在开始这一节前，先回顾前面的一些知识点，使得理解本节内容的时候更加系统性。
首先是可微，连续和存在极限三者的关系：
differentiable → \to → → \to →
但是反方向的推导是不成立的：
exist limit ↛ \nrightarrow ↛ ↛ \nrightarrow ↛
中间值定理(Intermediate value theorem)
在一个函数的连续区间[a, b]，如果存在一个c，使得f ( a ) ≤ f ( c ) ≤ f ( b ) f(a) \leq f(c) \leq f(b) f ( a ) ≤ f ( c ) ≤ f ( b ) a ≤ c ≤ b a \leq c \leq b a ≤ c ≤ b
罗尔定理(Rolle's Theorem)
Suppose that y = f(x) is continuous at every point of the closed interval [a, b]and differentiable at every point of its interior (a, b). If
f ( a ) = f ( b ) f(a) = f(b) f ( a ) = f ( b ) then there's at least one number c in (a, b) at which
证明这个定理需要用到上一节的连续定理 ：
“在函数f的一个连续闭区间[a, b]，存在最大和最小值。”
那么对于上述定理的函数f，它的极值存在于三种情况：
在区间内存在c，使得f'(c) = 0
f'没有定义域（即不存在）的区域
区间的边界处，即f(a)或f(b)
因为定理的前提说了，函数f在区间内均可微（differentiable），那么第二条不可能发生。假如第三条成立，那么最大值等于最小值等于f(a)或者f(b)，这意味着函数在这个区间是一条水平线，也就是说有无数导数为零的点。据此可证明以上定理成立。
中值定理/拉格朗日中值定理 (The mean value theorem)
Suppose y = f(x) is continuous on a close interval [a, b] and differentiable on the interval's interior (a, b). Then there is at least one point c in (a, b) at which
f ( b ) − f ( a ) b − a = f ′ ( c ) \frac{f(b) - f(a)}{b - a} = f'(c) b − a f ( b ) − f ( a ) = f ′ ( c ) 我们借助图表来来证明此定理
假设穿过a，b两点的割线函数为g(x)，根据点斜率公式可知它的表达式为：
g ( x ) = f ( a ) + f ( b ) − f ( a ) b − a ( x − a ) g(x) = f(a) + \frac{f(b) - f(a)}{b - a}(x - a) g ( x ) = f ( a ) + b − a f ( b ) − f ( a ) ( x − a ) 将f(x)和g(x)相减，构造一个新的函数h(x)，其表达式为：
h ( x ) = f ( x ) − g ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) b − a ( x − a ) h(x) = f(x) - g(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a) h ( x ) = f ( x ) − g ( x ) = f ( x ) − f ( a ) − b − a f ( b ) − f ( a ) ( x − a ) 因为g(x)过f(x)的a和b点，那么h(a) = 0, h(b) = 0。又因为f(x)和g(x)在a，b区间均可微，根据3.2节微分定理，h(x)也是可微的。这样就凑齐了罗尔定理的成立条件，根据此定理，在a，b区间中存在c，使得：
对h求导
h ′ ( x ) = f ′ ( x ) − f ( b ) − f ( a ) b − a h'(x) = f'(x) - \frac{f(b) - f(a)}{b - a} h ′ ( x ) = f ′ ( x ) − b − a f ( b ) − f ( a ) 代入c，得到
h ′ ( c ) = 0 = f ′ ( c ) − f ( b ) − f ( a ) b − a f ′ ( c ) = f ( b ) − f ( a ) b − a h'(c) = 0 = f'(c) - \frac{f(b) - f(a)}{b - a} \\
f'(c) = \frac{f(b) - f(a)}{b - a} h ′ ( c ) = 0 = f ′ ( c ) − b − a f ( b ) − f ( a ) f ′ ( c ) = b − a f ( b ) − f ( a ) 证毕。
