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\title{Cvi\v{c}en\'i - Matematick\'a anal\'yza 1, MFF}  % tyto dvě položky jsou zde v podstatě formálně, ve skutečnosti nejsou nikde 
\author{Marek C\'uth}

\begin{document}	
\centerline{\MakeUppercase{i. taylor\r{u}v polynom}}\ \\
\textbf{1. Najd\v{e}te Taylor\r{u}v polynom $k$-t\'eho \v{r}\'adu v bod\v{e} 0 pro funkce:}\\[5pt]
a) $\tg(x),\;k=3$\quad
b) $\sin(\sin x),\;k=5$\quad
c) $\frac{1+x+x^2}{1-x+x^2},\;k=4$\quad
d) $\cos(\sin x),\;k=5$\quad
e) $\sin(1- \cos x),\;k=3$\\
f) $e^{2x-x^2},\;k=5$\quad
g) $\frac{x}{e^x-1},\;k=4$\quad
h) $\log(\cos x),\;k=6$\quad\\[10pt]
\textbf{2. Odhadn\v{e}te absolutn\'i chybu aproximace $\sin x \approx x- \frac{x^3}{6}$ na intervalu $[-1/2,1/2]$.}\\[5pt]
\textbf{3. Spo\v{c}t\v{e}te $\sqrt{5}$ s p\v{r}esnost\'i $10^{-2}$}\\[5pt]
\textbf{4. Spo\v{c}t\v{e}te limity}\\[5pt] \resetpr
\pr $\lx \tfrac{\sin x - x + \tfrac{x^3}{6}}{x^5}$
\pr $\lp \Big(n^2-\cotg^2(\tfrac{1}{n})\Big)$
\pr $\lx \tfrac{2\sin(\sin x) - \sin(2x)\sqrt[3]{1+x^2}}{x^5}$
\pr $\lx \tfrac{\cos x - 1 + \tfrac{x^2}{2}}{x^4}$\\
\pr $\lp n^4\Big(\cos(\tfrac{1}{n}) - \exp(-\tfrac{1}{2n^2})\Big)$
\pr $\lx \tfrac{e^x\sin x - x(1+x)}{x^3}$
\pr $\lx \tfrac{\sin(e^{(x^2)} - 1) - 1 + \cos(\sqrt{2}x)}{x^4}$\\
\pr $\lx \tfrac{(1+\sin x)^x - \exp(x^2) + \tfrac{x^3}{2}}{x^4}$
\pr $\lp \sqrt[6]{n^5}\Big(\sin(\tfrac{1}{\sqrt{n}}) - \tfrac{1}{\sqrt[6]{n}}\log(1+\tfrac{1}{\sqrt[3]{n}})\Big)$\\[10pt]


\noindent Dal\v{s}\'i p\v{r}\'iklady k procvi\v{c}en\'i lze nal\'ezt nap\v{r}\'iklad:
\begin{itemize}
	\item v sekci I zde: \url{https://www.karlin.mff.cuni.cz/~cuth/doc/MFF/MA/ma2-2015-16.pdf}
	\item na webu kolegyně Kuncové zde: \url{https://www.karlin.mff.cuni.cz/~kuncova/historie28.php}
\end{itemize}
P\v{r}\'iklady zkou\v{s}kov\'e obt\'i\v{z}nosti lze nal\'ezt ve zkou\v{s}kov\'ych p\'isemk\'ach z Matematiky III, kter\'e jsou vystaveny zde:\\ \url{http://www.karlin.mff.cuni.cz/~kalenda/edu.php?edutype=archpis}\\[30pt]

\vysledky
\textbf{1.}\resetpr \pr $x+\frac{1}{3}x^3$
	\pr $x-\frac{1}{3}x^3+\frac{1}{10}x^5$
	\pr $1+2x+2x^2-2x^4$
	\pr $1-\frac{1}{2}x^2+\frac{5}{24}x^4$
	\pr $\frac{x^2}{2}$	
	\pr $1+2x+x^2-\frac{2}{3}x^3-\frac{5}{6}x^4-\frac{1}{15}x^5$
	\pr $1-\frac{1}{2}x+\frac{1}{12}x^2-\frac{1}{720}x^4$
	\pr $-\frac{x^2}{2}-\frac{1}{12}x^4-\frac{1}{45}x^6$\\[3pt]
 \textbf{2.} $\frac{1}{3840}$\\
 \textbf{3.} $2+\tfrac{1}{4}-\tfrac{1}{64}+\tfrac{1}{2^9} - \tfrac{5}{2^{14}} = 2.23602294922$\\
 \textbf{4.}\resetpr \pr $\tfrac{1}{120}$ \pr $\tfrac{2}{3}$ \pr $\tfrac{3}{5}$ \pr $\tfrac{1}{24}$ \pr $-\tfrac{1}{12}$ \pr $\tfrac{1}{3}$ \pr $\tfrac{2}{3}$ \pr $\tfrac{1}{6}$
\newpage
\centerline{\MakeUppercase{ii. hled\'an\'i primitivn\'i funkce, \'uvod}}\ \\
\textbf{Vyj\'ad\v{r}ete primitivn\'i funkce na maxim\'aln\'ich intervalech existence} \\[10pt]
