高数强化NO27|辅助函数的构造|双中值问题|泰勒中值问题|零点问题

辅助函数的构造

$$\begin{array}{rl}& \text{证：(1) 用反证法. 假设}\exists c\in (a,b)\text{满足}g(c)=0.\\ & \text{则由罗尔定理可得，}\exists {\xi }_1\in (a,c),{\xi }_2\in (c,b)\text{使得}{g}^{\prime }({\xi }_1)=g^{\prime }({\xi }_2)=0\\ & \text{再由罗尔定理可得：}\exists \xi \in ({\xi }_1,{\xi }_2),\text{使得}{g}^{\prime \prime }(\xi )=0.\text{与条件相矛盾.}\\ & \\ & \text{故在}(a,b)\text{内}g(x)\ne 0.\end{array}$$
$$\begin{array}{rl}& \text{分析：}f(\xi )g^{\prime \prime }(\xi )-f^{\prime \prime }(\xi )g(\xi )=0\\ & \\ & f(x)g^{\prime \prime }(x)+f^{\prime }(x)g^{\prime }(x)-(f^{\prime }(x)g^{\prime }(x)+f^{\prime \prime }(x)g(x))\end{array}$$
$$\begin{array}{rl}& (2)\text{设}F(x)=f(x)g^{\prime }(x)-f^{\prime }(x)g(x),F(a)=F(b)=0\\ & \exists \xi \in (a,b),\text{满足}{F}^{\prime }(\xi )=0,\text{即}\\ & f(\xi )g^{\prime \prime }(\xi )-f^{\prime \prime }(\xi )g(\xi )=0,\text{整理后可得}\frac{f(\xi )}{g(\xi )}=\frac{f^{\prime \prime }(\xi )}{g^{\prime \prime }(\xi )}.\end{array}$$
$$\begin{array}{rl}& \text{【小结】常见的构造辅助函数的方法有如下三种：}\\ & \\ & 1.\text{直接积分法：先移项把结论整理成}{F}^{\prime }(\xi )=0，\text{此时左端即为}{F}^{\prime }(\xi )，\text{可以得到}{F}^{\prime }(x)，\\ & \text{再积分求出一个原函数即可以得到辅助函数}F(x)；\\ & \\ & 2.\text{结论是两部分相加，考虑凑导数公式}(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }；\\ & \\ & 3.\text{形如}{f}^{(n)}(\xi )+p(\xi )f^{(n-1)}(\xi )=0\text{辅助函数的构造方法：两边同除}{f}^{(n-1)}(\xi )，\text{得到}\\ & \frac{f^{(n)}(\xi )}{f^{(n-1)}(\xi )}+p(\xi )=0\text{再积分即可得到}F(x)=e^{\int p(x)dx}{f}^{(n-1)}(x)。\\ & \\ & \text{还应注意，积分求辅助函数的时候，要会灵活运用分部积分等积分技巧.}\end{array}$$
$$\begin{array}{rl}& \text{辅助函数的构造}\\ & \\ & \text{数值型：}\\ & f^{\prime }(\xi )+f(\xi )=0.f^{\prime }(x)+f(x)=0,e^x{f}^{\prime }(x)+e^{x}f(x)=0,F(x)=e^{x}f(x)\\ & \\ & f^{\prime }(\xi )+g(\xi )f(\xi )=0.e^{\int g(x)dx}{f}^{\prime }(x)+{\left(e^{\int g(x)dx}\right)}^{\prime }f(x)=0,F(x)=e^{\int g(x)dx}f(x)\end{array}$$

$$\begin{array}{rl}& \text{证：(1) 令}F(x)=f(x)-x,F(1)=0,F(0)=f(0)-0=0,\\ & \text{则由罗尔定理可知：}\exists \xi \in (0,1)\text{使得}{f}^{\prime }(\xi )=1\end{array}$$
$$\begin{array}{rl}& (2)\text{令}F(x)=e^x\left(f^{\prime }(x)-1\right)\\ & F(\xi )=0,f(x)\text{为奇函数，}{f}^{\prime }(x)\text{为偶函数}\\ & f^{\prime }(-\xi )=1,F(-\xi )=0,\\ & \exists \eta \in (-\xi ,\xi )\subset (-1,1),\text{使得}{F}^{\prime }(\eta )=0\\ & \text{整理后可得}{f}^{\prime \prime }(\eta )+f^{\prime }(\eta )=1,\exists \eta \in (-1,1).\end{array}$$
$$\begin{array}{rl}& \text{【小结】}u\cdot v\text{型的辅助函数常用的构造技巧：}\\ & \\ & (1)\xi f^{\prime }+f\to F=xf；\\ & \\ & (2)f^{\prime }+\lambda f\to F=e^{\lambda x}f；\\ & \\ & (3)f\cdot f^{\prime }\to F=f^2。\end{array}$$

