Section02_单调性与极值

极值的判断

1. 定义：$y=f(x)\quad(x\in \mathbb{D})$
   • 若$\exists\ \delta>0$，当 $0<\vert x-x_{0}\vert <\delta$ 时，$f(x)<f(x_{0})$，$x_{0}$为极大值点，$f(x_{0})$为极大值
   • 若$\exists\ \delta>0$，当 $0<\vert x-x_{0}\vert <\delta$ 时，$f(x)>f(x_{0})$，$x_{0}$为极小值点，$f(x_{0})$为极小值

Note: 极值点定不为端点

1. 步骤
   1. 找出$x\in \mathbb{D}$
   2. 找出$f'(x)\begin{cases}=0\\\nexists\end{cases}$ 对应的$x_{0}$
   3. 判别法
      1. Th1 第一充分条件（两种条件均可使用）
         1. $\begin{cases}f(x)>0,x>x_{0}\\f(x)<0,x<x_{0}\end{cases}\Rightarrow x=x_{0}\text{为极小值点}$
         2. $\begin{cases}f(x)<0,x>x_{0}\\f(x)>0,x<x_{0}\end{cases}\Rightarrow x=x_{0}\text{为极大值点}$
      2. Th2 第二充分条件（仅适用于$f'(x_{0})=0,\exists\ f''(x_{0})\ne 0$）
         1. $f''(x_{0})<0\Rightarrow x=x_{0}\text{为极大值点}$
         2. $f''(x_{0})>0\Rightarrow x=x_{0}\text{为极大值点}$

问题类型

型一 极值点的判断

• 例1 $\displaystyle{f'(1)=0,\lim_{x\to 1}\frac{f'(x)}{\sin \pi x}=-1, x=1}$是什么点？ $$\begin{array}{ll} & \text{Method 1} \\ \because & \displaystyle{\lim_{x\to 1}\frac{f'(x)}{\sin \pi x}=-1} \\ \therefore & \exists \delta>0, \text{When }0<\vert x-1\vert<\delta, \frac{f'(x)}{\sin \pi x} = -1 <0 \\ \therefore & \begin{cases} f'(1) < 0, x\in(1-\delta,1) \\ f'(1) > 0, x\in(1, 1+\delta) \end{cases} \\ \therefore & x=1 \text{为函数}f(x) \text{的极小值点} \\\\ & \text{Method 2} \\ \because & \displaystyle{\lim_{x\to 1}\frac{f'(x)}{\sin (\pi + \pi(x-1))}=-1, f'(1) = 0}\\ \therefore & \displaystyle{\lim_{x\to 1}\frac{f'(x)-f'(1)}{- \sin (\pi(x-1))}=-1}\\ \because & \displaystyle{\lim_{x\to 1}\sin \pi(x-1) \sim \pi(x-1)} \\ \therefore & \text{左式}=\displaystyle{\lim_{x\to 1}\frac{f'(x)-f'(1)}{\pi(x-1)}} = 1 \\ & f''(1) = \pi > 0\\ \therefore & x=1 \text{为极小值点} \\ \end{array}$$
• 例2 $f(x)\in[-\infty,+\infty]$，$f'(x)$图像如下所示，$f(x)$有多少个极值点？ _01.jpg) $$\begin{array}{ll} & \text{当}x = a,b,0,c,d \text{时}, f'(x) \begin{cases} =0 \\ \nexists \end{cases} \\ \because & \begin{cases} f'(x) > 0, x<a \\ f'(x) < 0，x>a \end{cases} \\ \therefore & x = a \text{为极大值点} \\ \because & \begin{cases} f'(x) < 0, x<b \\ f'(x) > 0, x>b \end{cases} \\ \therefore & x = b \text{为极小值点} \\ \because & \begin{cases} f'(x)>0,x<0 \\ f'(x)<0, x>0 \end{cases} \\ \therefore & x=0 \text{为极大值点} \\ \because & \begin{cases} f'(x)<0, x<c \\ f'(x)<0, x>c \end{cases} \\ \therefore & x=c \text{不为极值点} \\ \because & \begin{cases} f'(x)<0, x<d \\ f'(x)>0, x>d \end{cases} \\ \therefore & x=d \text{为极小值点} \\ \end{array}$$
• 例3 $f(x)$在$x=0$处二阶可导，$f(0)=0$且$\displaystyle{\lim_{x\to 0}\frac{f(x)+f'(x)}{x} = 2}$，则： A. $f(0)$为极大值点 B. $f(0)$为极小值点 C. $(0,f(0))$为拐点 D. $f(0)$不为极值点，$(0,f(0))$也不为拐点 $$\begin{array}{ll} \because & \displaystyle{\lim_{x\to 0}\frac{f(x)+f'(x)}{x}=2} \\ \therefore & \displaystyle{\lim_{x\to 0}f(x)+f'(x)} = 0 \\ \therefore & f'(0) = 0 \\ \therefore & \displaystyle{\lim_{x\to 0}\frac{f(x)+f'(x)}{x}=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}+\lim_{x\to 0}\frac{f'(x)-f'(0)}{x-0}} \\ & = f'(0) +f''(0) \\ \therefore & f''(0) = 2 >0 \\ \therefore & x=0 \text{为极小值点}\\ \end{array}$$
• 例4 $f(x)$二阶可导，$\displaystyle{\lim_{x\to 2}\frac{f'(x)}{(x-2)^{3}}=\frac{2}{3}}$，则： A. $f(2)$为极小值 B. $f(2)$为极大值 C. $(2,f(2))$为拐点 D. $f(2)$不为极值，且$(2,f(2))$不为拐点 $$\begin{array}{ll} \because & \displaystyle{\lim_{x\to 2}\frac{f'(x)}{(x-2)^{3}}=\frac{2}{3}} \\ \therefore & f'(2)=0 \\ & \exists\ \delta>0, 0<\vert x-2\vert <\delta, \frac{f'(x)}{(x-2)^{3}} = \frac{2}{3} > 0 \\ \therefore & \begin{cases} f'(x)<0, x\in(2-\delta,2) \\ f'(x)>0, x\in(2,2+\delta) \end{cases} \\ \therefore & x=2 \text{为极小值点}\\ \end{array}$$