\textbf{1. P\v{r}\'iklady na integrov\'an\'i "p\v{r}\'imo":}\\
a) $\int x^9 + \frac{1}{x} - 5e^x + x^{-3} - \cos{x} \dx$ \quad
b) $\int 2e^{3x} - \sqrt[5]{5-x} \dx$ \quad
c) $\int \frac{x^2 + 3x + 6}{x^4} \dx$\quad
d) $\int x(1-x)^{10} \dx$\\[10pt]
\textbf{2. P\v{r}\'iklady na integrov\'an\'i "per partes":}\\
a) $\int x^3 \sin{x} \dx$ \quad
b) $\int e^x \cos{x} \dx$ \quad
c) $\int x^n e^x \dx\;, n\in\en$ \quad
d) $\int x \log{x} \dx$ \quad
e) $\int x e^x \cos{x} \dx$  \\[10pt]
\textbf{3. P\v{r}\'iklady na integrov\'an\'i pomoc\'i substituce:}\\ \resetpr
\pr $\int \cotg{x} \dx$
\pr $\int \frac{x^2}{\cos^2{x^3}} \dx$
\pr $\int \frac{x}{1+x^4} \dx$
\pr $\int \frac{1}{x\log{x}} \dx$
\pr $\int \frac{x}{\sqrt{x^2+5}} \dx$
\pr $\int \frac{1}{x\log{x}\log(\log{x})} \dx$ \\[10pt]
\textbf{4. Dal\v{s}\'i p\v{r}\'iklady k procvi\v{c}en\'i:}\\ \resetpr
\pr $\int \left(1 - \frac {1}{x^2}\right)\sqrt{x\sqrt{x}} \dx$
\pr $\int (2^x + 3^x)^2 \dx$
\pr $\int \frac{1}{x^2}\sin(\frac 1x) \dx$
\pr $\int \frac{\sin{x}}{\sqrt{\cos^3x}} \dx$
\pr $\int \frac{x^2}{(8x^3 + 27)^{2/3}} \dx$\\
\pr $\int \frac{\arctg{x}}{1 + x^2} \dx$
\pr $\int \frac{2^{2x}}{9^x - 4^x} \dx$
\pr $\int \arctg{x} \dx$
\pr $\int x^2\sin(2x) \dx$
\pr $\int \sqrt{x}\log^2x \dx$
\pr $\int x^2e^{-2x} \dx$\\
\pr $\int \left(\frac{\log{x}}{x}\right)^2 \dx$
\pr $\int x^5e^{x^3} \dx$
\pr $\int e^{\sqrt{x}} \dx$
\pr $\int x\sin{\sqrt{x}} \dx$
\pr $\int \cos^2x \dx$\\[10pt]

\noindent Dal\v{s}\'i p\v{r}\'iklady k procvi\v{c}en\'i primitivních funkcí lze nal\'ezt nap\v{r}\'iklad:
\begin{itemize}
\item na webu kolegyně Kuncové zde: \url{https://www.karlin.mff.cuni.cz/~kuncova/historie28.php}
\item ve skriptech Ilja Černý: Inteligentní kalkulus. Online zde: \url{http://matematika.cuni.cz/ikalkulus.html}
\end{itemize}
P\v{r}\'iklady zkou\v{s}kov\'e obt\'i\v{z}nosti lze nal\'ezt ve zkou\v{s}kov\'ych p\'isemk\'ach z Matematiky III, kter\'e jsou vystaveny zde:\\ \url{http://www.karlin.mff.cuni.cz/~kalenda/edu.php?edutype=archpis}

\vysledky
\textbf{1.}
  	a) $\frac{x^{10}}{10} + \log{|x|} - 5e^x - \frac{1}{2x^2} - \sin{x}$ na $(-\infty,0)$ a $(0,\infty)$\quad
  	b) $\frac 23 e^{3x} + \frac{5(5-x)^{\frac 65}}{6},\; x\in\er$\\
  	c) $-\frac{1}{x} - \frac{3}{2x^2} - \frac{2}{x^3}$ na $(-\infty,0)$ a $(0,\infty)$\quad
		d) $-\frac{(1-x)^{11}}{11} + \frac{(1-x)^{12}}{12}$\\
\textbf{2.}
  	a) $-x^3\cos{x} + 3x^2\sin{x} + 6x\cos{x} - 6\sin{x},\; x\in\er$\quad
  	b) $\frac 12(e^x\sin{x} + e^x\cos{x}),\; x\in\er$\quad
		c) $I_n:=\int x^n e^x \dx = x^n e^x - nI_{n-1};\,I_{1} := xe^x - e^x,\; x\in\er$\\		
  	d) $\frac 14(2x^2\log{x} - x^2),\; x\in(0,\infty)$\quad
  	e) $\frac 12e^x(x\sin{x} + x\cos{x} - \sin{x}),\; x\in\er$\\
		\textbf{3.}\resetpr