$$\begin{array}{rl}& \text{分析：}\\ & f^{\prime }(x)=\left(1-\frac{1}{x}\right)f(x)\\ & f^{\prime }(x)+\left(\frac{1}{x}-1\right)f(x)=0\\ & \\ & F(x)=e^{\int \left(\frac{1}{x}-1\right)dx}f(x)\\ & =x{e}^{-x+C}f(x)(C\in \mathbb{R})\\ & \\ & \text{令}C=1,\text{与题中函数类似}\end{array}$$
$$\begin{array}{rl}& \text{解：令}F(x)=x{e}^{1-x}f(x),F(1)=f(1)\\ & \text{由积分中值定理可知，}\\ & F(\eta )=k\cdot \frac{1}{k}\cdot \eta e^{1-\eta }f(\eta )=f(1)=F(1)\\ & \\ & \text{由罗尔定理可知，}\exists \xi \in (\eta ,1)\subset (0,1)\text{使得}\\ & F^{\prime }(\eta )=0,\text{即}{f}^{\prime }(\xi )=\left(1-\frac{1}{\xi }\right)f(\xi )\end{array}$$
$$\begin{array}{rl}& \text{【小结】积分的区间长度和前面乘的系数如果是1，这个时候被积函数直接取为辅助函}\\ & \text{数。}\end{array}$$

$$\begin{array}{rl}& 分析\\ & f(\xi )=(2-\xi )f^{\prime }(\xi )\\ & f(\xi )+(\xi -2)f^{\prime }(\xi )=0\\ & f^{\prime }(x)+\frac{1}{x-2}f(x)=0\\ & \\ & F(x)=e^{\int \frac{1}{x-2}dx}f(x)\\ & =e^{\ln |x-2|}f(x)\\ & =(x-2)f(x)\end{array}$$
$$\begin{array}{rl}& \text{证：(1) 设}F(x)=(x-2)f(x).F(1)=F(2)=0\\ & \text{故由罗尔定理可知，}\exists \xi \in (1,2)\text{使得}{F}^{\prime }(\xi )=0\\ & \text{即}f(\xi )=(2-\xi )e^{{\xi }^2}\end{array}$$
$$\begin{array}{rl}& 分析\\ & \frac{f(2)}{\eta }-\ln 2\cdot e^{{\eta }^2}=0\frac{f(2)}{x}-\ln 2\cdot e^{x^2}=0\\ & \frac{f(2)}{x}-\ln 2\cdot f^{\prime }(x)=0\\ & {\left(f(2)\ln x-\ln 2\cdot f(x)\right)}^{\prime }\end{array}$$
$$\begin{array}{rl}& (2)G(x)=f(x)\ln x-\ln 2\cdot f(x),\\ & G(1)=0,G(2)=0,\\ & \text{故由罗尔定理可知，}\exists \eta \in (1,2)\text{使得}{G}^{\prime }(\eta )=0,\text{即}f(\eta )=\ln 2\cdot \eta e^{{\eta }^2}\end{array}$$
$$\begin{array}{rl}& \text{【小结】运用柯西中值定理，一定要有两个函数，两个点，一个中值。如果题目中没有}\\ & \text{直接给出两个点，可以根据条件判断是否有隐含的点。}\end{array}$$

双中值问题

$$\begin{array}{rl}& \text{证：(1) 令}F(x)=f(x)+x-1,F(x)\text{在}[0,1]\text{上连续}.\\ & F(0)=-1<0,F(1)=1>0,\\ & \text{故由零点定理可知，}\exists \xi \in (0,1),\text{使得}f(\xi )=1-\xi .\end{array}$$
$$\begin{array}{rl}& (2)\text{由拉格朗日中值定理可知：}\\ & \exists \eta \in (0,\xi ),\frac{f(\xi )-f(0)}{\xi -0}=f^{\prime }(\eta ),\text{即}\frac{1-\xi }{\xi }=f^{\prime }(\eta )\\ & \exists \zeta \in (\xi ,1),\frac{f(1)-f(\xi )}{1-\xi }=f^{\prime }(\zeta ),\text{即}\frac{1-(1-\xi )}{1-\xi }=f^{\prime }(\zeta )\\ & \\ & \text{故}{f}^{\prime }(\eta )f^{\prime }(\zeta )=1\end{array}$$
$$\begin{array}{rl}& \text{【小结】}1.\text{当}\xi ,\eta \text{具有轮换对称性或题目中明确要求}\xi ,\eta \text{互不相同时一般使用拉格朗日}\\ & \text{中值定理。}\\ & \\ & 2.\text{具体解题步骤是：（1）寻找一点}c\text{，将区间}[a,b]\text{分成两个小区间}[a,c]\text{和}[c,b]；（\text{此}\\ & \text{步为做题的关键）；（2）分别在每个区间上使用拉格朗日中值定理；（3）合并两式.}\end{array}$$

