在应用中值定理前跳过验证两个前提假设，会在FRQ中失分。

## 寻找中值定理保证存在的点$c$

当你确认中值定理适用于某区间上的函数后，遵循以下4步流程找出定理保证存在的所有有效$c$值：

1. 计算区间上的平均变化率$\frac{f(b) - f(a)}{b-a}$
2. 计算函数的一阶导数$f'(x)$
3. Set $f'(c)$ equal to the average rate of change, then solve for $c$
4. Discard any solutions for $c$ that do not lie strictly inside the open interval $(a,b)$

**例题:** Let $f(x) = x^3 - 2x$ on $[0, 3]$. Confirm MVT applies, then find all values of $c$ guaranteed by the theorem.

1. Verify hypotheses: $f(x)$ is a polynomial, so it is continuous on $[0,3]$ and differentiable on $(0,3)$, so MVT applies.
2. Calculate average rate of change:
3. $$f(3) = 3^3 - 2(3) = 21, f(0) = 0, \frac{f(3)-f(0)}{3-0} = 7$$
4. Compute derivative and set equal to 7, then solve for $c$:
5. $$f'(x) = 3x^2 - 2, 3c^2 - 2 = 7 \implies 3c^2 = 9 \implies c^2 = 3 \implies c = \sqrt{3}, -\sqrt{3}$$
6. Filter solutions by interval: $-\sqrt{3} \approx -1.73$ is outside $(0,3)$, so the only valid solution is $c = \sqrt{3}$.

> **tip**
>
> AP考试题目几乎总会给出一个位于区间外的多余$c$解，你必须明确舍去它才能获得满分。

## 应用中值定理证明函数行为和解决问题

Beyond routine calculation, MVT is used to justify higher-order conclusions about function behavior, a common FRQ skill. If you know $m \leq f'(x) \leq M$ for all $x$ in $[a,b]$, MVT tells you $m(b-a) \leq f(b) - f(a) \leq M(b-a)$. This is also the theoretical foundation for the rule that a positive derivative everywhere on an interval implies the function is increasing on that interval.

**例题:** Let $f$ be differentiable for all real numbers, with $f(2) = 5$ and $f'(x) \leq 3$ for all $x$. What is the maximum possible value of $f(5)$? Justify your answer with MVT.

1. Apply MVT to $f$ on $[2,5]$: since $f$ is differentiable everywhere, it is continuous everywhere, so both MVT hypotheses are satisfied.
2. By MVT, there exists a $c \in (2,5)$ such that:
3. $$f'(c) = \frac{f(5) - f(2)}{5-2} = \frac{f(5) - 5}{3}$$
4. Substitute the bound $f'(c) \leq 3$ and solve:
5. $$\frac{f(5) - 5}{3} \leq 3 \implies f(5) - 5 \leq 9 \implies f(5) \leq 14$$
6. Conclusion: the maximum possible value of $f(5)$ is 14.

**例题:** A car entered a 62-mile highway stretch at 1:15 PM and exited at 2:00 PM. The speed limit is 70 mph. Use MVT to prove the car was speeding at some point.

1. Let $d(t)$ = distance traveled $t$ hours after 1:15 PM. Total elapsed time is 45 minutes = $\frac{3}{4}$ hours. $d(t)$ is continuous on $[0, 3/4]$ and differentiable on $(0, 3/4)$, so MVT applies.
2. By MVT, there exists a time $c \in (0, 3/4)$ where:
3. $$d'(c) = \frac{62 - 0}{\frac{3}{4} - 0} \approx 82.67 \text{ mph}$$
4. The car's instantaneous speed at $c$ is ~82.67 mph, which exceeds the 70 mph speed limit, so the car must have been speeding at some point.

**概念自测**

For which of the following functions on the given interval does the Mean Value Theorem NOT apply?

1. Select the correct answer

   - A) $f(x) = \sin x$ on $\left[0, \frac{\pi}{2}\right]$
   - B) $f(x) = x^3 + 2x^2$ on $[-2, 1]$
   - C) $f(x) = \frac{x}{x+1}$ on $[-2, 0]$
   - D) $f(x) = (x + 3) e^x$ on $[0, 2]$

   *解析:* Sine, polynomials, and products of polynomials and exponentials are continuous and differentiable everywhere. Option C has an infinite discontinuity at $x=-1$, which lies inside $[-2, 0]$, so the continuity hypothesis fails and MVT does not apply.