\pr $\log|\sin{x}|$ na ka\v{z}d\'em z interval\r{u} $(k\pi, (k+1)\pi), k\in\zet$\\
  	\pr $\frac 13 \tg{x^3}$ na ka\v{z}d\'em z interval\r{u} $(\sqrt[3]{- \frac \pi2 + k\pi}, \sqrt[3]{\frac \pi2 + k\pi}), k\in\zet$
  	\pr $\frac 12\arctg{x^2},\; x\in\er$\\
  	\pr $\log|\log{x}|$ na $(0,1)$ a $(1,\infty)$
  	\pr $\sqrt{x^2 + 5},\; x\in\er$
  	\pr $\log|\log(\log{x})|$ na $(1,e)$ a $(e,\infty)$\\
	\textbf{4.}\resetpr
	\pr $\frac{4(x^2 + 7)}{7\sqrt[4]{x}},\; D_f = (0,\infty)$
	\pr $\frac{4^x}{\log{4}} + 2\frac{6^x}{\log{6}} + \frac{9^x}{\log{9}},\; D_f = \er$
	\pr $\cos(\frac 1x),\; D_f = \er\setminus\{0\}$	\\
	\pr $\frac {2}{\sqrt{\cos{x}}},\; D_f = \bigcup_{k\in\zet}(-\frac \pi2 + 2k\pi,\frac \pi2 + 2k\pi)$
		\pr $\frac 18\sqrt[3]{8x^3 + 27},\; D_f = \er\setminus\{-\frac 32\}$\\
		\pr $\frac 12\arctg^2{x},\; D_f = \er$
		\pr $-\frac{1}{2\log{\frac 23}}\log|1 - (\frac 23)^{2x}|,\; D_f = \er\setminus\{0\}$\\
		\pr $x\arctg{x} - \frac 12\log(1 + x^2),\; D_f = \er$
		\pr $-\frac{2x^2 - 1}{4}\cos(2x) + \frac x2\sin(2x),\; D_f = \er$\\
		\pr $\frac 23 x^{3/2}\left(\log^2x - \frac 43\log{x} + \frac 89\right),\; D_f = (0,\infty)$
		\pr $-\frac{e^{-2x}}{2}(x^2 + x + \frac 12),\; D_f = \er$\\
		\pr $-\frac 1x(\log^2x + 2\log{x} + 2),\; D_f = (0,\infty)$
		\pr $\frac 13(x^3 - 1)e^{x^3},\; D_f = \er$
		\pr $2(\sqrt{x}-1)e^{\sqrt{x}},\; D_f = (0,\infty)$\\
		\pr $2(6-x)\sqrt{x}\cos{\sqrt{x}} - 6(2-x)\sin{\sqrt{x}},\; D_f = (0,\infty)$
		\pr  $\frac x2 + \frac{\sin(2x)}{4},\; D_f = \er$\\

\newpage
\centerline{\MakeUppercase{ii. hled\'an\'i primitivn\'i funkce, pokra\v{c}ov\'an\'i}}\ \\
\textbf{Vyj\'ad\v{r}ete primitivn\'i funkce na maxim\'aln\'ich intervalech existence} \\[10pt]
\textbf{5. P\v{r}\'iklady na integrování pomocí druhé věty o substituci:}\\ \resetpr
%\pr $\int \sqrt{4-x^2} \dx$
\pr $\int \tfrac{1}{(1-x^2)^{3/2}} \dx$
\pr $\int \tfrac{1}{(x^2+a^2)^{3/2}} \dx$ ($a>0$)\\[10pt]
\textbf{6. P\v{r}\'iklady, kde se mus\'i funkce "lepit":}\\ \resetpr
\pr $\int |2x+1| \dx$
\pr $\int |\cos{x}| \dx$
\pr $\int \max\{x,x^2\} \dx$
\pr $\int \sqrt{x^6} \dx$
\pr $\int \sin{|2x-1|} \dx$
\pr $\int |\sin{x} + \cos{x}| \dx$\\
\pr $\int e^{-|x|} \dx$\\[10pt]
\textbf{7. Integrace racion\'aln\'ich funkc\'i:}\\ \resetpr
\pr $\int \tfrac{5x^6-20x^5+20x^4+x^2-4x+11}{(x-2)^2} \dx$
\pr $\int \tfrac{2x+1}{x^2+4x+5} \dx$
\pr $\int \tfrac{2x+1}{x^2+4x-5} \dx$
\pr $\int \tfrac{8x^3 - 5x^2 - 15x + 24}{(x+1)^2(2x^2 - 6x + 5)} \dx$
\pr $\int \tfrac{3x+1}{(9x^2-12x+6)^2} \dx$\\
\pr $\int \frac{1}{e^{2x} + e^x - 2} \dx$
\pr $\int \tfrac{1}{x}\tfrac{\log(2x)}{\log^2(x)+3}$\\[10pt]

\vysledky
\textbf{5.}\resetpr
	%\pr $2\arcsin \frac x2 + \sin(2\arcsin \frac x2) = 2\arcsin \frac x2 + \frac x2\sqrt{4-x^2},\;D_f = (-2,2)$
	\pr $\tg(\arcsin x) = \frac{x}{\sqrt{1-x^2}},\;D_f = (-1,1)$
