中值定理 - 高级微分 | 12 年级数学

如果 y = f(x) 是定义在 [a, b] 上的实值函数，使得

• f(x) 在 [a, b] 上是连续的
• f(x) 在 (a, b) 上可微，则至少存在一个点 c ∈ (a, b) 使得 f(b) – f(a) = f'(c)(b – a)

也称为拉格朗日中值定理或扩展中值定理。

几何意义

我们画曲线 y = f(x) 并取曲线上的点 A(a, f(a)) 和 B(b, f(b))，然后

弦的斜率 AB = (f(b) – f(a))/(b – a)

自从，

f'(c) = (f(b) – f(a))/(b – a) （中值定理的陈述）

f'(c) = 弦 AB 的斜率

这表明曲线 y = f(x) 在点 x = c 处的切线平行于弦 AB。

罗尔定理

如果 f(a) = f(b) 那么，

存在一个实数 c ∈ (a, b) 使得 f'(c) = 0，这种特殊情况称为罗尔定理。

罗尔定理本质上指出，任何在其上的两个不同点处获得相等值的实值微分函数必须在它们之间的某处至少有一个静止点，即一阶导数（切线的斜率到函数的图形 ) 为零。

几何意义

令 f(x) 是定义在 [a, b] 上的实值函数，它在 [a, b] 上是连续的。这意味着我们可以在 x = a 和 x = b 的值之间绘制 f(x) 的图，而且 f(x) 在 (a, b) 上是可微的，这意味着 f(x) 的图在(a, b) 的每个点。现在实数 c ∈ (ab) 的存在使得 f'(c) = 0 表明曲线在 x = c 处的切线具有斜率 0，即切线平行于 x 轴，因为 f(a) = f(b)。

罗尔定理的示例问题

问题 1：验证以下函数的罗尔定理： f (x) = x2 – 5x + 9 on [1, 4]

解决方案：

Since, f(x) is Polynomial in  x

f(x) is continuous on [1. 4]

f(x) is differentiable on (1, 4)

f(1) = 5, f(4) = 42 – 5 * 4 + 9 = 5

Hence, f(0) = f(4) = 5

Thus, all the conditions of Rolle’s Theorem are satisfied.

The derivative of f(x) should vanish for at least one point in c in (0, 4). To obtain the value of c,  

We proceed as follows,  

f (x) = x2 – 5x + 9

f'(x) =  2x-5⇢

if x =5/2  then f'(x) = 0

Hence, ∃ c = 5/2 in (1, 4)

we know that , 5/2 ∈1, 4)

Thus, Rolle’s Theorem is  verified.

问题 2：讨论罗尔定理对函数f(x) = x ^2/3 , x ∈ [-1, 1] 的适用性

解决方案：

f(x) = x^2/3 continuous at every  point of [-1, 1] 

Now, f(x) = x^2/3 

f'(x) = 2/3(x^-1/3)

As, x ⇢ 0⁺ f'(x) ⇢ ∞ 

f'(0) does not exist.

f (x) is not differentiable at  x=0

f (x) is not differentiable in (-1. 1)

 Rolle’s theorem is not applicable for the given function. 

柯西中值定理

令 f(x) 和 g(x) 是定义在 [a, b] 上的函数，使得，

f(x) 和 g(x) 在 [a, b] 上都是连续的

f(x) 和 g(x) 在 (a, b) 上都是可微的，则至少存在一个点 c ∈ (a, b) 使得

f'(c) {g(b) – g(a)}= g'(c){ f(b) – f(a)}

The case that g(a) = g(b) is easy. So, assume that g(a) ≠ g(b). Define

h(x) = f'(x) {g(b) – g(a)}= g'(x){f(b) – f(a)}

Clearly, h(a) = h(b). Applying Rolle’s Theorem we have that there is a c with a < c < b

Such that h'(x) = f'(c) {g(b) – g(a)}= g'(c){f(b) – f(a)}

For this c we have that,

f'(c) {g(b) – g(a)} = g'(c){f(b) – f(a)}

The classical Mean Value Theorem is a special case of Cauchy’s Mean Value Theorem.

It is the case when g(x) ≡ x. The Cauchy Mean Value Theorem can be used to prove L’Hospital’s Theorem.

