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Update perpare.tex
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@@ -6,15 +6,25 @@
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% 因为所以
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\usepackage{amsmath}
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% 数学公式
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\usepackage{geometry}
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\setcounter{tocdepth}{4}
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\setcounter{secnumdepth}{4}
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% 设置四级目录
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\usepackage{geometry}
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\geometry{papersize={21cm,29.7cm}}
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\geometry{left=3.18cm,right=3.18cm,top=2.54cm,bottom=2.54cm}
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% 设置页边距
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\usepackage{indentfirst}
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\setlength{\parindent}{2.45em}
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% 设置首行缩进
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\usepackage{setspace}
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\renewcommand{\baselinestretch}{1.5}
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% 设置行距
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\usepackage{tikz}
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% 绘图
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\usetikzlibrary{positioning}
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% 为了实现相对位置的设定
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\usepackage{xcolor}
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% 为了实现不同的颜色
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\author{Didnelpsun}
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\title{考研数学准备}
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\begin{document}
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@@ -35,7 +45,7 @@
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$y=f(x)$,定义域为$D$,值域为$R$,若对于每一个$y\in R$,必然存在$x\in D$使$y=f(x)$成立,则可以定义一个新函数$x=\psi(y)$,这个函数就是$y=f(x)$的\textbf{反函数},一般记作$x=f^{-1}(y)$,其定义域为$R$,值域为$D$,对于反函数,原来的函数称为\textbf{直接函数}。
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\begin{enumerate}
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\item \textcolor{red}{严格单调}函数必然有反函数,即函数导数恒正或恒负必然有反函数。
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\item $x=f^{-1}(y)$与$y=f(x)$在同一坐标系中完全重合
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\item $x=f^{-1}(y)$与$y=f(x)$在同一坐标系中完全重合。
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\item $y=f^{-1}(x)$与$y=f(x)$关于$y=x$对称。
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\item $f[f^{-1}(x)]$或$f[\psi(x)]$变为x,称为湮灭。
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\end{enumerate}
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@@ -50,7 +60,7 @@ $y=f(x)$,定义域为$D$,值域为$R$,若对于每一个$y\in R$,必然
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$\therefore x\in[-1,3]$
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$\therefore\frac{d\psi(x)}{dx}=(-x^2+2x+3)'=-2x+2=0$
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$\therefore\frac{\rm{d}\psi(x)}{\rm{d}x}=(-x^2+2x+3)'=-2x+2=0$
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$\therefore x=1$,驻点为1
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@@ -78,34 +88,34 @@ $f(-x)=\ln(-x+\sqrt{x^2+1})=\ln(\frac{1}{\sqrt{x^2+1}+x})=-\ln(x+\sqrt{x^2+1})=-
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对其求单调性,即通过链式法则求导:
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$\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+1}}\cdot (1+\frac{2x}{2\sqrt{x^2+1}})=\frac{1}{\sqrt{x^2+1}}>0$
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$\frac{\rm{d}y}{\rm{d}x}=\frac{1}{x+\sqrt{x^2+1}}\cdot (1+\frac{2x}{2\sqrt{x^2+1}})=\frac{1}{\sqrt{x^2+1}}>0$
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所以该函数严格单调增。
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然后求$y$的反函数。
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然后求$y$的反函数:
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$$
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\begin{aligned}
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\because y&=\ln(x+\sqrt{x^2+1}) \\
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e^y&=e^{\ln(x+\sqrt{x^2+1})} \\
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&=x+\sqrt{x^2+1}
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\end{aligned}
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\begin{aligned}
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\because y & =\ln(x+\sqrt{x^2+1}) \\
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e^y & =e^{\ln(x+\sqrt{x^2+1})} \\
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& =x+\sqrt{x^2+1}
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\end{aligned}
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$$
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$$
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\begin{aligned}
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\because -y&=-\ln(x+\sqrt{x^2+1}) \\
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&=\ln(\frac{1}{x+\sqrt{x^2+1}}) \\
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&=\ln(\sqrt{x^2+1}-x) \\
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e^{-y}&=\sqrt{x^2+1}-x
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\end{aligned}
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\begin{aligned}
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\because -y & =-\ln(x+\sqrt{x^2+1}) \\
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& =\ln(\frac{1}{x+\sqrt{x^2+1}}) \\
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& =\ln(\sqrt{x^2+1}-x) \\
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e^{-y} & =\sqrt{x^2+1}-x
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\end{aligned}
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$$
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$$
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\begin{aligned}
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\therefore e^y-e^{-y}&=2x \\
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x&=\frac{e^y-e^{-y}}{2}
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\end{aligned}
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\begin{aligned}
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\therefore e^y-e^{-y} & =2x \\
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x & =\frac{e^y-e^{-y}}{2}
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\end{aligned}
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$$
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解出了用x表示y的函数表达$x=f^{-1}(y)$,即反函数,则$f^{-1}(x)=\frac{e^x-e^{-x}}{2}$
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@@ -120,21 +130,126 @@ $$
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\end{itemize}
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\textbf{例题3:}设$
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f(x)=\left\{
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\begin{array}{rcl}
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\ln\sqrt{x} & & {x\geqslant 1}\\
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2x-1 & & {x< 1}
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\end{array} \right.