	\pr $\frac{1}{a^2}\sin(\arctg \frac xa) = \frac{x}{a^2\sqrt{a^2+x^2}},\;D_f = \er$\\
	\textbf{6.}\resetpr
	\pr $F(x) = \left\{ 
  		\begin{array}{ll} -(x^2+x) & x < -\frac 12\\
			x^2+x+\frac 12 & x \geq -\frac 12\\ \end{array} \right.$
	\pr $F(x) = \left\{
  		\begin{array}{ll} \sin{x} + 4k & x\in[-\frac \pi2 + 2k\pi, \frac \pi2 + 2k\pi], k\in\zet\\
  		
			-\sin{x}+4k+2 & x\in(\frac \pi2 + 2k\pi, 3\frac \pi2 + 2k\pi), k\in\zet\\ \end{array} \right.$\\
		\pr $F(x) = \left\{ 
  		\begin{array}{ll} \frac {x^3}{3} & x\in(-\infty,0)\\
			 \frac {x^2}{2}& x\in[0,1]\\
			 \frac {x^3}{3} + \frac 16& x\in(1,\infty) \end{array} \right.$
		\pr	$\frac 14 |x|x^3,\; x\in\er$
  	\pr $F(x) = \left\{ 
  		\begin{array}{ll} -\frac 12\cos(2x-1) & x\geq \frac 12\\
			\frac 12\cos(2x-1) - 1 & x < \frac 12\\ \end{array} \right.$\\
		\pr $F(x) = \begin{cases}
        \cos x - \sin x + 4k\sqrt{2} - 2\sqrt{2} & x\in [-\pi+2k\pi,-\tfrac{\pi}{4}+2k\pi], k\in\zet\\
        \sin x - \cos x + 4k\sqrt{2} & x\in [-\tfrac{\pi}{4}+2k\pi,\tfrac{3\pi}{4} + 2k\pi], k\in\zet\\
        \cos x - \sin x + 4k\sqrt{2} + 2\sqrt{2} & x\in [\tfrac{3\pi}{4}+2k\pi,\pi+2k\pi], k\in\zet.
        \end{cases}$
		\pr $F(x) = \left\{ 
  		\begin{array}{ll} e^x - 2 & x < 0\\
			-e^{-x} & x \geq 0\\ \end{array} \right.$ \\
	\textbf{7.}\resetpr
	\pr $x^5 + x - \tfrac{7}{x-2},\;D_f = \er\setminus\{2\}$
  \pr $\log(x^2+4x+5)-3\arctg(x+2),\;D_f = \er$\\
	\pr $\tfrac{3}{2}\log|x+5| + \tfrac{1}{2}\log|x-1|,\;D_f = \er\setminus\{1,-5\}$
	\pr $3\log|x+1| - \tfrac{2}{x+1} + \tfrac{1}{2}\log(2x^2-6x+5) + 2\arctg(2x-3),\;D_f = \er\setminus\{-1\}$\\
	\pr $-\tfrac{1}{6(9x^2-12x+6)} + \tfrac{\sqrt{2}}{8}\arctg(\tfrac{3x-2}{\sqrt{2}}) + \tfrac{3x-2}{4(9x^2-12x+6)} = \tfrac{\sqrt{2}}{8}\arctg(\tfrac{3x-2}{\sqrt{2}}) + \tfrac{9x-8}{12(9x^2-12x+6)},\;D_f = \er$\\
	\pr $-\frac x2 + \frac 13\log{|e^x-1|} + \frac 16\log(e^x+2),\;D_f = \er\setminus\{0\}$
	\pr $\tfrac{1}{2}\log(\log^2 x + 3) + \tfrac{\log 2}{\sqrt{3}}\arctg(\tfrac{\log x}{\sqrt{3}}),\;D_f = (0,\infty)$\\
	\newpage

\centerline{\MakeUppercase{ii. hled\'an\'i primitivn\'i funkce, pokra\v{c}ov\'an\'i II}}\ \\
\textbf{Vyj\'ad\v{r}ete primitivn\'i funkce na maxim\'aln\'ich intervalech existence} \\[10pt]
\textbf{8. Goniometrick\'e substituce:}\\ \resetpr
\pr $\int \frac{\sin^3x + \sin{x}}{\cos^3x + \cos{x}} \dx$
\pr $\int \frac{1}{(2+\cos x)\sin x} \dx$
\pr $\int \frac{1}{\cos^2x(4\sin^2x - 1)} \dx$
\pr $\int \frac{\cos x - 1}{\sin x - 1} \dx$
\pr $\int \frac{\sin^2x}{1+\sin^2x} \dx$
\pr $\int \frac{1}{5+ \cos x} \dx$\\[20pt]
\textbf{9. P\v{r}\'iklady s odmocninou:}\\ \resetpr
\pr $\int \tfrac{1}{x}\sqrt{\tfrac{x+1}{x-1}} \dx$
\pr $\int \tfrac{x}{\sqrt{-x^2+2x+8}} \dx$
\pr $\int \tfrac{1}{x + \sqrt{x^2+x+1}}\dx$
\pr $\int \sqrt{x^2+2x-3}\dx$\\[10pt]
\vysledky
\textbf{8.}\resetpr