朗格朗日中值定理十分重要，它可以推导出很多推论，是分析函数性质的核心定理之一。 下面有两个重要的推论。
推论一： 函数的导函数为零的，是常数函数。
引用下英文表述： If f'(x) = 0 at each point x of an open interval (a, b), then f(x) = C for all x ∈ ( a , b ) x \in (a, b) x ∈ ( a , b )
证明：
在a，b区间内取任意不相等的两点x 1 x_1 x 1 x 2 x_2 x 2
f ( x 2 ) − f ( x 1 ) x 2 − x 1 = f ′ ( c ) = 0 \frac{f(x_2) - f(x_1)}{x_2 - x_1} = f'(c) = 0 x 2 − x 1 f ( x 2 ) − f ( x 1 ) = f ′ ( c ) = 0 因为x 2 ≠ x 1 x_2 \neq x_1 x 2 = x 1 f ( x 2 ) = f ( x 1 ) f(x_2) = f(x_1) f ( x 2 ) = f ( x 1 )
证毕。
推论二： 导函数值相等的函数，它们之间的差值为一常数。
引用下英文表述：If f'(x) = g'(x) at each point x in an open interval (a, b), then there exist a constant C such that f(x) = g(x) + C for all x ∈ ( a , b ) x \in (a, b) x ∈ ( a , b )
证明：
在a，b区间内取任意不相等的两点x 1 x_1 x 1 x 2 x_2 x 2
f ( x 2 ) − f ( x 1 ) x 2 − x 1 = f ′ ( c 1 ) g ( x 2 ) − g ( x 1 ) x 2 − x 1 = g ′ ( c 2 ) \frac{f(x_2) - f(x_1)}{x_2 - x_1} = f'(c_1) \\
\frac{g(x_2) - g(x_1)}{x_2 - x_1} = g'(c_2) x 2 − x 1 f ( x 2 ) − f ( x 1 ) = f ′ ( c 1 ) x 2 − x 1 g ( x 2 ) − g ( x 1 ) = g ′ ( c 2 ) 因为
f ′ ( c 1 ) = g ′ ( c 2 ) f'(c_1) = g'(c_2) f ′ ( c 1 ) = g ′ ( c 2 ) 所以
f ( x 2 ) − f ( x 1 ) x 2 − x 1 = g ( x 2 ) − g ( x 1 ) x 2 − x 1 f ( x 2 ) − f ( x 1 ) = g ( x 2 ) − g ( x 1 ) f ( x 2 ) − g ( x 2 ) = f ( x 1 ) − g ( x 1 ) \frac{f(x_2) - f(x_1)}{x_2 - x_1} = \frac{g(x_2) - g(x_1)}{x_2 - x_1} \\
f(x_2) - f(x_1) = g(x_2) - g(x_1) \\
f(x_2) - g(x_2) = f(x_1) - g(x_1) x 2 − x 1 f ( x 2 ) − f ( x 1 ) = x 2 − x 1 g ( x 2 ) − g ( x 1 ) f ( x 2 ) − f ( x 1 ) = g ( x 2 ) − g ( x 1 ) f ( x 2 ) − g ( x 2 ) = f ( x 1 ) − g ( x 1 ) 即f(x) - g(x) = C。
证毕。
注意：推论二可以推广到无限区间，比如( a , ∞ ) (a, \infty) ( a , ∞ ) ( − ∞ , b ) (-\infty, b) ( − ∞ , b ) ( − ∞ , ∞ ) (-\infty, \infty) ( − ∞ , ∞ )
应用举例：
已知地球的重力加速度是9.8 m / s e c 2 9.8m/sec^2 9.8 m / se c 2 υ ( t ) \upsilon(t) υ ( t )
υ ′ ( t ) = 9.8 \upsilon'(t) = 9.8 υ ′ ( t ) = 9.8 根据上面推论二，可知
υ ( t ) = 9.8 t + C \upsilon(t) = 9.8t + C υ ( t ) = 9.8 t + C 当t = 0时，速度为零，即C = 0。
假设位移和时间的函数为s(t)，则
s ′ ( t ) = υ ( t ) = 9.8 t s'(t) = \upsilon(t) = 9.8t s ′ ( t ) = υ ( t ) = 9.8 t 那么，根据推论二，可知