型二 函数零点或方程的根

常用方法

• 零点定理 $f(x)\in C[a,b],f(a)f(b)<0 \Rightarrow \exists\ c\in(a,b), f(c)=0$
• Rolle中值定理
  1. $f(x)\Rightarrow F(x)\quad(F'(x)=f(x))$
  2. 若$F(a)=F(b),\ \exists\ \zeta\in(a,b), F'(\zeta)=f(\zeta)=0$
• 单调性
  1. $f(x)=(\cdots)\quad x\in(m,M)$
  2. $f'(x)\begin{cases}=0\\\nexists\end{cases}$
  3. 关注两端的趋势$\displaystyle{\lim_{x\to m}f(x), \lim_{x\to M}f(x)}$

例题

• 例1 $\displaystyle{a_{0}+\frac{a_{1}}{2}+\cdots +\frac{a_{n}}{n} = 0}$，证明 $a_{0}+a_{1}x+\cdots + a_{n}x^{n} = 0$ 至少有一正根

$$\begin{array}{ll} & \text{令}f(x) = a_{0} + a_{1}x +\cdots +a_{n}x^{n} \\ & \text{则}F(x) = \int f(x)\cdot dx \\ \therefore & F(x) = a_{0}x + \frac{a_{1}}{2}x^{2} + \cdots + \frac{a_{n}}{n}x^{n+1} \\ \because & F(0) = 0, F(1) = a_{0} + \frac{a_{1}}{2} + \cdots + \frac{a_{n}}{n} \\ \because & a_{0}+\frac{a_{1}}{2}+\cdots + \frac{a_{n}}{n} = 0\\ \therefore & F(0) = F(1) \\ \therefore & \exists\ \zeta\in(0,1), F'(\zeta) = 0 \\ & \text{即}f(\zeta) = 0 \\ \therefore & \text{原方程至少有一个正根} \\ \end{array}$$