	\pr $\frac 32\log(\cos^2x + 1) - 2\log|\cos{x}|,\; D_f = \er\setminus\bigcup_{k\in\zet}\{\frac \pi2 + k\pi\}$
	\pr $\frac 13\log(\cos x + 2) - \frac 12\log(\cos x + 1) + \frac 16\log(1-\cos x) = \frac 16\log\left(\frac{(1-\cos x)(\cos x + 2)^2}{(1+\cos x)^3}\right),\; D_f = \er\setminus\bigcup_{k\in\zet}\{k\pi\}$
	\pr  $\frac 13\tg{x} + \frac{2}{3\sqrt{3}}\log\left|\frac{\sqrt{3}\tg{x} - 1}{\sqrt{3}\tg{x} + 1}\right|,\; D_f = \er\setminus\bigcup_{k\in\zet}\{\frac \pi2 + k\pi, \frac \pi6 + k\pi, -\frac \pi6 + k\pi\}$
	\pr Pro ka\v{z}d\'e $k\in\zet$ je primitivní funkce na $(\frac \pi2 + 2k\pi,\frac \pi2 + 2\pi + 2k\pi)$ d\'ana p\v{r}edpisem $F(x) = \left\{ 
  		\begin{array}{ll} -\log(\tg^2\tfrac{x}{2}+1) + 2\log|\tg\tfrac{x}{2}-1|  - \frac{2}{\tg\tfrac{x}{2}-1} & x\in(\frac \pi2 + 2k\pi,\frac \pi2 + 2\pi + 2k\pi)\setminus\{2k\pi + \pi\}\\
			0 & x = 2k\pi + \pi\\ \end{array} \right.$\\
	\pr $F(x) = \left\{ 
  		\begin{array}{ll} \arctg(\tg x) - \frac{1}{\sqrt{2}}\arctg(\sqrt{2}\tg x) + k\pi(1-1/\sqrt{2}) & x\in(-\frac \pi2 + k\pi,\frac \pi2 + k\pi),\quad k\in\zet\\
			\frac \pi2 - \frac{\pi}{2\sqrt{2}} + k\pi(1-1/\sqrt{2}) & x = \frac \pi2 + k\pi,\quad k\in\zet\\ \end{array} \right.$\\
			\pr $F(x) = \left\{
  		\begin{array}{ll} \frac {1}{\sqrt{6}} \arctg\left(\sqrt{\frac 23}\tg{(\frac x2)}\right)  + k\frac{\pi}{\sqrt{6}} & \text{ pro }x\in(-\pi + 2k\pi, \pi + 2k\pi), k\in\zet\\
			\frac {\pi}{2\sqrt{6}} + k\frac{\pi}{\sqrt{6}} & \text{ pro }x = \pi + 2k\pi, k\in\zet\\ \end{array} \right.$\\[10pt]
			\textbf{9.}\resetpr \pr $\log(\sqrt{\tfrac{x+1}{x-1}} + 1) - \log\Big|\sqrt{\tfrac{x+1}{x-1}} - 1\Big| - 2\arctg\sqrt{\tfrac{x+1}{x-1}},\;D_f = \er\setminus(-1,1]$\\
			\pr $2\arctg\sqrt{\tfrac{x+2}{4-x}} - \sqrt{-x^2+2x+8},\;D_f=(-2,4)$\\
			\pr $-\tfrac{3}{2}\log\Big|2(x+\sqrt{x^2+x+1})+1\Big| + \tfrac{3}{4(x+\sqrt{x^2+x+1}) + 2} + 2\log\Big|x+\sqrt{x^2+x+1}\Big|,\; D_f=\er\setminus\{-1\}$\\
			\pr $\tfrac{1}{2}(x+1)\sqrt{x^2+2x-3} - 2\log(x+1+\sqrt{x^2+2x-3}),\; D_f=\er\setminus(-3,-1)$.\\
   Jiné možnosti jak výsledek posledního příkladu zapsat jsoun například:
   \begin{itemize}
        \item $- 2\log(x+1+\sqrt{x^2+2x-3}) - \tfrac{2}{(x+1+\sqrt{x^2+2x-3})^2} + \tfrac{1}{4}(x+\sqrt{x^2+2x-3}) + \tfrac{1}{8}(x+\sqrt{x^2+2x-3})^2$
        \item  $\sgn(x+3)\Big(-2\log|t+1| + 2\log|t+1| + \tfrac{2}{t+1} + \tfrac{2}{t-1} - \tfrac{2}{(t+1)^2} + \tfrac{2}{(t-1)^2}\Big)$, kde $t = \sqrt{\tfrac{x-1}{x+3}}$
   \end{itemize}

   \newpage
    
		\centerline{\MakeUppercase{iii. newton\r{u}v integr\'al}}\ \\
\textbf{Vypo\v{c}t\v{e}te n\'asleduj\'ic\'i integr\'aly}\\ \resetpr
	\textbf{1.}
	\pr $\int_0^2 |1-x| \dx$
	\pr $\int_0^\infty \frac{1}{x^2 + 3x + 2} \dx$
	\pr $\int_0^{100\pi}\sqrt{1-\cos 2x} \dx$
	\pr $\int_{\frac1e}^e |\log x| \dx$