s ( t ) = 4.9 t 2 + C s(t) = 4.9t^2 + C s ( t ) = 4.9 t 2 + C 当t = 0，s等于自由落体的初始高度h，那么
s ( t ) = 4.9 t 2 + h s(t) = 4.9t^2 + h s ( t ) = 4.9 t 2 + h 练习
f ( x ) = x 2 + 2 x − 1 , [ 0 , 1 ] f(x) = x^2 + 2x - 1, \ [0, 1] f ( x ) = x 2 + 2 x − 1 , [ 0 , 1 ]
f ( b ) − f ( a ) b − a = f ( 1 ) − f ( 0 ) 1 − 0 = 3 \frac{f(b) - f(a)}{b - a} = \frac{f(1) - f(0)}{1 - 0} = 3 b − a f ( b ) − f ( a ) = 1 − 0 f ( 1 ) − f ( 0 ) = 3
f ′ ( x ) = 2 x + 2 , f ′ ( c ) = 3 , c = 1 / 2 f'(x) = 2x + 2, \ f'(c) = 3, \ c = 1/2 f ′ ( x ) = 2 x + 2 , f ′ ( c ) = 3 , c = 1/2
f ( x ) = x 2 / 3 , [ 0 , 1 ] f(x) = x^{2/3}, \ [0, 1] f ( x ) = x 2/3 , [ 0 , 1 ]
f ( b ) − f ( a ) b − a = 1 2 / 3 − 0 2 / 3 1 − 0 = 1 \frac{f(b) - f(a)}{b - a} = \frac{1^{2/3} - 0^{2/3}}{1 - 0} = 1 b − a f ( b ) − f ( a ) = 1 − 0 1 2/3 − 0 2/3 = 1
f ′ ( x ) = 2 3 x − 1 / 3 , f ′ ( c ) = 1 , c = 8 / 27 f'(x) = \frac{2}{3}x^{-1/3}, \ f'(c) = 1, \ c = 8/27 f ′ ( x ) = 3 2 x − 1/3 , f ′ ( c ) = 1 , c = 8/27
f ( x ) = x + 1 x , [ 1 2 , 2 ] f(x) = x + \frac{1}{x}, \ [\frac{1}{2}, 2] f ( x ) = x + x 1 , [ 2 1 , 2 ]
f ( b ) − f ( a ) b − a = 2 + 1 2 − 1 2 − 2 2 − 1 2 = 0 \frac{f(b) - f(a)}{b - a} = \frac{2 + \frac{1}{2} - \frac{1}{2} - 2}{2 - \frac{1}{2}} = 0 b − a f ( b ) − f ( a ) = 2 − 2 1 2 + 2 1 − 2 1 − 2 = 0
f ′ ( x ) = 1 − 1 x 2 , f ′ ( c ) = 0 , c = 1 f'(x) = 1 - \frac{1}{x^2}, \ f'(c) = 0, \ c = 1 f ′ ( x ) = 1 − x 2 1 , f ′ ( c ) = 0 , c = 1
f ( x ) = x − 1 , [ 1 , 3 ] f(x) = \sqrt{x - 1}, \ [1, 3] f ( x ) = x − 1 , [ 1 , 3 ]
f ( b ) − f ( a ) b − a = 2 − 0 3 − 1 = 2 2 \frac{f(b) - f(a)}{b - a} = \frac{\sqrt{2} - 0}{3 - 1} = \frac{\sqrt{2}}{2} b − a f ( b ) − f ( a ) = 3 − 1 2 − 0 = 2 2
f ′ ( x ) = 1 2 x − 1 , f ′ ( c ) = 2 2 , c = 3 / 2 f'(x) = \frac{1}{2\sqrt{x - 1}}, \ f'(c) = \frac{\sqrt{2}}{2}, \ c = 3/2 f ′ ( x ) = 2 x − 1 1 , f ′ ( c ) = 2 2 , c = 3/2
f ( x ) = x 2 / 3 , [ − 1 , 8 ] f(x) = x^{2/3}, \ [-1, 8] f ( x ) = x 2/3 , [ − 1 , 8 ]
f ( b ) − f ( a ) b − a = 8 2 / 3 − ( − 1 ) 2 / 3 8 + 1 = 1 / 3 \frac{f(b) - f(a)}{b - a} = \frac{8^{2/3} - (-1)^{2/3}}{8 + 1} = 1/3 b − a f ( b ) − f ( a ) = 8 + 1 8 2/3 − ( − 1 ) 2/3 = 1/3