$$\begin{array}{rl}& \text{证：令}{g}_1(x)=x,g_2(x)=e^x\\ & \\ & \text{则将}{g}_1(x)\text{与}f(x)\text{在}[a,b]\text{上用柯西中值定理：}\\ & \exists \xi \in (a,b),\frac{f(b)-f(a)}{g_1(b)-g_1(a)}=\frac{f^{\prime }(\xi )}{g_1^{\prime }(\xi )},\text{即}\frac{f(b)-f(a)}{b-a}=\frac{f^{\prime }(\xi )}{1}①\\ & \\ & \text{再将}{g}_2(x)\text{与}f(x)\text{在}[a,b]\text{上用柯西中值定理：}\\ & \exists \eta \in (a,b),\frac{f(b)-f(a)}{g_2(b)-g_2(a)}=\frac{f^{\prime }(\eta )}{g_2^{\prime }(\eta )},\text{即}\frac{f(b)-f(a)}{e^b-e^a}=\frac{f^{\prime }(\eta )}{e^{\eta }}②\\ & \\ & ①②\text{相比得：}\frac{f^{\prime }(\xi )}{f^{\prime }(\eta )}=\frac{e^b-e^a}{b-a}\cdot e^{-\eta }\end{array}$$
$$\begin{array}{rl}& \text{【小结】}1.\text{当}\xi ,\eta \text{不具有轮换对称性，题目中也未要求}\xi ,\eta \text{互不相同时使用柯西中值定}\\ & \text{理，证明思路如下：（1）寻找两个函数}{g}_1(x),g_2(x)；（2）g_1(x),g_2(x)\text{分别和}f(x)\\ & \text{搭配使用柯西中值定理：}\frac{f(b)-f(a)}{g_1(b)-g_1(a)}=\frac{f^{\prime }(\xi )}{g_1^{\prime }(\xi )},\frac{f(b)-f(a)}{g_2(b)-g_2(a)}=\frac{f^{\prime }(\eta )}{g_2^{\prime }(\eta )}；（3）\text{合并上}\\ & \text{述两式。}\\ & \\ & 2、\text{函数}{g}_1(x),g_2(x)\text{的构造，基本思路如下：}\\ & （1）\text{把结论中}\xi \text{有关的放一起，}\eta \text{有关的放一起，即整理成}\frac{f^{\prime }(\xi )}{g_1^{\prime }(\xi )},\frac{f^{\prime }(\eta )}{g_2^{\prime }(\eta )}\text{形式；}\\ & （2）\text{从中读出}{g}_1^{\prime }(\xi ),g_2^{\prime }(\eta )\text{进而得出两个辅助函数}{g}_1(x),g_2(x)。\end{array}$$

泰勒中值定理

$$\begin{array}{rl}& \text{ Taylor（泰勒）公式：}\\ & f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+\cdots +\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n+\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}\\ & (\xi \text{介于}x\text{与}{x}_0\text{之间})\\ & \\ & \text{例：}f(x)=x^2+2x+1,f(1)=1^2+2\cdot 1+1,f(2)=2^2+2\cdot 2+1\\ & \\ & \text{泰勒中值定理：表达式中的}\xi \text{是随着}x\text{的变化而变化的.}\end{array}$$
$$\begin{array}{rl}& \text{证：在}x=0\text{处用泰勒中值定理}\\ & f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(\xi )}{3!}{x}^3(\xi \text{介于}0\text{与}x\text{之间})\\ & \\ & f(1)=f(0)+f^{\prime }(0)\cdot 1+\frac{f^{\prime \prime }(0)}{2!}\cdot 1^2+\frac{f^{\prime \prime \prime }({\xi }_1)}{3!}\cdot 1^3{\xi }_1\in (0,1)①\\ & f(-1)=f(0)+f^{\prime }(0)\cdot (-1)+\frac{f^{\prime \prime }(0)}{2}\cdot (-1{)}^2+\frac{f^{\prime \prime \prime }({\xi }_2)}{3}\cdot (-1{)}^3{\xi }_2\in (-1,0)②\\ & \\ & ①-②\text{相减得：}{f}^{\prime \prime \prime }({\xi }_1)+f^{\prime \prime \prime }({\xi }_2)=6\\ & \\ & \text{由}{f}^{\prime \prime \prime }(x)\text{的连续性，设}{f}^{\prime \prime \prime }(x)\text{在}[-1,1]\text{上的最小值为}m,\text{最大值为}M.\\ & 2m\le f^{\prime \prime \prime }({\xi }_1)+f^{\prime \prime \prime }({\xi }_2)=6\le 2M,\\ & \text{由介值定理可知，}\exists \xi \in ({\xi }_2,{\xi }_1)\subset [-1,1],\text{使得}{f}^{\prime \prime \prime }(\xi )=3.\end{array}$$
$$\begin{array}{rl}& \text{【小结】}1.\text{对于泰勒中值定理，首先要会识别，当题目中出现二阶及二阶以上的高阶导}\\ & \text{数时，考虑泰勒中值定理；}\\ & \\ & 2.\text{如果函数}n+1\text{可导，则使用泰勒中值定理一般展开到}n\text{阶；}\\ & \\ & 3.\text{关于泰勒中值定理在哪一点展开是这部分的难点，基本思路是看题目中函数在哪一点}\\ & \text{给的信息最多，就在哪一点展开。特别的，如果题目中已知函数}f(x)\text{在某一点}x=x_0\text{处}\\ & \text{的导数值，使用泰勒公式时一般就直接在该点展开。}\end{array}$$