	\pr $\int_0^{\log 2} \sqrt{e^x-1} \dx$\\
	\pr $\int_0^{2\pi} x^2\cos x \dx$\\[3pt]
	\textbf{2.}\resetpr
	\pr $\int_{-1}^{\infty} \frac{1}{\sqrt{x+1} + \sqrt{(x+1)^3}}$
	\pr $\int_0^{5\pi} \frac{1}{\sin^2x + 2\cos^2x} \dx$
	\pr $\int_{1}^3 \frac{1}{x\sqrt{-x^2+4x-3}} \dx$
	\pr $\int_0^{2\pi} \frac{1}{2+\sin x} \dx$\\
	\pr $\int_{-\infty}^\infty \frac{e^x}{e^{2x}+e^x+1} \dx$
	\pr $\int_0^1 x^2\arctg x \dx$
	\\[20pt]
	
	Dal\v{s}\'i p\v{r}\'iklady k procvi\v{c}en\'i lze nal\'ezt nap\v{r}\'iklad v sekci V zde:\\ \url{https://www.karlin.mff.cuni.cz/~cuth/doc/MFF/MA/ma2-2016-17.pdf}\\
a tak\'e v sekc\'ich VII, IX zde:\\
\url{http://www.karlin.mff.cuni.cz/~kalenda/data/ma215-cv.pdf}\\
a tak\'e v kapitole 10 zde:\\ \url{http://matematika.cuni.cz/ikalkulus.html}\\
a tak\'e na webu kolegyně Kuncové zde:\\
\url{https://www.karlin.mff.cuni.cz/~kuncova/historie28.php}
\\[20pt]
P\v{r}\'iklady zkou\v{s}kov\'e obt\'i\v{z}nosti lze nal\'ezt ve zkou\v{s}kov\'ych p\'isemk\'ach z Matematiky III, kter\'e jsou vystaveny zde:\\ \url{http://www.karlin.mff.cuni.cz/~kalenda/edu.php?edutype=archpis}\\[30pt]
	\vysledky
	\textbf{1.}\resetpr \pr $1$
	\pr $\log 2$
	\pr $200\sqrt{2}$
	\pr $2-\frac2e$
	\pr $2 - \frac{\pi}{2}$
	\pr $4\pi$\\
	\textbf{2.}\resetpr
	\pr $\pi$
	\pr $5\frac{\sqrt{2}}{2}\pi$
	\pr $\tfrac{\pi}{\sqrt{3}}$
	\pr $\frac{2\sqrt{3}}{3}\pi$
	\pr $\frac{2}{3\sqrt{3}}\pi$
	\pr $\frac{1}{6}(\log 2 - 1 + \pi/2)$
\newpage	
\centerline{\MakeUppercase{iv. konvergence newtonova integr\'alu}}\ \\
\textbf{1. Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'ich integr\'al\r{u}}\\ \resetpr
\pr $\int_0^1 \frac{1}{e^{\sqrt{x}-1}} \dx$
\pr $\int_0^1 \frac{\sqrt{x}}{\sqrt{1-x^4}} \dx$
	\pr $\int_0^1 \frac{\log(1-\cos x)}{\sqrt[3]{x(x-1)}} \dx$
	\pr $\int_1^{\infty} \frac{e^x}{x^4} \dx$
	\pr $\int_8^\infty \frac{x+1}{(x+4)^3}\sin x \dx$\\[5pt]
	\textbf{2. Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'ich integr\'al\r{u}}\\ \resetpr
	\pr $\int_0^1 \tfrac{1}{\sqrt[3]{x(e^x - e^{-x})}} \dx$
	\pr $\int_0^1 (\log x)e^{-x^2} \dx$
	\pr $\int_0^\infty \tfrac{\arctan x}{x^2} \dx$
	\pr $\int_1^\infty \sin^2\tfrac 1x \dx$
	\pr $\int_0^1 \frac {1}{e^x-\cos x} \dx$\\
	\pr $\int_0^{\frac\pi2} \frac{1}{\sqrt{\sin x}\cos x} \dx$
	\pr $\int_0^{\frac\pi2} \frac{1}{\sqrt{\sin x}\sqrt{\cos x}} \dx$
	\pr $\int_0^{\frac\pi2} \tg x \dx$
	\pr $\int_0^{\frac\pi2} \frac{1}{\sqrt{\tg x}} \dx$
	\pr $\int_0^1 x^{-10 x} \dx$
	\pr $\int_0^1 \tfrac{\log(1-x^2)}{x^2\sqrt{1-x^2}} \dx$\\
	\pr $\int_0^\infty \pi - 2\arctan x \dx$
	\pr $\int_0^1 \tfrac{\arccos x}{\log^2(1/x)} \dx$
	\pr $\int_1^2 \frac{\sqrt{x-1}}{\sqrt{x}-\sqrt[3]{x}} \dx$\\[5pt]
	\textbf{3. Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'ich integr\'al\r{u} (p\v{r}\'iklady zkou\v{s}kov\'e obt\'i\v{z}nosti)}\\ \resetpr