f ′ ( x ) = 2 3 x 1 / 3 = 1 / 3 , x = 8 f'(x) = \frac{2}{3x^{1/3}} = 1/3, \ x = 8 f ′ ( x ) = 3 x 1/3 2 = 1/3 , x = 8
f ( x ) = x 4 / 5 , [ 0 , 1 ] f(x) = x^{4/5}, \ [0, 1] f ( x ) = x 4/5 , [ 0 , 1 ]
f ( b ) − f ( a ) b − a = 1 4 / 5 − 0 4 / 5 = 1 \frac{f(b) - f(a)}{b - a} = 1^{4/5} - 0^{4/5} = 1 b − a f ( b ) − f ( a ) = 1 4/5 − 0 4/5 = 1
f ′ ( x ) = 4 5 x 1 / 5 = 1 , x = ( 4 5 ) 5 f'(x) = \frac{4}{5x^{1/5}} = 1, \ x = (\frac{4}{5})^5 f ′ ( x ) = 5 x 1/5 4 = 1 , x = ( 5 4 ) 5
f ( x ) = x ( 1 − x ) , [ 0 , 1 ] f(x) = \sqrt{x(1-x)}, \ [0, 1] f ( x ) = x ( 1 − x ) , [ 0 , 1 ]
f ( b ) − f ( a ) b − a = 0 \frac{f(b) - f(a)}{b - a} = 0 b − a f ( b ) − f ( a ) = 0
f ′ ( x ) = 1 2 [ x ( 1 − x ) ] − 1 / 2 ⋅ [ ( 1 − x ) − x ] = 1 − 2 x 2 [ x ( 1 − x ) ] 1 / 2 = 0 , x = 1 / 2 f'(x) = \frac{1}{2}[x(1 - x)]^{-1/2}\cdot[(1 - x) - x] = \frac{1 - 2x}{2[x(1 - x)]^{1/2}} = 0, \ x = 1/2 f ′ ( x ) = 2 1 [ x ( 1 − x ) ] − 1/2 ⋅ [( 1 − x ) − x ] = 2 [ x ( 1 − x ) ] 1/2 1 − 2 x = 0 , x = 1/2
f ( x ) = { sin x x , − π ≤ x < 0 0 , x = 0 f(x) = \begin{cases}
\frac{\sin x}{x}, & -\pi \leq x < 0 \\
0, & x = 0
\end{cases} f ( x ) = { x s i n x , 0 , − π ≤ x < 0 x = 0 对于sin x x \frac{\sin x}{x} x s i n x x ≠ 0 x \neq 0 x = 0
lim x → 0 sin x x = 1 ≠ f ( 0 ) \lim_{x \to 0}\frac{\sin x}{x} = 1 \neq f(0) x → 0 lim x sin x = 1 = f ( 0 ) 因此f在x = 0处不连续，也就不满足拉格朗日中值定理的前提条件。
f ( x ) = { x , 0 ≤ x < 1 0 , x = 1 f(x) = \begin{cases}
x, & 0 \leq x < 1 \\
0, & x = 1
\end{cases} f ( x ) = { x , 0 , 0 ≤ x < 1 x = 1 函数确实在(0, 1)区间可微且其导数为1，但是它并不在区间[0, 1]内连续，因为它在x = 1处不连续，因此也就不满足罗尔定理的前提条件。
f ( x ) = { 3 , x = 0 − x 2 + 3 x + a , 0 < x < 1 m x + b , 1 ≤ x ≤ 2 f(x) = \begin{cases}
3, & x = 0 \\
-x^2 + 3x + a, & 0 < x < 1 \\
mx + b, & 1 \leq x \leq 2
\end{cases} f ( x ) = ⎩ ⎨ ⎧ 3 , − x 2 + 3 x + a , m x + b , x = 0 0 < x < 1 1 ≤ x ≤ 2 需要在x = 0处连续，则：
lim x → 0 + ( − x 2 + 3 x + a ) = a = 3 \lim_{x \to 0^+}(-x^2 + 3x + a) = a = 3 x → 0 + lim ( − x 2 + 3 x + a ) = a = 3 需要在x = 1处连续，则：
lim x → 1 − ( − x 2 + 3 x + 3 ) = lim x → 1 + ( m x + b ) m + b = 5 \lim_{x \to 1^-}(-x^2 + 3x + 3) = \lim_{x \to 1^+}(mx + b) \\
m + b = 5 x → 1 − lim ( − x 2 + 3 x + 3 ) = x → 1 + lim ( m x + b ) m + b = 5 需要在x = 1处可微，则：
