高数强化NO11|极值和拐点|单调性和凹凸性|最值计算|结合方程和图像|结合极限

单调性和凹凸性

$$\begin{array}{rl}& x<0\text{时，}f(x)\nearrow ,f^{\prime }(x)>0,\text{排除}AC.\\ & 0<x<x_0\text{时，}f(x)\nearrow ,f^{\prime }(x)>0,\text{排除}B.\\ & \text{故选}D.\end{array}$$
$$\begin{array}{rl}& \text{改题：}\\ & \text{已知函数}y=f^{\prime }(x)\text{在其定义域内可导，它的图像如图所示，}\\ & \text{则其导函数}y=f^{\prime \prime }(x)\text{的图像为（）}.\end{array}$$
$$逻辑一样，还是选D。$$
$$\begin{array}{rl}& \boxed{【小结】}\text{本题需要理解}f(x)\text{与}{f}^{\prime }(x)\text{的关系：函数单调递增，}{f}^{\prime }(x)>0;\text{函数单调递}\\ & \text{减，}{f}^{\prime }(x)<0。\end{array}$$

$$\begin{array}{rl}& \frac{\partial f(x,y)}{\partial x}>0,\frac{\partial f(x,y)}{\partial y}<0,\\ & \text{关于}x\nearrow \text{关于}y\searrow x_1<x_2,y_1>y_2⟹f(x_1,y_1)<f(x_2,y_1)<f(x_2,y_2)\end{array}$$

$$\begin{array}{rl}& {\int }_0^{1}h(x)dx>0\\ & {\int }_0^{1}f(x)dx=0,f\left(\frac{1}{2}\right)<0\\ & 选D\end{array}$$
$$\begin{array}{rl}& \text{法2: 考试, 特殊值法.}\\ & \text{令}f(x)=\frac{1}{2}-x,{\int }_0^{1}f(x)dx=0,\text{但}f\left(\frac{1}{2}\right)=0,\text{排除}A.\\ & \text{再令}f(x)=x-\frac{1}{2},{\int }_0^1\left(x-\frac{1}{2}\right)dx=0,\text{但}f\left(\frac{1}{2}\right)=0,\text{排除}C.\\ & \text{再令}f(x)=-x^2+\frac{1}{3},{\int }_0^1\left(\frac{1}{3}-x^2\right)dx=0,\text{但}f\left(\frac{1}{2}\right)=\frac{1}{12}>0,\text{排除}B.\\ & \text{选}D.\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{选择题一般考查凹凸性的几何意义，几何意义主要有两个方面，第一，与切线}\\ & \text{的关系，若曲线是凹的，则曲线在切线之上；若曲线是凸的，曲线在切线之下。第二，}\\ & \text{与两点连线的关系，}\\ & \text{若曲线是凹的，则两点连线在曲线之上；若曲线是凸的，两点连线在曲线之下。}\end{array}$$

极值和拐点

直接考查

$$\begin{array}{rl}& \text{解: }y=x\sin x+2\cos x,\\ & y^{\prime }=\sin x+x\cos x-2\sin x=x\cos x-\sin x,\\ & y^{\prime \prime }=\cos x-x\sin x-\cos x=-x\sin x,\\ & y^{\prime \prime \prime }=-\sin x-x\cos x.\\ & \\ & (A)y^{\prime \prime }(0)=0,y^{\prime \prime \prime }(0)=0,x=0\text{的右侧附近}{y}^{\prime \prime }<0,\text{左侧附近}<0,\text{不是拐点};\\ & (B)y^{\prime \prime }(\pi )=0,y^{\prime \prime \prime }(\pi )\ne 0,\text{故}(\pi ,-2)\text{为曲线的拐点};\\ & (C)y^{\prime \prime }\left(\frac{\pi }{2}\right)\ne 0,\text{不是拐点};\\ & (D)y^{\prime \prime }\left(\frac{3\pi }{2}\right)\ne 0,\text{也不是拐点}.\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\\ & 1.\text{此题考查的是函数取拐点的第一、第二充分条件，利用第一充分条件判断时，}\\ & \text{应先求出该函数的二阶导数为}0\text{的点，再依次判断这些点两侧每个区间上二阶导数的符}\\ & \text{号，根据第一充分条件判断它们是否取拐点，或者也可以根据第二充分条件，计算函数}\\ & \text{在二阶导数为零的点处的三阶导数来判断。}\\ & \\ & 2.\text{不同形式函数极值点的判别}\\ & 1)\text{如果能够求出函数的单调区间，一般考虑使用第一充分条件进行判断。特别地，对于}\\ & \text{分段函数，基本都需要运用第一充分条件，使用时，对于分段点，不需要求导数，只需}\\ & \text{要求出左右两边的导数，再结合单调性进行分析；}\\ & 2)\text{不易求单调区间的，比如对于抽象函数，或者类型比较复杂的函数（隐函数或者参数}\\ & \text{方程），一般考虑使用第二充分条件。}\end{array}$$