	\pr $\int_1^\infty \sqrt[4]{e^{1/x^2} - e^{-1/x^2}}\frac{\log(x+1)}{x+1} \dx$
	\pr $\int_0^1 \log(\arctg x)\frac{\pi/2 - \arcsin x}{(e^{1-x}-1)^2} \dx$
	\pr $\int_0^5 \frac{\log(x^2-10x+26)}{xe^{1/x}(5-x)^{5/2}}\dx$\\
	\pr $\int_1^\infty \frac{\big(1 - \cos\tfrac{1}{x}\big)^{3/4}}{\sqrt{\sin \tfrac{1}{x}}}\arctg\Big(3+\tfrac{\log x}{x}\Big)\dx$
	\pr $\int_0^1 \frac{e^{2x^2} - e^{x^2}}{x^3\sqrt{\sin x}}\log(2 + \arctg x)\dx$
	\\[20pt]
 \noindent Dal\v{s}\'i p\v{r}\'iklady k procvi\v{c}en\'i lze nal\'ezt nap\v{r}\'iklad:
\begin{itemize}
	\item ve skriptech Ilja Černý: Inteligentní kalkulus (kapitola 10, vyšetřování absolutní konvergence integrálů z Příkladů 10.101-10.145). Online zde: \url{http://matematika.cuni.cz/ikalkulus.html}
	\item na webu kolegyně Kuncové zde: \url{https://www.karlin.mff.cuni.cz/~kuncova/historie28.php}
    \item příklady ze zkouškových písemek Kalkulu 1 v semestru 2019-2020: \url{https://www.karlin.mff.cuni.cz/~cuth/doc/MFF/Kalkulus/kalkulus1_zkPis_web.pdf}
\end{itemize}
	\vysledky
	\textbf{1.}\resetpr \pr K (konverguje) \pr K \pr K \pr D (diverguje) \pr K\\
	\textbf{2.}\resetpr \pr K \pr K \pr D \pr K \pr D \pr D \pr K \pr D \pr K \pr K \pr K \pr D \pr D \pr K\\
	\textbf{3.}\resetpr \pr K \pr D \pr K \pr D \pr D\\[40pt]
  
  \newpage
	\centerline{\MakeUppercase{v. Riemann\r{u}v-Stieltjes\r{u}v integr\'al}}\ \\[5pt]	
\textbf{1. Spo\v{c}\'itejte hodnotu n\'asleduj\'ic\'ich Riemannov\'ych-Stieltjesov\'ych integr\'al\r{u}}\\ \resetpr
\pr $\int_0^1 x^2 \dgx{x^3}$
\pr $\int_0^1 x^2 \dgx{e^x}$
\pr $\int_1^e (x+4) \dgx{(e^x+\log x)}$\\
\pr $\int_0^3 e^x \dgx{g(x)}$, kde $g(x):=\begin{cases}1 & \text{pokud }0\leq x < 1\\ 3 & \text{pokud }1\leq x \leq 2\\ -1 & \text{pokud }2<x\leq 3\end{cases}$
\pr $\int_1^3 x^2 \dgx{g(x)}$, kde $g(x):=\begin{cases}1 & \text{pokud }x=1\\ 2 & \text{pokud }1<x<2\\ 3 & \text{pokud }2\leq x\leq 3\end{cases}$\\
\pr $\int_0^{100\pi} |\sin x| \dgx{\sin(x)}$
\pr $\int_0^2 |x^2-1| \dgx{([x]+x)}$ (kde $[\cdot]$ značí dolní celou část)\\
\pr $\int_1^4 f(x) \dgx{g(x)}$, kde $f(x):=\begin{cases}-1 & \text{pokud }x\in [1,2)\\ 2 & \text{pokud }x\in [2,3)\\ -2 & \text{pokud }x\in [3,4]\end{cases}\quad$ a $\quad g(x):=\begin{cases}x+1 & \text{pokud }x\in [1,2]\\ x^2 & \text{pokud } x\in (2,3]\\ x^3 & \text{pokud }x\in (3,4]\end{cases}$\\
\pr $\int_{-3}^0 \big([x+2] - x\big) \dgx{g(x)}$, kde $g(x) = |e^x-e^{-2}|$ (a $[\cdot]$ značí dolní celou část)
\\[20pt]
\vysledky
	\textbf{1.}\resetpr \pr $\frac{3}{5}$ \pr $e-2$ \pr $(e+3)e^e + 3 - 3e$ \pr $2e-4e^2$ \pr $5$ \pr $0$ \pr $5$ \pr $-99$
    \pr $ 3e^{-3} - 5e^{-2} - e^{-1} + 2$
    \newpage
    \centerline{\MakeUppercase{vi. oby\v{c}ejn\'e diferenci\'aln\'i rovnice prvn\'iho \v{r}\'adu}}\ \\

\textbf{Najd\v{e}te v\v{s}echna maxim\'aln\'i \v{r}e\v{s}en\'i n\'asleduj\'ic\'ich diferenci\'aln\'ich rovnic}\\[5pt]
