中值定理（MVT）

AP 微积分 AB· AP Calculus AB CED — Analytical Applications of Differentiation· 14 分钟阅读

1. 什么是中值定理（MVT）？★★☆☆☆⏱ 3 min

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。

📘 定义

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}

2. 中值定理的前提假设与罗尔定理（特殊情况）★★☆☆☆⏱ 4 min

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）。

📘 定义

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 $f$ 。

3. 寻找中值定理保证存在的点$c$★★★☆☆⏱ 4 min

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：

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

📐 例题

Let $f (x) = x^{3} - 2 x$ 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: $- 3 \approx - 1.73$ is outside $(0, 3)$ , so the only valid solution is $c = 3$ .

4. 应用中值定理证明函数行为和解决问题★★★☆☆⏱ 3 min

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.

✓ 快速检测

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

Select the correct answer
- A) $f (x) = sin x$ on $[0, \frac{π}{2}]$
- B) $f (x) = x^{3} + 2 x^{2}$ on $[- 2, 1]$
- C) $f (x) = \frac{x}{x + 1}$ on $[- 2, 0]$
- D) $f (x) = (x + 3) e^{x}$ on $[0, 2]$
显示答案
C —
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.

5. 常见陷阱

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.

6. 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	$f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ 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 $c$	Calculate average rate 2. Compute $f^{'} (x)$ 3. Set $f^{'} (c) =$ average rate 4. Keep only $c \in (a, b)$	Discard all solutions outside the open interval
Bounding Function Values	$m \leq f^{'} (x) \leq M ⟹ m (b - a) \leq f (b) - f (a) \leq M (b - a)$	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

真题中的出现

AI 根据考纲规律估算的考点位置,请对照官方真题核实准确性。仅作复习重点参考。

2023 · MCQ
判断中值定理何时适用
2022 · FRQ
用中值定理确定函数值的范围

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.