$$\begin{array}{rl}& \text{解: 记方程}2y^3-2y^2+2xy-x^2=1\text{为}①\\ & \text{在方程}①\text{两边同时对}x\text{求导:}\\ & 6y^2\cdot y^{\prime }-4y\cdot y^{\prime }+2y+2x{y}^{\prime }-2x=0②\\ & \text{在}②\text{中, 令}{y}^{\prime }=0,\text{可得}y=x.\text{代入}①:\\ & 2y^3-y^2-1=0,(\text{令}y\text{依次代入}2,1,-2,-1,0\text{五个值试, 一个一个试})\\ & \text{显然, }y=1\text{为唯一一个根.}\\ & \text{法1：在}②\text{两边同时对}x\text{求导:}\\ & 12y\cdot y^{\prime }\cdot y^{\prime }+6y^2\cdot y^{\prime \prime }-4y^{\prime }\cdot y^{\prime }-4y\cdot y^{\prime \prime }+2y^{\prime }+2y^{\prime }+2x{y}^{\prime \prime }-2=0③\\ & \text{将}x=1,y=1,y^{\prime }=0\text{代入}③\text{可得 }{y}^{\prime \prime }=\frac{1}{2}>0\\ & \text{故}x=1\text{为极小值点.}\end{array}$$

最值的计算

$$\begin{array}{rl}& \text{① 求：}\\ & (\text{端点, 驻点, 和不可导点})②\text{比较函数值.}\\ & \\ & \text{③ 没必要判断}\\ & \text{驻点是否为极值点}\end{array}$$