lim x → 1 − f ′ ( x ) = lim x → 1 − ( − 2 x + 3 ) = 1 lim x → 1 + f ′ ( x ) = lim x → 1 + ( m ) = m ∴ m = 1 b = 4 \lim_{x \to 1^-}f'(x) = \lim_{x \to 1^-}(-2x + 3) = 1 \\
\lim_{x \to 1^+}f'(x) = \lim_{x \to 1^+}(m) = m \\
\therefore m = 1 \\
b = 4 x → 1 − lim f ′ ( x ) = x → 1 − lim ( − 2 x + 3 ) = 1 x → 1 + lim f ′ ( x ) = x → 1 + lim ( m ) = m ∴ m = 1 b = 4
b Use Rolle's Theorem to prove that between every two zeros of x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 x^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0 x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 n x n − 1 + ( n − 1 ) a n − 1 x n − 2 + ⋯ + a 1 nx^{n - 1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1 n x n − 1 + ( n − 1 ) a n − 1 x n − 2 + ⋯ + a 1
假设零点分别位于a，b两点，f ( x ) = x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 f(x) = x^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0 f ( x ) = x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0
f(a) = f(b) = 0
根据罗尔定理，存在x = c，使得f'(c) = 0，求f'(x)可得：
f ′ ( x ) = n x n − 1 + ( n − 1 ) a n − 1 x n − 2 + ⋯ + a 1 f'(x) = nx^{n - 1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1 f ′ ( x ) = n x n − 1 + ( n − 1 ) a n − 1 x n − 2 + ⋯ + a 1
即f'(x)存在一个零点。
证毕。
因为函数f在区间[a, b]有三个零点，假设其为x 1 , x 2 , x 3 x_1, x_2, x_3 x 1 , x 2 , x 3
f ( x 1 ) = f ( x 2 ) = f ( x 3 ) = 0 f(x_1) = f(x_2) = f(x_3) = 0 f ( x 1 ) = f ( x 2 ) = f ( x 3 ) = 0
根据罗尔定理，x 1 x_1 x 1 x 2 x_2 x 2 c 1 c_1 c 1 f ′ ( c 1 ) = 0 f'(c_1)=0 f ′ ( c 1 ) = 0 c 2 c_2 c 2 f ′ ( c 2 ) = 0 f'(c_2) = 0 f ′ ( c 2 ) = 0
f ′ ( c 1 ) = f ′ ( c 2 ) = 0 f'(c_1) = f'(c_2) = 0 f ′ ( c 1 ) = f ′ ( c 2 ) = 0
再根据罗尔定理，c 1 c_1 c 1 c 2 c_2 c 2 c 3 c_3 c 3 f ′ ′ ( c 3 ) = 0 f''(c_3)=0 f ′′ ( c 3 ) = 0 f ′ ′ f'' f ′′
因为f''在区间[a, b]大于零，即连续。根据罗尔定理，如果f'存在两个零点x 1 x_1 x 1 x 2 x_2 x 2
f ′ ( x 1 ) = f ′ ( x 2 ) = 0 f'(x_1) = f'(x_2) = 0 f ′ ( x 1 ) = f ′ ( x 2 ) = 0
那么在x 1 x_1 x 1 x 2 x_2 x 2 c c c f ′ ′ ( c ) = 0 f''(c) = 0 f ′′ ( c ) = 0 f ′ ′ > 0 f'' > 0 f ′′ > 0 f ′ ′ < 0 f'' < 0 f ′′ < 0
假设三次多项式的导函数有四个或更多的零点（解），那么根据罗尔定理，f'就有至少3个零点（解）。但是f'是二次函数，它最多只有两个解，显然矛盾。所以函数f最多有三个零点。
