Section01b_中值定理 (2)

型一 证明$f^{(n)}(\xi)=0$

常用方法

1. Case 1 $f'(\xi)=0$，寻找$f(?)=f(?)$
2. Case 2 $f''(\xi)=0$，寻找$f'(?)=f'(?)$ 或 $f(?)=f(?)=f(?)$

例题

• 例1 $f(x)\in C[0,2]$，在$(0,2)$内可导且$3f(0)=f(1)+2f(2)$。证明：$\exists\ \xi\in(a,b)$，使得$f'(\xi)=0$ $$\begin{array}{ll} \because & f(x)\in C[0,2] \\ & \text{设}f(x)\in C[1,2],\ \exists\ m,M \\ \therefore & 3m\le f(1) + 2f(2)\le 3M\\ \because & 3f(0) = f(1)+2f(2) \\ \therefore & m\le f(0)\le M \\ \therefore & \exists\ c\in(1,2),\ f(c) = f(0) \\ \therefore & \exists\ \xi\in(0,c)\subset(0,1),\ f'(\xi)=0 \\ \end{array}$$
• 例2 $f(x)$二阶可导，且$\displaystyle{\lim_{x\to 0}\frac{f(x)-1}{x}=0}$又$f(1)=1$，证明：$\exists\ \xi\in(0,1),\ f''(\xi)=0$ $$\begin{array}{ll} \because & \displaystyle{\lim_{x\to 0}\frac{f(x)-1}{x}} = 0\\ \therefore & f(0) = 1, f'(0) = 0 \\ \because & f(0)=f(1)=1 \\ \therefore & \exists\ c\in(0,1),\ f'(c) = 0 \\ \because & f'(c)=f'(0) \\ \therefore & \exists\ \xi\in(0,c)\subset(0,1),\ f''(\xi) = 0\\ \end{array}$$
• 例3 $f(x)$二阶可导，且$\displaystyle{\lim_{x\to 0}\frac{f(x)}{x}=1}$又$f(1)=1$，证明：$\exists\ \xi\in(0,1),\ f''(\xi)=0$ $$\begin{array}{ll} \because & \displaystyle{\lim_{x\to 0}\frac{f(x)}{x}} = 1\\ \therefore & f(0)=0, f'(0)=1 \\ \because & f(0)=0, f(1)=1 \\ \therefore & \exists\ c\in(0,1), f'(c)= \frac{f(1)-f(0)}{1-0} = 1 \\ \because & f'(c) = f'(0) \\ \therefore & \exists\ \xi\in(0,c)\subset(0,1), f'(\xi)=0 \\ \end{array}$$
• 例4 $f(x)$二阶可导，存在 $\text{L: }y=f(x)$，连$\text{A}(a,f(a)), \text{B}(b,f(b))$ 交 $\text{L}$ 于 $\text{C}(c,f(c))\quad a<c<b$，证明 $\exists\ \xi\in(a,b), f''(\xi)=0$ $$\begin{array}{ll} \because & \text{连A}(a, f(a)), \text{B}(b, f(b))\text{交L于C}(c, f(c)) \\ \therefore & \exists\ \xi_{1}\in(a,c), \xi_{2}\in(c,b),\ f'(\xi_{1}) = \frac{f(c)-f(a)}{c-a}, f'(\xi_{2})=\frac{f(b)-f(c)}{b-c} \\ \because & \text{A,B,C三点共线} \\ \therefore & f'(\xi_{1})=f'(\xi_{2}) \\ \therefore & \exists\ \xi\in(\xi_{1},\xi_{2})\subset(a,b),\ f''(\xi)=0 \\ \end{array}$$

型二 仅有$\xi$

常用方法

1. 还原法
   1. 条件$\begin{cases}\text{仅有}\xi\\\text{导数差一阶}\\\text{两项}\end{cases}$
   2. 工具 $\displaystyle{\frac{f'}{f} = (\ln f)',\ \frac{f''}{f'} = (\ln f')'}$
2. 分组法
   1. 条件$\begin{cases}\text{仅有}\xi\\ \text{导数差一阶}\\ \text{非两项}\end{cases}$
   2. 思想：将非两项凑成两项，如：$g^{(n)}(x)\pm g^{(n-1)}(x)\quad n=1,2,\cdots$