$$\begin{array}{rl}& \text{解: (1) }f(x+\pi )={\int }_{x+\pi }^{x+\pi +\frac{\pi }{2}}|\sin t|dt\stackrel{u=t-\pi }{=}{\int }_x^{x+\frac{\pi }{2}}|\sin (u+\pi )|d(u+\pi )\\ & ={\int }_x^{x+\frac{\pi }{2}}|-\sin u|du={\int }_x^{x+\frac{\pi }{2}}|\sin u|du=f(x).\text{故}f(x)\text{是以}\pi \text{为周期的周期函数.}\\ & \\ & (2)\text{求}f(x)\text{的值域基本等同于求}f(x)\text{的最值. 又}f(x)\text{以}\pi \text{为周期, 故}\\ & \text{求}f(x)\text{在}x\in \mathbb{R}\text{上的最值等同于求}f(x)\text{在}[0,\pi ]\text{上的最值.}\\ & 因为|\sin x|连续，故f(x)可导\\ & f^{\prime }(x)=\left|\sin \left(x+\frac{\pi }{2}\right)\right|-|\sin x|=|\cos x|-|\sin x|=0,x=\frac{\pi }{4},x=\frac{3\pi }{4}.\\ & \text{在}x=\frac{\pi }{4}\text{的附近: }{f}^{\prime \prime }(x)=(\cos x-\sin x{)}^{\prime }=-\sin x-\cos x,f^{\prime \prime }\left(\frac{\pi }{4}\right)=-\sqrt{2}<0,x=\frac{\pi }{4}\text{为极大值点;}\\ & \text{在}x=\frac{3\pi }{4}\text{的附近: }{f}^{\prime }(x)=(-\cos x-\sin x{)}^{\prime }=\sin x-\cos x,f^{\prime \prime }\left(\frac{3\pi }{4}\right)=\sqrt{2}>0,x=\frac{3\pi }{4}\text{为极小值点.}\\ & f(0)={\int }_0^{\frac{\pi }{2}}|\sin t|dt={\int }_0^{\frac{\pi }{2}}\sin tdt=1,f(\pi )={\int }_{\pi }^{\frac{3\pi }{2}}|\sin t|dt=1,\\ & f\left(\frac{\pi }{4}\right)={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}|\sin t|dt=\sqrt{2},f\left(\frac{3\pi }{4}\right)={\int }_{\frac{3\pi }{4}}^{\frac{5\pi }{4}}|\sin t|dt=2-\sqrt{2}.\\ & \text{则}f(x)\text{的值域为}[2-\sqrt{2},\sqrt{2}].\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\\ & 1.\text{遇到求函数值域问题，可以转化为求函数的最值问题。}\\ & \\ & 2.\text{求函数在闭区间上最值的基本思路：先求导，找到所有驻点及不可导点，再求出上述}\\ & \text{所有点及区间端点处的函数值，这些函数值中，最大的就是函数的最大值，最小的就是}\\ & \text{函数的最小值。}\\ & \\ & \text{如果要求函数在开区间上的最值，则将区间端点值用极限值代替即可。此时，如果区间}\\ & \text{端点值是最大或最小的，那么函数的最值是取不到的。}\end{array}$$
$$\begin{array}{rl}& F(x)={\int }_a^{x}f(t)dt\\ & ①f(x)\text{可积，则}F(x)\text{连续}.\\ & ②f(x)\text{连续，则}F(x)\text{可导}.\\ & ③f(x)\text{可导，则}F(x)\text{二阶可导}.\end{array}$$