中值定理的应用

• 中值定理是微积分的基本定理。
• 物理解释（如速度分析）。
• 如果导数大于零，则 f 是严格递增函数。
• 泰勒级数和数论。

增减函数

借助中值定理，我们可以找到

增加函数

函数y = f(x) 被称为 x 在区间 (a, b) 中的增函数：

如果 f'(x) > 0 对于 ax ∈ (a, b)

示例问题

问题1：测试以下函数是否在增加。

f(x) = x ³ – 3x ² + 3x – 100, x ∈ R

解决方案：

f(x) = x³ – 3x² + 3x – 100, x ∈ R

f'(x) = 3x² – 6x + 3

= 3(x – 1)²

Since (x – 1)² is always positive x ≠1

f'(x) >0, for all x ∈ R.

Hence, f(x) is an increasing function,  for all x ∈ R.

问题 2：求函数f(x) = x ³ +12x ² +36x+6, x 增加的 x 的值。

解决方案：

f(x) = x³ + 12x² + 36x + 6

f'(x) = 3x² + 24x + 36 = 3(x + 2)(x + 6)

Now, f'(x) > 0, as f(x) is increasing

3(x + 2)(x + 6) > 0

Case 1: x + 2 < 0 and x + 6 > 0

x > -2 and x > -6

Case 2: x + 2 < 0 and x + 6 > 0

x < -2 and x < -6

f(x) = x³ + 12x² + 36x + 6 is increasing if and only if x < -6 or x > -2

Hence, x∈ (-∞, -6) or x∈ (-2, ∞)

递减函数

函数y = f(x) 被称为 x 在区间 (a, b) 中的减函数

如果对于 ax ∈ (a, b)，f'(x) < 0。

示例问题

问题 1：测试函数f (x) = 2 – 3x + 3x ² – x ³对于所有 x ∈ R 是递增还是递减

解决方案：

f(x) = 2 – 3x + 3x² – x³  

f'(x) = -3 + 6x – 3x²

= -3(x² – 2x + 1)

= -3(x – 1)²

Since (x – 1)² is always positive, x ≠1

f'(x) < 0, for all x ∈ R.

Hence, f(x) is a decreasing function, for all x ∈ R.

问题 2：求函数f(x) = 2x ³ – 9x ² + 12x + 2, x 递减的 x 值。

解决方案：

f(x) = 2x³ – 9x² + 12x + 2,

f'(x) = 6x² – 18x + 12

= 6(x – 1)(x – 2)

Now, f'(x) < 0, 

As f(x) is decreasing

6(x – 1)(x – 2) < 0

Case 1: (x – 1) < 0 and x – 2 > 0

x < 1 and x > 2 which is absurd

Case 2: (x  – 1) > 0 and x – 2 < 0

x > 1 and x < 2  

1 < x < 2

f(x) = x³ + 12x² + 36x + 6 is decreasing if and only if x ∈ (1, 2).

中值定理示例问题

问题 1：f(x) = (x – 1)(x – 2)(x – 3), x ∈ [0, 4]。如果可以应用均值定理，则求 c。

解决方案：

f(x) = (x – 1)(x – 2)(x – 3),

x ∈ [0, 4]

f(x) = x³ – 6x² + 11x – 6

As f(x) is Polynomial in x

• f(x) is continuous on [0. 4]
• f(x) is differentiable on (0, 4)

Thus, all the conditions of Mean Value theorem are satisfied.

To verify theorem we have to find c ∈ (0, 4) such that

f’(c) = (f(4) – f(0))/(4 – 0)                                …(1)

Now, f(4) = (4 – 1)(4 – 2)(4 – 3) = 6

f(0) = (0 – 1)(0 – 2)(0 – 3) = – 6 and

f'(x) = 3x² – 12x + 11

f'(c) = 3c² – 12c + 11

(f(4) – f(0))/(4 – 0) = 3c² – 12c + 11         [From 1]

3 = 3c² – 12c + 11

3c² – 12c + 8 = 0

c = 2 ± 2/√3

Both values of c lie between 0 and 4.

问题 2：验证 [1, e] 上函数f(x) = log(x) 的拉格朗日均值定理

解决方案：

f(x) = log(x) 

f(x) is continuous on [1, e]

f(x) is differentiable on (1, e), Thus all the conditions of Mean Value theorem are satisfied.

We want to find c ∈ (1, e) such that 

f(e) – f(1) = f'(c) {e – 1}           … from (1) [By mean value theorem]

log e -log 1 = f'(c) {e – 1}   

1 – 0 = f'(c) {e – 1}   

f'(c) = 1/(e – 1)                      …[2]

Now,

f(x) = log(x)     

Hence, f'(x) = 1/x

f'(c) = 1/c           …[3]

From [2] and [3]      

1/(e – 1) = 1/c     

Hence,     

c = e – 1 which is lies in interval (1, e) 

Thus, LaGrange’s mean value theorem verified.