例题

• 例1 设$f(x)\in C[0,1]$，在$(0,1)$内可导，$f(1)=0$ 证明：$\exists\ \xi\in(0,1)$，使 $2f(\xi)+\xi f'(\xi)=0$ $$\begin{array}{ll} & \text{Analysis} \\ & 2f(x) + xf'(x) = 0 \\ \Rightarrow & \frac{2}{x}+\frac{f'(x)}{x}=0 \\ \Rightarrow & (\ln x^{2} + \ln f(x))' = 0 \\ \Rightarrow & [\ln(x^{2}f(x))]' = 0 \\\\ & \text{Proof} \\ & \text{令}\phi(x) = x^{2}f(x) \\ \because & \phi(0) =0, f(1)=0 \\ \therefore & \phi(0) = \phi(1) = 0 \\ \therefore & \exists\ \xi\in(0,1), \phi'(\xi)=0 \\ & \text{而}\phi'(x) = 2xf(x)+x^{2}f'(x) \\ \because & \xi\in(0,1) \\ \therefore & 2f(\xi) +\xi f'(\xi) = 0 \\ \end{array}$$
• 例2 $f(x)\in C[a,b]$，在$(a,b)$内可导，$f(a)=f(b)=0$，证明：$\exists\ \xi\in(a,b)$，使$f'(\xi)-2f(\xi)=0$ $$\begin{array}{ll} & \text{Analysis} \\ & f'(x)-2f(x) = 0 \\ \Rightarrow & \frac{f'(x)}{f(x)} - 2 = 0 \\ \Rightarrow & [\ln f(x) - \ln e^{2x}]' = 0 \\ \Rightarrow & \big[\ln \frac{f(x)}{e^{2x}}\big]' = 0 \\\\ & \text{Proof} \\ & \text{令}\phi(x) = \frac{f(x)}{e^{2x}} \\ \because & f(a) = f(b) = 0 \\ \therefore & \phi(a) = \phi(b) = 0 \\ \therefore & \exists\ \xi\in(a,b), \phi'(\xi)=0 \\ & \text{而}\phi'(x) = \frac{e^{2x}[f'(x)-2f(x)]}{(e^{2x})^{2}} = \frac{f'(x)-2f(x)}{e^{2x}}\\ \because & e^{2x}\neq 0 \\ \therefore & f'(\xi) - 2f(\xi) = 0 \\ \end{array}$$
• 例3 $f(x)$ 在 $[0,1]$ 上二阶可导，且$f(0)=f(1)$ 证明：$\exists\ \xi\in(0,1)$，使$\displaystyle{f''(\xi) = \frac{2f'(\xi)}{1-\xi}}$ $$\begin{array}{ll} & \text{Analysis} \\ & f''(x) = \frac{2f'(x)}{1-x} \\ \Rightarrow & \frac{f''(x)}{f'(x)} + \frac{2}{x-1} = 0 \\ \Rightarrow & [\ln