$$\begin{array}{rl}& \text{解: 当}x\ge 1\text{时, }0\le t\le 1,\text{故}t\le x\\ & f(x)={\int }_0^1|x^2-t^2|dt=x^2-\frac{1}{3}\\ & \\ & \text{当}0<x<1\text{时, }\\ & f(x)={\int }_0^x|t^2-x^2|dt+{\int }_x^1|t^2-x^2|dt\\ & ={\int }_0^x(x^2-t^2)dt+{\int }_x^1(t^2-x^2)dt\\ & =x^3-\frac{x^3}{3}+\frac{1^2-x^3}{3}-x^2(1-x)\\ & =\frac{4}{3}{x}^3-x^2+\frac{1}{3}\\ & \\ & f(x)=\{\begin{array}{ll}\frac{4}{3}{x}^3-x^2+\frac{1}{3},& 0<x<1\\ x^2-\frac{1}{3},& 1\le x\end{array},f^{\prime }(x)=\{\begin{array}{ll}4x^2-2x,& 0<x<1\\ 2x,& x>1\end{array}\\ & \\ & f_{+}^{\prime }(1)=\underset{x\to 1^{+}}{\lim }\frac{x^2-\frac{1}{3}-\frac{2}{3}}{x-1}=2,f_{-}^{\prime }(1)=\underset{x\to 1^{-}}{\lim }\frac{\frac{4}{3}{x}^3-x^2+\frac{1}{3}-\frac{2}{3}}{x-1}=2\\ & \text{故}{f}^{\prime }(1)=2.f^{\prime }(x)=\{\begin{array}{ll}4x^2-2x,& 0<x\le 1\\ 2x,& x>1\end{array},\text{令}{f}^{\prime }(x)=0,x=\frac{1}{2}\\ & \\ & \text{法1: }f(0)={\int }_0^1|t^2|dt=\frac{1}{3}\boxed{\times }f(0)=\underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }{\int }_0^x|t^2|dt=\frac{1}{3}\boxed{\times }\\ & \text{错误: 本题}f(x)\text{和}f(0)\text{无任何关系, 只要提到}f(x)\text{就是错的.}\\ & \\ & \underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }{\int }_0^1|t^2|dt=\frac{1}{3},\underset{x\to +\infty }{\lim }f(x)=\underset{x\to +\infty }{\lim }{x}^2-\frac{1}{3}=+\infty \\ & f\left(\frac{1}{2}\right)=\frac{4}{3}{\left(\frac{1}{2}\right)}^3-{\left(\frac{1}{2}\right)}^2+\frac{1}{3}=\frac{1}{4},\text{则最小值为}f\left(\frac{1}{2}\right)=\frac{1}{4},\text{故最小值为}f\left(\frac{1}{2}\right)=\frac{1}{4}\end{array}$$
$$\begin{array}{rl}& \text{法2: }x>\frac{1}{2}\text{时, }{f}^{\prime }(x)>0,f(x)\text{单调递增；}0<x<\frac{1}{2}\text{时, }{f}^{\prime }(x)<0,f(x)\text{单调递减.}\\ & \text{故最小值为}f\left(\frac{1}{2}\right)=\frac{1}{4}.\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{形如}{\int }_a^b|f(x,t)|dt\text{，令}f(x,t)=0\text{，解出}t\text{，按照}t\text{的取值和}a,b\text{的关系进行讨论。}\end{array}$$
$$\begin{array}{rl}& \text{拓: }y=\arctan x\text{在}x\in \mathbb{R}\text{上是否存在最值?}\\ & \text{答: 不存在。因为最值一定是函数值，}x\text{取不到}\pm \infty ,\\ & \text{故}y\text{也就取不到最值，但能求其值域，此时取“开”区间。}\end{array}$$
$$\begin{array}{rl}& \text{改题:}\\ & f(u,v)={\int }_0^1|x^2-(u+v)x+uv|dx,(u>0,v>0)\\ & ={\int }_0^1|(x-u)(x-v)|dx\\ & \\ & ①u\ge 1,v\ge 1,\text{此时}0<x<1,x<u,x<v\\ & f(u,v)={\int }_0^1(u-x)(v-x)dx\\ & \\ & ②0\le u<1,v>1,\text{此时}0<x<1,x<v\\ & f(u,v)={\int }_0^1(v-x)|(x-u)|dx\\ & ={\int }_0^u(v-x)(u-x)dx+{\int }_u^1(v-x)(x-u)dx\\ & \\ & ③0\le v<1,u>1,\text{此时}0<x<1,x<u\\ & f(u,v)={\int }_0^1(u-x)|(x-v)|dx\\ & ={\int }_0^v(u-x)(v-x)dx+{\int }_v^1(u-x)(x-v)dx\\ & \\ & ④0\le u<1,0\le v<1,u<v,\text{此时}0<x<1\\ & ={\int }_0^u(u-x)(v-x)dx+{\int }_u^v(x-u)(v-x)dx+{\int }_v^1(x-u)(x-v)dx\\ & \\ & ⑤0\le u<1,0\le v<1,v<u,\text{此时}0<x<1\\ & ={\int }_0^v(u-x)(v-x)dx+{\int }_v^u(x-v)(u-x)dx+{\int }_u^1(x-v)(x-u)dx\end{array}$$
$$\begin{array}{rl}& \text{题目让求}{f}^{\prime }(x),\text{故}x=1\text{处的导数也要求.}\\ & \\ & \text{问: 如果不需要求f’(x)是否要求}x=1\text{处的导数.}\\ & \\ & \text{答: 可以不求. 顶多它是不可导点, 只要多算一下}f(1)\text{即可.}\\ & \\ & \underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }{\int }_0^x|t^2-x^2|dt=\frac{1}{3},\underset{x\to +\infty }{\lim }f(x)=\underset{x\to +\infty }{\lim }{x}^2-\frac{1}{3}=+\infty ,f(1)=\frac{2}{3}\\ & f\left(\frac{1}{2}\right)=\frac{4}{3}{\left(\frac{1}{2}\right)}^3-{\left(\frac{1}{2}\right)}^2+\frac{1}{3}=\frac{1}{4},\text{则最小值为}f\left(\frac{1}{2}\right)=\frac{1}{4},\text{故最小值为}f\left(\frac{1}{2}\right)=\frac{1}{4}\end{array}$$