• 例2 证明$\displaystyle{e^{x} = -x^{2}+4+a}$ 不可能有3个不同解 $$\begin{array}{ll} & \text{令}f(x) = e^{x} + x^{2} - 4x -a \\ & \text{设}x_{1}<x_{2}<x_{3} \text{为}f(x)\text{的3个零点} \\ \therefore & \exists\ \zeta_{1}\in(x_{1},x_{2}), \zeta_{2}\in(x_{2},x_{3}) \\ & f'(\zeta_{1}) = f'(\zeta_{2}) = 0 \\ \therefore & \exists\ \zeta\in(\zeta_{1},\zeta_{2}), f''(\zeta)=0 \\ \because & f''(x) = e^{x} + 2 > 0 \\ \therefore & \text{原方程不可能存在3个不同解} \\ \end{array}$$
• 例3 讨论$xe^{-x}=a\quad (a>0)$的根的个数 $$\begin{array}{ll} & \text{令}f(x) = xe^{-x} - a \quad x\in[0,+\infty) \\ \therefore & f'(x) = e^{-x}(1-x) \\ \because & f'(1) = 0 \text{且} \begin{cases} f'(x)>0, x<1 \\ f'(x)<0, x>1 \end{cases} \\ \therefore & x=1 \text{为函数}f(x)\text{的最大值点} \\ \because & f(0) = -a<0, \displaystyle{\lim_{n\to +\infty}xe^{-x}-a} = -a<0\\ \therefore & \text{若函数有零点} f(1) \ge 0 \\ & a\le \frac{1}{e} \\ \therefore & a > \frac{1}{e} \text{时，函数}f(x)\text{无零点，即原方程无解} \\ & a = \frac{1}{e} \text{时，函数}f(x)\text{仅有一零点，即原方程仅有一个根} \\ & a < \frac{1}{e} \text{时，函数}f(x)\text{仅有两零点，即原方程有两个不同根} \\ \end{array}$$
• 例4 讨论 $\displaystyle{\ln x = \frac{x}{e}-2}$ 的根 $$\begin{array}{ll} & \text{令}f(x) = \ln x - \frac{x}{e} + 2\quad x\in(0,+\infty) \\ \therefore & f'(x) = \frac{1}{x} - \frac{1}{e} \\ \because & f'(e) = 0 \text{且} \begin{cases} f'(x)>0, x<e \\ f'(x)<0, x>e \end{cases} \\ \therefore & x=e \text{为函数}f(x)\text{的最大值}, f(e) = 1 > 0 \\ \because & \displaystyle{\lim_{x\to 0^{+}}f(x) = -\infty < 0} \\ & \displaystyle{\lim_{x\to +\infty}f(x) = -\infty < 0} \\ \therefore & f(x)\text{有两个零点，即原方程有两个不同根} \\ \end{array}$$