f'(x) + \ln (x-1)^{2}]' = 0 \\ \Rightarrow & [\ln (x-1)^{2}f'(x)]' = 0 \\\\ & \text{Proof} \\ & \text{令}\phi(x) = (x-1)^{2}f'(x) \\ \because & f(0) = f(1) \\ \therefore & \exists c\in(0,1), f'(c) = 0 \\ \therefore & \phi(c) = \phi(1) = 0 \\ \therefore & \exists\ \xi\in(c,1)\subset(0,1), \phi'(\xi)=0 \\ & \text{而}\phi'(x) = 2(x-1)f'(x) + (x-1)^{2}f''(x) \\ \because & \xi\in(0,1) \\ \therefore & \frac{2f'(\xi)}{\xi-1}+f''(\xi) = 0 \\ & f''(\xi) = \frac{2f'(\xi)}{1-\xi} \end{array}$$
• 例4 $f(x)$ 二阶可导，$\displaystyle{\lim_{x\to 0}\frac{f(x)}{x} = 1, f(1)=1}$ 证：$\exists\ \xi\in(0,1),\ f''(\xi)-f'(\xi) +1 = 0$ $$\begin{array}{ll} & \text{Analysis} \\ & f''(x) - f'(x) + 1 = 0 \\ \Rightarrow & [f'(x)-1]' - [f(x)-x]' = 0 \\ \Rightarrow & \text{令}g(x) = f(x)-x \\ \Rightarrow & g''(x)-g'(x) = 0 \\ \Rightarrow & \frac{g''(x)}{g'(x)} - 1= 0 \\ \Rightarrow & [\ln g'(x)]' - [\ln e^{x}]' = 0 \\ \Rightarrow & \big[\ln \frac{g'(x)}{e^{x}}\big]' = 0 \\\\ & \text{Proof} \\ & \text{令}\phi(x) \big(= \frac{g'(x)}{e^{x}}\big) = \frac{f'(x)-1}{e^{x}} \\ \because & \displaystyle{\lim_{x\to 0}\frac{f(x)}{x} = 1} \\ \therefore & f(0) = 0, f'(0) = 1 \\ \because & f(0) = 0, f(1) = 1 \\ \therefore & \exists\ c\in(0,1), f'(c) = \frac{f(1)-f(1)}{1-0} = 1 \\ \therefore & \phi(c) = 0 = \phi(0) \\ \therefore & \exists\ \xi\in(0,c)\subset(0,1), \phi'(\xi)=0 \\ & \text{而}\phi'(x) = \frac{e^{x}[f''(x)-f'(x)+1]}{(e^{x})^{2}} = \frac{f''(x)-f'(x)+1}{e^{x}} \\ \because & e^{x}\neq 0 \\ \therefore & f''(\xi)-f'(\xi)+1 = 0 \end{array}$$