结合方程与图像

1. 结合方程
   

$$\begin{array}{rl}& \text{解: }{f}^{\prime \prime }(0)+{\left[f^{\prime }(0)\right]}^2=0,f^{\prime \prime }(0)=0\\ & f^{\prime \prime }(x)=-{\left[f^{\prime }(x)\right]}^2+x,\text{因}-{\left[f^{\prime }(x)\right]}^2+x\text{可导, 故}{f}^{\prime \prime }(x)\text{可导, }{f}^{\prime \prime \prime }(x)\text{存在}.\\ & f^{\prime \prime \prime }(x)+2f^{\prime }(x)f^{\prime \prime }(x)=1.\\ & f^{\prime \prime \prime }(0)+2f^{\prime }(0)f^{\prime \prime }(0)=1,f^{\prime \prime \prime }(0)=1,\\ & \\ & \text{则}(0,f(0))\text{为曲线的拐点. 选}(C)\end{array}$$
$$\begin{array}{rl}& \text{结论: 可导函数, 拐点和极值点, 不能在同一点处取到.排除AB}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\\ & 1.\text{将驻点代入方程的目的是为了求出二阶导数，如果二阶方程中二阶导数的系}\\ & \text{数中含有零因子，则需要先在方程两边同时除以零因子，然后再代入（此时一般不能直}\\ & \text{接代入，可在等式两边同时取极限）。}\\ & \\ & 2.\text{将驻点代入后得到的二阶导数的值为零，可以继续求解三阶导数的值，从而判断是否}\\ & \text{为拐点。}\end{array}$$
2. 结合图像

选C
$$\begin{array}{rl}& \boxed{【小结】}\\ & 1.\text{若题目中给出了导函数的图像，要判断函数的极值点，基本思路是先找出所}\\ & \text{有可能的极值点（即驻点和导数不存在的点），再逐一通过检验该点左、右两边函数的单}\\ & \text{调性进行判断是否为极值点。此外，如果要求函数的拐点，要找到导函数的极值点和导}\\ & \text{数不存在的点，通过判断左右两边的单调性进行判断。}\\ & \\ & 2.\text{若题目中给出了二阶导数图像，要判断函数的拐点，基本思路是先找出所有可能的拐}\\ & \text{点（即二阶导函数的极值点及二阶导函数不存在的点），再逐一通过检验该点左、右两边}\\ & \text{的单调性进行判断是否为拐点。}\\ & \\ & \text{至此可以将本类题型小结如下，}\\ & 1.\text{极值点是}f(x)\text{单调性改变的点，}{f}^{\prime }(x)\text{符号改变的点；}\\ & 2.\text{拐点是}f(x)\text{凹凸性改变的点，}{f}^{\prime }(x)\text{单调性改变的点，}{f}^{\prime \prime }(x)\text{符号改变的点。}\end{array}$$

结合极限

$$\begin{array}{rl}& \text{解: 法1: 特殊值. 令}f(x)=-(x-a{)}^2,\text{排除}A,C,D,\text{由排除法只能选}(B)\\ & \\ & \text{法2: 保号性. }f(x)-f(a)\text{在}x=a\text{的附近恒小于0, }f(x)\text{取得极大值.}\\ & \\ & \text{法3: 泰勒公式: }\underset{x\to a}{\lim }\frac{f(a)+f^{\prime }(a)(x-a)+\frac{f^{\prime \prime }(a)}{2}(x-a{)}^2+o((x-a{)}^2)-f(a)}{(x-a{)}^2}=-1\\ & \\ & f^{\prime }(a)=0,f^{\prime \prime }(a)=-2<0,\text{故}f(x)\text{在}x=a\text{处取得极大值.}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{若题目中以极限式作为已知条件求极值点与拐点时，通常思路有三种：}\\ & ①凑导数定义；②考虑保号性；③使用泰勒公式。\\ & 本题运用到了凑导数定义以及保号性。\end{array}$$