型三 不等式证明

常用方法

1. 根据单调性
2. 中值定理

例题

• 例1 $x>0$，证明 $\displaystyle{\frac{x}{1+x}<\ln (1+x)<x}$ $$\begin{array}{ll} & \text{Method 1} \\ & \text{令}f(x) = \frac{x}{1+x} - \ln(1+x)\quad x\in(0,+\infty)\\ & \text{令}g(x) = x - \ln(1+x)\quad x\in(0,+\infty) \\ \because & f'(x) = \frac{1}{(1+x)^{2}} - \frac{1}{1+x} = \frac{-x}{(1+x)^{2}} \text{且} \begin{cases} f'(0)=0 \\ f'(x)<0, x>0 \end{cases} \\ & g'(x) = \frac{x}{1+x} \text{且}\begin{cases} g'(0) = 0 \\ g'(x) > 0, x>0 \end{cases} \\ \therefore & \displaystyle{f(x) < \lim_{x\to 0^{+}}f(x), g(x)>\lim_{x\to 0^{+}}g(x)} \\ \therefore & f(x) < 0, g(x) > 0 \\ & \frac{x}{1+x} - \ln(1+x) < 0 \\ & x - \ln(1+x) > 0 \\ \therefore & \frac{x}{1+x} < \ln(1+x) < x \\ \\ & \text{Method 2} \\ & \text{令}f(x) = \ln(1+x) \\ \therefore & \exists\ \zeta\in(0,x), f'(\zeta) = \frac{\ln(1+x)-\ln(1+0)}{x-0} = \frac{\ln(1+x)}{x} \\ \because & f'(\zeta) = \frac{1}{1+\zeta} \\ \therefore & \frac{1}{1+x} < f'(\zeta) < \frac{1}{1+0}\\ & \frac{1}{1+x} < \frac{\ln(1+x)}{x} < \frac{1}{1+0} \\ \because & x > 0 \\ \therefore & \frac{x}{1+x} < \ln(1+x) < x \\ \end{array}$$
• 例2 $e<a<b$，证明 $a^{b}>b^{a}$ $$\begin{array}{ll} \because & a^{b} > b^{a} \Leftrightarrow e^{b\ln a} > e^{a\ln b} \Leftrightarrow b\ln a > a\ln b \\ & \text{令}f(x) = x\ln a - a\ln x\quad x\in(a,+\infty) \\ \therefore & f'(x) = \ln a - \frac{a}{x} \\ \because & x>a>e \\ \therefore & f'(x) = \ln a - \frac{a}{x} > 0 \\ \because & f(a) = 0 \\ \therefore & f(x) > f(a) = 0\quad x>a \\ \because & b > a \\ \therefore & f(b) > 0 \\ & b\ln a - a\ln b >0 \\ & b\ln a > a\ln b \\ & a^{b} > b^{a} \end{array}$$
• 例3 $\displaystyle{0<a<b}$，证明 $\displaystyle{\frac{\ln b - \ln a}{b-a}>\frac{2a}{a^{2}+b^{2}}}$ $$\begin{array}{ll} & \text{令}f(x) = \ln x\quad x\in(0,+\infty) \\ \therefore & \exists\ \zeta \in(a,b)\subset(0,+\infty), f'(\zeta) = \frac{\ln b - \ln a}{b-a} \\ & \text{而}f'(x) = \frac{1}{x} \\ \because & a^{2}+b^{2} > 2ab \quad 0<a<b \\ \therefore & \frac{2a}{a^{2}+b^{2}} < \frac{2a}{2ab} = \frac{1}{b} \\ \because & f'(\zeta) = \frac{1}{\zeta} > \frac{1}{b} > \frac{2a}{a^{2}+b^{2}} \\ \therefore & f'(\zeta) = \frac{\ln b - \ln a}{b-a} > \frac{2a}{a^{2}+b^{2}} \\ & \text{原不等式成立} \end{array}$$
• 例4 $b>a>0$，证明 $\displaystyle{\ln\frac{b}{a}>\frac{2(b-a)}{a+b}}$ $$\begin{array}{ll} & \ln \frac{b}{a} > \frac{2(b-a)}{a+b}\Leftrightarrow (a+b)(\ln b - \ln a) - 2(b-a) > 0 \\ & \text{令}f(x) = (a+x)(\ln x - \ln a) - 2(x-a) \quad x> a\\ \therefore & f(a) = 0, f'(x) = (\ln x -\ln a) + \frac{a+x}{x} - 2 \\ \therefore & f'(a) = 0, f''(x) = \frac{1}{x} + \frac{-a}{x^{2}} \\ \because & x>a>0 \\ \therefore & f''(x) > 0 \\ \therefore & f'(x) > f'(a) = 0,\quad x>a \\ \therefore & f(x) > f(a) = 0,\quad x>a \\ \because & b > a \\ \therefore & f(b) = (a+b)(\ln b - \ln a) - 2(b-a) > 0 \\ & \ln \frac{b}{a} > \frac{2(b-a)}{a+b} \end{array}$$