型三 有$\xi$ 有$a,b$

常用方法

1. Case 1 $\xi$ 与 $a,b$ 可分离
   1. 将 $\xi$ 与 $a,b$ 分离
   2. 对 $a,b$ 一侧$\displaystyle{\begin{cases}f(b)-f(a)\text{ or }\frac{f(b)-f(a)}{b-a}\Rightarrow \text{Lagrange}\\\frac{f(b)-f(a)}{g(b)-g(a)}\Rightarrow \text{Cauchy}\end{cases}}$
2. Case 2 $\xi$ 与 $a,b$ 不可分离
   1. 将 $\xi$ 换为 $x$
   2. 去分母，移项$\Rightarrow (\cdots) = 0\Rightarrow(\cdots)' = 0$

例题

• 例1 $ab>0\quad(a<b)$，证：$\exists\ \xi\in(a,b)$ 使得 $ae^{b}-be^{a} = (a-b)(1-\xi)e^{\xi}$ $$\begin{array}{ll} & \text{Analysis} \\ & ae^{b}-be^{a}=(a-b)(1-\xi)e^{\xi} \\ \Rightarrow & \frac{ae^{b}-be^{a}}{a-b} = (1-\xi)e^{\xi} \\ \Rightarrow & \frac{\frac{e^{b}}{b}-\frac{e^{a}}{a}}{\frac{1}{b}-\frac{1}{a}} \\ \Rightarrow & \text{令}g(x) = \frac{e^{x}}{x}, h(x) = \frac{1}{x} \\\\ & \text{Proof} \\ & \text{令}g(x) = \frac{e^{x}}{x}, h(x) = \frac{1}{x} \\ \because & h'(x) \neq 0 \\ \therefore & \exists\ \xi\in(a,b), \frac{g'(\xi)}{h'(\xi)} = \frac{g(b)-g(a)}{h(b)-h(a)} \\ & \text{而}g'(x) = \frac{e^{x}(x-1)}{x^{2}}, h'(x) = -\frac{1}{x^{2}}\\ \therefore & -e^{\xi}(\xi -1) = \frac{g(b)-g(a)}{h(b)-h(a)}\\ & e^{\xi}(1-\xi) = \frac{ae^{b}-be^{a}}{a-b} \\ & ae^{b}-be^{a} = (a-b)e^{\xi}(1-\xi) \end{array}$$
• 例2 $f(x),g(x)\in C[a,b]$，在$(a,b)$内可导且$g'(x)\neq 0$，证明 $\exists\ \xi\in(a,b)$，使得 $\displaystyle{\frac{f(x)-f(\xi)}{g(\xi)-g(b)}=\frac{f'(\xi)}{g'(\xi)}}$ $$\begin{array}{ll} & \text{Analysis} \\ & \frac{f(a) - f(x)}{g(x)-g(b)} = \frac{f'(x)}{g'(x)} \\ \Rightarrow & g'(x)[f(a)-f(x)] - [g(x)-g(b)]f'(x) = 0 \\ \Rightarrow & \{[f(a)-f(x)][g(x)-g(b)]\}' = 0\\\\ & \text{Proof} \\ & \text{令}\phi(x) = [f(a)-f(x)][g(b)-g(x)] \\ \therefore & \phi(a) = \phi(b) = 0 \\ \therefore & \exists\ \xi\in(a,b), \phi'(\xi) = 0 \\ & \text{而}\phi'(x) = g'(x)[f(a)-f(x)]-[g(x)-g(b)]f'(x) \\ \therefore & g'(\xi)[f(a)-f(\xi)] - [g(\xi)-g(b)]f'(\xi) = 0\\ \because & g'(x) \neq 0 \\ \therefore & g(x) - g(b) \neq 0\quad x\ne b \\ \therefore & \frac{f(a)-f(\xi)}{g(\xi)-g(b)} = \frac{f'(\xi)}{g'(\xi)} \\ \end{array}$$

型四 多中值

常用方法

1. Case 1
   1. 条件：仅有$f'(\xi), f'(\eta)$
   2. 方法
      1. 找三点
      2. 两次$\text{Lagrange}$
2. Case 2
   1. 含 $\xi,\eta$ 项的复杂程度不同
   2. 方法
      1. 保留复杂中值项
      2. 并将复杂项转化为$\displaystyle{\begin{cases}(\cdots)'\Rightarrow \text{Lagrange}\\\frac{(\cdots)'}{(\cdots)'}\Rightarrow \text{Cauchy}\end{cases}}$