零点问题

$$\begin{array}{rl}& \text{解：}x=0\text{显然为一个根.}\\ & x\ne 0\text{时，令}F(x)=\frac{x}{\arctan x}-k,\\ & F^{\prime }(x)=\frac{\arctan x-\frac{x}{1+x^2}}{{\arctan }^{2}x}\\ & \\ & \text{令}g(x)=\arctan x-\frac{x}{1+x^2},\\ & g^{\prime }(x)=\frac{1}{1+x^2}-\frac{1}{1+x^2}-x\cdot \left(-\frac{2x}{(1+x^2{)}^2}\right)=\frac{2x^2}{(1+x^2{)}^2}>0\\ & \\ & g(x)\text{单调递增，}g(0)=0.\\ & x<0\text{时，}{F}^{\prime }(x)<0,F(x)\text{单调递减；}\\ & x>0\text{时，}{F}^{\prime }(x)>0,F(x)\text{单调递增.}\\ & \\ & \underset{x\to 0}{\lim }F(x)=1-k.\\ & \text{当}1-k>0\text{时，方程有一个根}x=0;\\ & \text{当}1-k=0\text{时，方程有一个根}x=0;\\ & \text{当}1-k<0\text{时，方程有三个根.}\end{array}$$
$$\begin{array}{rl}& \text{【小结】讨论或证明方程根的个数用到的理论有两个：单调性与零点存在定理。}\\ & \text{基本步骤如下：}\\ & ①\text{确定函数的单调区间；}\\ & ②\text{确定单调区间分界点函数值的符号（端点不在定义域内的用极限代替）；}\\ & ③\text{在每个单调区间上使用零点存在定理。}\end{array}$$

$$\begin{array}{rl}& \text{解：}f(0)=f(1)=f(2)=0,\\ & \exists {\xi }_1\in (0,1),{\xi }_2\in (1,2),\text{使得}{f}^{\prime }({\xi }_1)=0,f^{\prime }({\xi }_2)=0,\text{故}{f}^{\prime }(x)\text{有两个零点；}\\ & \\ & f(x)\text{中含}{x}^2,x\text{的次数为}2,f^{\prime }(x)=2x(x-1)(x-2)+x^2{\left[(x-1)(x-2)\right]}^{\prime },f^{\prime }(0)=0\\ & \text{故}{f}^{\prime }(x)\text{至少有三个零点.}\end{array}$$
$$\begin{array}{rl}& \text{选(D)}\\ & f(x)\text{是四次多项式，}{f}^{\prime }(x)\text{是三次多项式. 至多有三个零点，则}{f}^{\prime }(x)\text{仅有3个零点.}\end{array}$$
$$\begin{array}{rl}& \text{【小结】讨论导数的零点问题一般要用到三方面的知识：}\\ & 1.\text{罗尔定理：如果}f(x)\text{具有}n\text{个零点，那么}{f}^{\prime }(x)\text{至少有}n-1\text{个零点；}\\ & \\ & 2.\text{泰勒公式：如果}x=a\text{为}f(x)\text{的}n\text{重根（}f(x)\text{与}(x-a{)}^n\text{同阶），则}x=a\text{为}{f}^{\prime }(x)\text{的}\\ & n-1\text{重根；}\\ & \\ & 3.\text{代数基本定理：}n\text{次多项式至多有}n\text{个零点.}\end{array}$$