> **exam_tip**
>
> When using MVT for justification on FRQs, you must explicitly name the theorem and confirm its hypotheses to earn full credit.

## 常见错误

- **错误做法:** Writing MVT hypotheses as 'continuous on $(a,b)$ and differentiable on $[a,b]$' (swapping open/closed intervals)
  - 原因: Students mix up interval requirements because derivatives are rarely discussed at endpoints.
  - 正确做法: Always state explicitly: 'continuous on the closed interval $[a,b]$, differentiable on the open interval $(a,b)$'.
- **错误做法:** Applying MVT to a function with a corner, cusp, or vertical tangent inside $(a,b)$
  - 原因: Students only check continuity and forget that non-differentiability at an interior point violates the second hypothesis.
  - 正确做法: After checking continuity, explicitly verify differentiability at all interior points before applying MVT.
- **错误做法:** Keeping solutions for $c$ that are at endpoints or outside $(a,b)$
  - 原因: Students misremember MVT as guaranteeing $c$ in $[a,b]$ instead of $(a,b)$.
  - 正确做法: After solving for $c$, discard any solution that is $\leq a$ or $\geq b$, only keep values strictly between $a$ and $b$.
- **错误做法:** Claiming MVT guarantees exactly one $c$ in $(a,b)$
  - 原因: Students misread 'at least one' as 'exactly one'.
  - 正确做法: Find all solutions of $f'(c) = \frac{f(b)-f(a)}{b-a}$ in $(a,b)$, and list all valid solutions.
- **错误做法:** Applying Rolle's Theorem when $f(a)=f(b)$ but core MVT hypotheses are violated
  - 原因: Students focus on the extra $f(a)=f(b)$ condition and forget to check core hypotheses first.
  - 正确做法: Always check continuity and differentiability first, even when applying Rolle's Theorem.

## 速查表

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| MVT Core Hypotheses | Continuous on closed $[a,b]$, differentiable on open $(a,b)$ | Both must be satisfied to apply the theorem |
| MVT Core Formula | $\displaystyle f'(c) = \frac{f(b)-f(a)}{b-a}$ for some $c \in (a,b)$ | $f'(c)$ = instantaneous rate, RHS = average rate |
| Rolle's Theorem | If MVT hypotheses hold *and* $f(a)=f(b)$, then $\exists c \in (a,b)$ with $f'(c)=0$ | Special case of MVT, same core hypotheses |
| Step Process to Find $c$ | 1. Calculate average rate<br>2. Compute $f'(x)$<br>3. Set $f'(c) = $ average rate<br>4. Keep only $c \in (a,b)$ | Discard all solutions outside the open interval |
| Bounding Function Values | $m \leq f'(x) \leq M \implies m(b-a) \leq f(b)-f(a) \leq M(b-a)$ | Used to find maximum/minimum possible function values |
| AP Justification Rule | Always name MVT and verify hypotheses before use | Required for full credit on FRQs |

## 下一步

The Mean Value Theorem is the foundational theoretical result for all of Unit 5, Analytical Applications of Differentiation. Immediately after mastering MVT, you will apply it to justify conclusions about intervals of increase and decrease, the First Derivative Test for local extrema, and the identification of critical points for absolute extrema. Without understanding how MVT connects the sign of the derivative to overall function behavior, you cannot earn full credit for FRQ justifications, which make up a large share of AP Calculus AB exam points. Long-term, MVT also underpins core results like the Fundamental Theorem of Calculus Part 1 and error bounds for linear approximation.

- [极值定理、全局与局部极值、临界点](https://www.owlsprep.com/zh/study/ap-calculus-ab-u5-extreme-value-theorem-global-vs/)
- [确定函数递增/递减的区间](https://www.owlsprep.com/zh/study/ap-calculus-ab-u5-determining-intervals-where-a-function/)
- [相对极值的一阶导数测试](https://www.owlsprep.com/zh/study/ap-calculus-ab-u5-first-derivative-test-for-relative/)

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