例题

• 例1 $f(x)\in C[0,1]$，在$(0,1)$内可导，$f(0)=0, f(1)=1$
  1. 证明 $\exists\ c\in(0,1)$，使得$f(c) = 1-c$ $$\begin{array}{ll} & \text{令}\phi(x) = f(x) - 1 + x \\ \because & \phi(0) = -1, \phi(1) = 1 \\ \therefore & \phi(0)\phi(1) = -1 < 0 \\ \therefore & \exists\ c \in (0,1), \phi(c) = 0 \\ \therefore & f(c) = 1-c \\ \end{array}$$
  2. 证明 $\exists\ \xi,\eta\in(0,1)\quad(\xi\neq\eta)$，使得 $f'(\xi)f'(\eta)=1$ $$\begin{array}{ll} & \text{由题1得}f(c) = 1-c \\ \therefore & \exists\ \xi\in(0,c),\eta\in(c,1), \\ & f'(\xi) = \frac{f(c)-f(0)}{c-0} = \frac{1-c}{c}, \\ & f'(\eta) = \frac{f(1)-f(c)}{1-c} = \frac{c}{1-c} \\ \therefore & f'(\xi)f'(\eta) = 1 \\ \end{array}$$
• 例2 $f(x)\in C[0,1]$，在$(0,1)$内可导，且$f(0)=0,f(1)=1$
  1. 证明：$\exists\ c\in(0,1)$，使$f(c)=\frac{1}{2}$ $$\begin{array}{ll} & \text{令}\phi(x) = f(x)-\frac{1}{2} \\ \therefore & \phi(0) = -\frac{1}{2}, \phi(1) = \frac{1}{2} \\ \therefore & \phi(0)\phi(1)<0 \\ \therefore & \exists c\in(0,1), \phi(c) = 0 \\ \therefore & f(c) = \frac{1}{2} \\ \end{array}$$
  2. 证明：$\exists\ \xi\in(0,c), \eta\in(c,1)$，使$\displaystyle{\frac{1}{f'(\xi)}+\frac{1}{f'(\eta)}=2}$ $$\begin{array}{ll} \because & f(0) = 0, f(1)=1, f(c) = \frac{1}{2} \\ \therefore & \exists\ \xi\in (0,c), \eta\in (c,1) \\ & f'(\xi) = \frac{f(c)-f(0)}{c-0} = \frac{1}{2c} \\ & f'(\eta) = \frac{f(1)-f(c)}{1-c} = \frac{1}{2(1-c)} \\ \therefore & \frac{1}{f'(\xi)} + \frac{1}{f'(\eta)} = 2c + 2(1-c) = 2 \\ \end{array}$$
• 例3 $f(x)\in C[a,b]$，在$(a,b)$内可导，$f(a)=f(b), f'_{+}(a) >0$，证明：$\exists\ \xi,\eta\in(a,b)$，使得$f'(\xi)>0,f'(\eta)<0$ $$\begin{array}{ll} \because & f'_{+}(a) > 0 \\ \therefore & \exists\ c\in(a,b), f(c) > f(a) \\ \therefore & \exists\ \xi\in(a,c), \eta\in(c,b) \\ & f'(\xi) = \frac{f(c)-f(a)}{c-a} \\ & f'(\eta) = \frac{f(b)-f(c)}{b-c} \\ \because & f(c)>f(a), f(a) =f(b), a<c<b \\ \therefore & f'(\xi) > 0, f'(\eta) <0 \\ \end{array}$$
• 例4 $f(x)\in C[a,b]$，在$(a,b)$内可导，且$f(a)=f(b)=1$，证明：$\exists\ \xi,\eta\in(a,b)$，使$e^{\eta-\xi}[f'(\eta)+f'(\eta)] = 1$ $$\begin{array}{ll} & \text{Analysis} \\ & e^{\eta-\xi}[f'(\eta)+f(\eta)] = 1 \\ \Rightarrow & e^{\eta}[f'(\eta)+f(\eta)] = e^{\xi} \\ \Rightarrow & \text{左式}=(e^{\eta}f(\eta))' \\\\ & \text{Proof} \\ & \text{令}\phi(x) = e^{x}f(x) \\ \because & f(a) = f(b) =1 \\ \therefore & \phi(a) = e^{a},\phi(b) = e^{b} \\ \therefore & \exists\ \eta\in(a,b), \phi'(\eta) = \frac{e^{b}-e^{a}}{b-a} \\ & \text{而}\phi'(x) = e^{x}[f'(x)+f(x)] \\ \therefore & e^{\eta}[f'(\eta)+f(\eta)] = \frac{e^{b}-e^{a}}{b-a} \\ & \text{令}h(x) = e^{x} \\ \because & \exists\ \xi\in(a,b), h'(\xi) = \frac{h(b)-h(a)}{b-a} = \frac{e^{b}-e^{a}}{b-a};\text{且} h'(x)=e^{x} \\ \therefore & \exists\ \xi\in(a,b), e^{\xi} = \frac{e^{b}-e^{a}}{b-a} \\ \therefore & e^{\eta}[f'(\eta)+f(\eta)] = e^{\xi} \\ & e^{\eta-\xi}[f'(\eta)+'(\eta)]=1 \end{array}$$
• 例5 $f(x)\in C[a,b]$，在$(a,b)$内可导$(a>0)$ 证明：$\exists\ \xi,\eta\in(a,b)$，使得$f'(\xi)=(a+b)\frac{f'(\eta)}{2\eta}$ $$\begin{array}{ll} & \text{Analysis} \\ & f'(\xi)=(a+b)\frac{f'(\eta)}{2\eta} \\ \Rightarrow & \frac{f'(\xi)}{(a+b)} = \frac{f'(\eta)}{(x^{2})'} \\\\ & \text{Proof} \\ & \text{令}g(x) = x^{2}\quad x\in[a,b] \\ \because & a > 0 \\ \therefore & g'(x) \neq 0\\ \therefore & \exists\ \eta\in(a,b), \frac{f'(\eta)}{g'(\eta)} = \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f(b)-f(a)}{b^{2}-a^{2}} \\ & \text{而}g'(x) = 2x \\ \therefore & \frac{f'(\eta)}{2\eta} = \frac{f(b)-f(a)}{(b-a)(b+a)} \\ \because & \exists\ \xi\in(a,b), f'(\xi) = \frac{f(b)-f(a)}{b-a} \\ \therefore & \frac{f'(\eta)}{2\eta} = \frac{f'(\xi)}{b+a} \\ & f'(\xi) = (a+b)\frac{f'(\eta)}{2\eta} \end{array}$$
• 例6 $f(x)\in C[a,b]$，在$(a,b)$内可导$(a>0)$ 证明$\exists\ \xi,\eta\in(a,b)$，使得$ab\cdot f'(\xi) = \eta^{2}f'(\eta)$ $$\begin{array}{ll} & \text{Analysis} \\ & ab\cdot f'(\xi) = \eta^{2}f'(\eta) \\ \Rightarrow & ab\cdot f'(\xi) = \frac{f'(\eta)}{\frac{1}{\eta^{2}}} \\ \Rightarrow & \text{右式}=\frac{f'(\eta)}{(-\frac{1}{x})'} \\\\ & \text{Proof} \\ & \text{令}g(x) = -\frac{1}{x}\quad x\in[a,b] \\ \because & a > 0 \\ \therefore & g'(x) \neq 0\\ \therefore & \exists\ \eta\in(a,b), \frac{f'(\eta)}{g'(\eta)} = \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f(b)-f(a)}{\frac{1}{a}-\frac{1}{b}} = ab\cdot \frac{f(b)-f(a)}{b-a} \\ & \text{而}g'(x) = \frac{1}{x^{2}} \\ \therefore & \eta^{2}f'(\eta) = ab\cdot \frac{f(b)-f(a)}{b-a} \\ \because & \exists\ \xi\in(a,b), f'(\xi) = \frac{f(b)-f(a)}{b-a} \\ \therefore & \eta^{2}f'(\eta) = ab\cdot f'(\xi) \end{array}$$