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f(x)=\left\{
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\begin{array}{rcl}
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\ln\sqrt{x}, & & x\geqslant 1 \\
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2x-1, & & x< 1
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\end{array}
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\right.
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$,求$f[f(x)]$
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首先广义化:$
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f[f(x)]=\left\{
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\begin{array}{rcl}
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\ln\sqrt{f(x)}, & & f(x)\geqslant 1 \\
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2f(x)-1, & & x<1
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\end{array}
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\right.
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$
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然后画图:\\
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\begin{tikzpicture}[domain=-1:9.5]
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\draw[-latex](-1.5,0) -- (9.5,0) node[below]{$x-axis$};
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\draw[-latex](0,-1.5) -- (0, 1.5) node[above]{$y-aixs$};
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\draw[very thin, gray, densely dashed](-1.5,1.5)grid(9.5,-1.5);
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\draw [black, thick](-0.25,-1.5) -- (1,1);
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\draw[black, thick,domain=1:9.5] plot (\x, {ln(sqrt(\x))});
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\draw [red, densely dashed](-1.5,1) -- (9.5,1) node[below]{$x=1$};
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\filldraw [black] (1,1) circle (2pt) node[above]{$(1,1)$};
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\filldraw [black] (e^2,1) circle (2pt) node[above]{$(e^2,1)$};
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\draw[densely dashed](1,1) -- (1, 0) node[below]{$1$};
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\draw[densely dashed](e^2,1) -- (e^2,0) node[below]{$e^2$};
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\end{tikzpicture}
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所以将定义域分为三段:$[-\infty ,1],[1,e^2],[e^2, +\infty]$,然后根据不同定义域对应的不同函数再代回$f[f(x)]$:
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$$
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f[f(x)]=\left\{
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\begin{array}{rcl}
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\ln\sqrt{\ln\sqrt{x}}, & & x\geqslant e^2 \\
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\ln x-2, & & 1\geqslant x<e^2 \\
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4x-3, & & x<1
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\end{array}
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\right.
