错题集-极值

发表: 4/26/2026 更新: 4/27/2026 字数: 0 字 时长: 0 分钟

判断极值的方法：极值判别法

$\mathrm{例}1.f^{\prime }(1)=0.\underset{x\to 1}{lim}\frac{f(x)}{\sin \pi x}=-2,\mathrm{问}x=1\mathrm{是}\mathrm{极}\mathrm{大}\mathrm{点}\mathrm{还}\mathrm{是}\mathrm{极}\mathrm{小}\mathrm{点}?$

解:

法一:

$\mathrm{\exists }\delta >0,\mathrm{当}0<|x-1|<\delta \mathrm{时}.\frac{f^{\prime }(x)}{\sin \pi x}<0.$
$\left\{\begin{array}{c}{f}^{\prime }(x)<0,x\in (1-\delta ,1)\\ f^{\prime }(x)>0,x\in (1,1+\delta )\end{array}\right\}\Rightarrow x=1\mathrm{为}\mathrm{极}\mathrm{小}\mathrm{点}.$(极限保号性)

法二:

$f^{\prime }(1)=0.$
$-2=\underset{x\to 1}{lim}\frac{f^{\prime }(x)}{\sin [\pi +\pi (x-1)]}=-\underset{x\to 1}{lim}\frac{f^{\prime }(x)}{\pi (x-1)}=-\frac{1}{\pi }\underset{x\to 1}{lim}\frac{f^{\prime }(x)-f^{\prime }(1)}{x-1}=-\frac{1}{\pi }{f}^{″}(1)$
$\Rightarrow f^{″}(1)=2\pi >0$
$\therefore x=1\mathrm{为}\mathrm{极}\mathrm{小}\mathrm{点}.$

$\mathrm{例}2.0<a<b.\mathrm{证}:\ln \frac{b}{a}>\frac{2(b-a)}{a+b}.$

解:

法一:

$\mathrm{证}:\ln \frac{b}{a}>\frac{2(b-a)}{a+b}\iff \ln \frac{b}{a}>\frac{2(\frac{b}{a}-1)}{1+\frac{b}{a}}\iff \ln \frac{b}{a}-\frac{2(\frac{b}{a}-1)}{1+\frac{b}{a}}>0,\frac{b}{a}>1.$
$\mathrm{令}f(x)=\ln x-\frac{2(x-1)}{1+x}(x>1).$
$f(1)=0-0=0,f^{\prime }(x)=\frac{1}{x}-\frac{4}{(1+x{)}^2}=\frac{(x-1)(x+3)}{x(1+x{)}^2}>0(x>1).$
$\left\{\begin{array}{c}f(1)=0\\ f^{\prime }(x)>0(x>1)\end{array}\right\}\Rightarrow f(x)>0(x>1).$
$\therefore \ln x-\frac{2(x-1)}{1+x}>0(x>1)\Rightarrow \ln \frac{b}{a}-\frac{2(\frac{b}{a}-1)}{1+\frac{b}{a}}>0.$
$\therefore \ln \frac{b}{a}>\frac{2(b-a)}{a+b}.$

法二:

$\mathrm{证}:\ln \frac{b}{a}>\frac{2(b-a)}{a+b}\iff (a+b)(\ln b-\ln a)-2(b-a)>0.$
$\mathrm{令}\phi (x)=(a+x)(\ln x-\ln a)-2(x-a)(x>a).\phi (a)=0$
${\phi }^{\prime }(x)=\ln x-\ln a+\frac{a}{x}-1,{\phi }^{\prime }(a)=0;$
${\phi }^{″}(x)=\frac{1}{x}-\frac{a}{x^2}=\frac{x-a}{x^2}>0(x>a).$
$\left\{\begin{array}{c}{\phi }^{\prime }(a)=0\\ {\phi }^{″}(x)>0(x>a)\end{array}\right\}\Rightarrow {\phi }^{\prime }(x)>0(x>a).$
$\left\{\begin{array}{c}\phi (a)=0\\ {\phi }^{\prime }(x)>0(x>a)\end{array}\right\}\Rightarrow \phi (x)>0(x>a).$
$\because b>a$
$\therefore \phi (b)>0.$