型六 Lagrange的应用

常见情形

1. $f(b)-f(a)\Rightarrow \text{Lagrange}$
2. $f(a),f(b),f(c)\text{ or }f'(a),f'(b),f'(c)\Rightarrow \text{Lagrange}\times 2$
3. $f\Leftrightarrow f'$ 且无积分

例题

• 例1 $\displaystyle{\lim_{x\to +\infty}x^{2}\bigg(\sin \frac{1}{x-1}-\sin\frac{1}{x+1}\bigg)}$ $$\begin{array}{ll} & \text{令}f(t) = \sin t \\ \therefore & \exists\ \xi\in(\frac{1}{x-1},\frac{1}{x+1}), f'(\xi)=\frac{\sin \frac{1}{x-1}-\sin\frac{1}{x+1}}{\frac{1}{x-1}-\frac{1}{x+1}} \\ & \sin \frac{1}{x-1}- \sin \frac{1}{x+1} = \frac{2}{x^{2}-1}f'(\xi)\\ \therefore & \displaystyle{\text{原式}=\lim_{x\to +\infty}\frac{2x^{2}}{x^{2}-1}f'(\xi)} \\ \because & \displaystyle{\lim_{x\to +\infty}\frac{1}{x+1} = \lim_{x\to +\infty}\frac{1}{x+1 = 0}} \\ \therefore & \text{原式}= \displaystyle{\lim_{x\to +\infty}\frac{2x^{2}}{2x^{2}-1}f'(0) = \lim_{x\to +\infty}\frac{2x^{2}}{2x^{2}-1}\cos 0 = 2} \\ \end{array}$$
• 例2 $\displaystyle{\lim_{x\to\infty}f'(x)=e}$，且$\displaystyle{\lim_{x\to \infty}[f(x)-f(x-1)]=\lim_{x\to\infty}\bigg(\frac{x+c}{x-c}\bigg)^{x}}$，求$c$ $$\begin{array}{ll} \because & \exists\ \xi\in(x-1,x),\ f'(\xi) = \frac{f(x)-f(x-1)}{x-(x-1)} = f(x)-f(x-1) \\ \therefore & \text{左式}=\displaystyle{\lim_{x\to \infty}f'(\xi)} \\ \because & \displaystyle{\lim_{x\to \infty}x=\lim_{x\to \infty}x-1 = \infty} \\ \therefore & \text{左式} = \displaystyle{\lim_{\xi\to \infty}f'(\xi) = e} \\ \therefore & \displaystyle{\lim_{x\to \infty}\bigg(\frac{x+c}{x-c}\bigg)^{x} = e} \\ & \displaystyle{\lim_{x\to \infty}\bigg[\bigg(1+\frac{2c}{x-c}\bigg)^{\frac{x-c}{2c}}}\bigg]^{x\frac{2c}{x-c}} = e^{\lim_{x\to \infty}\frac{2cx}{x-c}} = e^{2c} \\ \therefore & 2c = 1 \\ & c=\frac{1}{2} \end{array}$$
• 例3 $f(0)=0,f''(x)>0$，证：$2f(1)<f(2)$ $$\begin{array}{ll} \because & f(0) = 0 \\ \therefore & \exists\ \xi_{1}\in(0,1),\ \xi_{2}\in(1,2) \\ & f'(\xi_{1}) = \frac{f(1)-f(0)}{1-0} = f(1) \\ & f'(\xi_{2}) = \frac{f(2)-f(1)}{2-1} = f(2)-f(1) \\ \because & f''(x)>0 \text{ 且 } \xi_{2}>\xi_{1} \\ \therefore & f'(\xi_{2})>f'(\xi_{1}) \\ & f(2)-f(1)>f(1) \\ & f(2) > 2f(1) \end{array}$$
• 例4 $f(x)$在$[a,b]$上可导，且$\vert f'(x)\vert\le M$，$f(x)$在$(a,b)$内至少有一个零点，证明 $\vert f(a)\vert + \vert f(b)\vert \le M(b-a)$ $$\begin{array}{ll} \because & f(x)\text{在}(a,b)\text{内至少有一个零点} \\ & \text{设}x=c \quad (a<c<b)\text{时，有}f(c) = 0 \\ \therefore & \exists\ \xi_{1}\in(a,c),\ \xi_{2}\in(c,b) \\ & f'(\xi_{1}) = \frac{f(c)-f(a)}{c-a} = \frac{-f(a)}{c-a} \\ & f'(\xi_{2}) = \frac{f(b)-f(c)}{b-c} = \frac{f(b)}{b-c} \\ \because & \vert f'(x)\vert \le M \\ \therefore & \vert f'(\xi_{1})\vert = \frac{\vert f(a)\vert}{c-a} \le M \\ & \vert f'(\xi_{2})\vert = \frac{\vert f(b)\vert}{b-c} \le M \\ \therefore & \vert f(a)\vert +\vert f(b)\vert \le M[(c-a)+(b-c)] \\ & \vert f(a)\vert +\vert f(b)\vert \le M(b-a) \\ \end{array}$$