$$f^{\prime \prime }(x)在0的两边大于0，所以不是拐点$$

$$有极小值。选B$$
$$\begin{array}{rl}& \boxed{【小结】}\text{如果所给极限式中分子或分母的符号确定，则优先考虑保号性。}\\ & 本题中分母恒正，用保号性之后可将分母忽略。\end{array}$$
$$\begin{array}{rl}& \text{极值和拐点，与阶数的关系.}\\ & \\ & f^{\prime }(x_0)=0,f^{\prime \prime }(x_0)\ne 0,\text{此时}x=x_0\text{为极值点.}\\ & \\ & f^{\prime }(x_0)=f^{\prime \prime }(x_0)=0,f^{\prime \prime \prime }(x_0)\ne 0,\text{此时}(x_0,f(x_0))\text{为拐点.}\\ & \\ & f^{\prime }(x_0)=f^{\prime \prime }(x_0)=f^{\prime \prime \prime }(x_0)=0,f^{(4)}(x_0)\ne 0,\text{此时}x=x_0\text{为极值点.}\\ & \\ & f^{\prime }(x_0)=f^{\prime \prime }(x_0)=f^{\prime \prime \prime }(x_0)=f^{(4)}(x_0)=0,f^{(5)}(x_0)\ne 0,\text{此时}(x_0,f(x_0))\text{为拐点.}\\ & \\ & \text{若}{f}^{\prime }(x_0)=f^{\prime \prime }(x_0)=\cdots =f^{(2n-1)}(x_0)=0,f^{(2n)}(x_0)\ne 0,\text{此时}x=x_0\text{为极值点.}\\ & f(x)=\frac{f^{(2n)}(x_0)}{(2n)!}(x-x_0{)}^{2n}+o\left((x-x_0{)}^{2n}\right)\text{和}(x-x_0{)}^{2n}\text{同阶. 阶数为偶}\\ & \text{数时，该点为极值点.（类比}y=x^2\text{）}\\ & \\ & \text{若}{f}^{\prime }(x_0)=f^{\prime \prime }(x_0)=\cdots =f^{(2n)}(x_0)=0,f^{(2n+1)}(x_0)\ne 0,\text{此时}(x_0,f(x_0))\text{为拐点.}\\ & f(x)=\frac{f^{(2n+1)}(x_0)}{(2n+1)!}(x-x_0{)}^{2n+1}+o\left((x-x_0{)}^{2n+1}\right)\text{和}(x-x_0{)}^{2n+1}\text{同阶. 阶数为奇}\\ & \text{数时，该点为拐点.（类比}y=x^3\text{）}\end{array}$$
$$\begin{array}{rl}& \text{练习: }f(x)=(e^x-1-x)\left(x-\ln (1+x)\right)(1+\cos x)(x-\sin x)(\tan x-\arcsin x),\text{求阶数.}\\ & \sim \frac{x^2}{2}\cdot \frac{x^2}{2}\cdot \frac{x^2}{2}\cdot \frac{x^3}{6}\cdot \frac{x^3}{6}=\frac{1}{8\cdot 6^2}{x}^{12}>0\\ & \\ & ①(0,0)\text{为曲线}f(x)\text{的拐点.}\\ & ②x=0\text{为}f(x)\text{的极值点.}\boxed{\surd }\text{极小值.}\end{array}$$
$$\begin{array}{rl}& \boxed{\tan x}\boxed{\sin x}\boxed{x}\boxed{\arcsin x}\boxed{\arctan x}\end{array}$$