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$$
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\subsection{有界性}
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函数指明定义域区间才能讨论函数是否有界。
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证明有界性:函数$f(x)$的定义域$D$,数集$I\in D$,如果存在某正数$M$,对于任一$x\in I$,有$\vert f(x)\vert\leqslant M$,则$f(x)$在$I$上有界,否则无界。
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\subsection{单调性}
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$\begin{matrix}
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\frac{\rm{d}y}{\rm{d}x}>0 & \Rightarrow & (x_1-x_2)[f(x_1)-f(x_2)]>0 & \Rightarrow & f(x)\nearrow \\
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\frac{\rm{d}y}{\rm{d}x}<0 & \Rightarrow & (x_1-x_2)[f(x_1)-f(x_2)]<0 & \Rightarrow &f(x)\searrow
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\end{matrix}
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$
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\subsection{奇偶性}
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\begin{enumerate}
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\item 奇函数:关于原点对称,$f(-x)=-f(x)$。
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\item 偶函数:关于y轴对称,$f(-x)=f(x)$。
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\item 对于定义在$[-l,l]$上的任意函数$f(x)$,$F_1(x)=f(x)-f(-x)$必为奇函数,$F_2(x)=f(x)+f(-x)$必为偶函数。可以参考上面所说的双曲正弦与双曲余弦函数。
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\item 若奇函数在0处有定义,那么$f(0)=0$。
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\item 若偶函数在0处存在导数,那么$f'(0)=0$,即x=0,曲线必然水平,即导数为0。
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\item 若函数$y=f(x)$的函数关于直线$x=T$对称的充分必要条件是$f(x)=f(2T-x)/f(x+T)=f(x-T)$。(令$T-x=t$进行换元计算得到)
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\end{enumerate}
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\textcolor{orange}{注意}:0和1处的函数定义应该注意。
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如当a为0时:$f(b)-f(a)=f'(\xi )(b-a)=f(b)=bf'(\xi)$
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如$f(x)>xf(1)$变形为$\frac{f(x)}{x}>f(1)$,辅助函数$F(x)=\frac{f(x)}{x}$
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所以加减法警惕0,乘除法警惕1。
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\subsection{周期性}
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$f(x+T)=f(x)$,其中T为周期。 \\
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\textcolor{red}{重要结论:}
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\begin{enumerate}
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\item 若$f(x)$为可导的偶函数,则$f'(x)$为奇函数。(1.4.1.1)
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\item 若$f(x)$为可导的奇函数,则$f'(x)$为偶函数。(1.4.1.2)
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\item 若$f(x)$为周期函数,则$f'(x)$也为周期函数且周期不变。(1.4.2)
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\item 连续的奇函数的一切原函数都是偶函数。(1.8.6)
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\item 连续的偶函数的原函数中仅有一个原函数是奇函数。(1.8.6)
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\item 若连续函数$f(x)$以T为周期且$\int_{0}^{T}f(x)\rm{d}x=0$,则$f(x)$的一切原函数也以T为周期。(1.8.8)
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\item 若$f(x)$在有限区间$(a,b)$中可导且$f'(x)$有界,则$f(x)$在$(a,b)$有界。(某一函数在固定区间内变化率是有界的,则变化范围是有界的)
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\end{enumerate}
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\section{函数的图像}
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\subsection{直角坐标系图像}
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\subsubsection{常见图像}
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\paragraph{基本初等函数与初等函数}
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基本初等函数包括:常数函数、幂函数、指数函数、对数函数、三角函数、反三角函数。
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1. 常数函数:$y=A$,A为常数,图像平行于x轴:
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\begin{tikzpicture}[domain=-1:5]
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\draw[-latex](-1,0) -- (5,0) node[below]{$x-axis$};
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\draw[-latex](0,-0.5) -- (0, 1.5) node[above]{$y-aixs$};
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\draw[black, thick](-1,1) -- (5,1) node[below]{$y=A$};
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\end{tikzpicture}
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2. 幂函数:$y=x^{\mu}$,$\mu$为实数,当$x>0$,$y=x^{\mu}$都有定义:
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\begin{tikzpicture}[scale=0.9]
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\draw[-latex](-2,0) -- (2,0) node[below]{$x-axis$};
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\draw[-latex](0,-2) -- (0,4) node[above]{$y-aixs$};
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\draw[black, thick,domain=0.3:2] plot (\x,1/\x) node[below]{$\mu =-1$};
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\draw[black, thick,domain=-2:-0.5] plot (\x,1/\x) node[above]{$\mu =-1$};
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\draw[black, thick,domain=0.01:2] plot (\x, {sqrt(\x)}) node[below]{$\mu =\frac{1}{2}$};
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\draw[black, thick,domain=-2:2] plot (\x,\x) node[above]{$\mu =1$};
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\draw[black, thick,domain=-2:2] plot (\x, {\x*\x}) node[above]{$\mu =2$};
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\end{tikzpicture}
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\paragraph{分段函数}
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\subsubsection{图像变换}
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\paragraph{平移变换}
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