\textbf{1.}\resetpr
	\pr $y'=3x^2y^2$
	\pr $y'=\sqrt{y}e^{-x}$
	\pr $yy'=\frac{1-2x}{y}$
	\pr $xy'=-2-y$, $y(1)=1$
 \pr $y' = xe^xy$, $y(1)=1$\\
 \pr $y' = x\sqrt[3]{1-y}$, $y(0)=0$
 \pr $y' = \sin x\sin y$, $y(0)=\tfrac{\pi}{2}$
 \pr $y' = \sqrt{4-y^2}\cdot x$, $y(0)=1$\\[5pt]
 \textbf{2.}\resetpr
	\pr $xy' = -(x+y)$
	\pr $xyy' = y^2 - x^2$
	\\[5pt]
 \textbf{3.}\resetpr
	\pr $y' = \tfrac{y}{x} - 1$
	\pr $xy' + y = \log x + 1$
	\pr $y' + \tfrac{x+1}{x}y = 3xe^{-x}$, $y(1)=-\tfrac{1}{e}$\\[5pt]
\textbf{4.}\resetpr
\pr $y' = \frac{y\log y}{\sin x}$, $y(\tfrac{\pi}{2}) = \tfrac{1}{e}$
\pr $y' + y\cos x = \sin 2x$, $y(0)=3$
\pr $xy' = y\log \tfrac{y}{x}$, $y(1)=e^3$\\[3pt]
\pr $y' = \frac{2x+y}{x}\log(\frac{2x+y}{x}) + \frac{y}{x}$
\pr $y' + |x|y = x^5$, $y(0)=0$
\pr $y' = \tfrac{1}{y^3}\sqrt[3]{y^4-16}\cdot e^x$, $y(0)=-\sqrt[4]{16 + \big(\tfrac{4}{3}\big)^3}$
	\\[20pt]
 \noindent Dal\v{s}\'i p\v{r}\'iklady k procvi\v{c}en\'i lze nal\'ezt nap\v{r}\'iklad:
\begin{itemize}
	\item v online sbírce Tomáše Bárty zde: \url{http://www.karlin.mff.cuni.cz/~barta/pcODR/}
        \item na webu kolegyně Kuncové zde: \url{https://www.karlin.mff.cuni.cz/~kuncova/historie28.php}
\end{itemize}
P\v{r}\'iklady zkou\v{s}kov\'e obt\'i\v{z}nosti lze nal\'ezt ve zkou\v{s}kov\'ych p\'isemk\'ach z Matematiky IV, kter\'e jsou vystaveny zde:\\ \url{http://www.karlin.mff.cuni.cz/~kalenda/edu.php?edutype=archpis}\\
případně se dají použít příklady ze zkouškových písemek Kalkulu 1 v semestru 2019-2020: \url{https://www.karlin.mff.cuni.cz/~cuth/doc/MFF/Kalkulus/kalkulus1_zkPis_web.pdf}\\[20pt]
\vysledky
\textbf{1.}\resetpr
\pr Stacion\'arn\'i \v{r}e\v{s}en\'i je $y_{s}(x)=0$, $x\in\er$. Pro ka\v{z}d\'e $c\in\er$ jsou maxim\'aln\'imi \v{r}e\v{s}en\'imi funkce 
\[
y^1_{c}(x)=\tfrac{-1}{x^3+c},\; x\in (-\infty,-\sqrt[3]{c})\quad \text{a}\quad y^2_{c}(x)=\tfrac{-1}{x^3+c},\; x\in (-\sqrt[3]{c},\infty)
\]
\pr Stacion\'arn\'i \v{r}e\v{s}en\'i je $y_{s}(x)=0$, $x\in\er$. Pro ka\v{z}d\'e $c>0$ je maxim\'aln\'im \v{r}e\v{s}en\'im funkce 
\[
y_{c}(x)=\begin{cases}0 & x\in(\infty,-\log c]\\ \tfrac{1}{4}(c-e^{-x})^2 & x\in(-\log c,\infty)\end{cases}
\]
\pr Stacion\'arn\'i \v{r}e\v{s}en\'i nejsou. Pro ka\v{z}d\'e $c\in\er$ je $y_c(x) = \sqrt[3]{3(x-x^2+c)}$ maxim\'aln\'i \v{r}e\v{s}en\'i definovan\'e na intervalech: pokud $c<-\tfrac{1}{4}$ pak $x\in\er$; pokud $c=-\tfrac{1}{4}$ pak $x\in(-\infty,\tfrac{1}{2})$ nebo $x\in(\tfrac{1}{2},\infty)$; pokud $c>-\tfrac{1}{4}$ pak $x\in(-\infty,\tfrac{1}{2} - \sqrt{\tfrac{1}{4}+c})$ nebo $x\in(\tfrac{1}{2} - \sqrt{\tfrac{1}{4}+c},\tfrac{1}{2} + \sqrt{\tfrac{1}{4}+c})$ nebo $x\in(\tfrac{1}{2} + \sqrt{\tfrac{1}{4}+c},\infty)$.\\
\pr $y(x) = \tfrac{3}{x} - 2$, $x\in (0,\infty)$
\pr $y(x) = \exp((x-1)e^x)$, $x\in \er$
\pr $y(x) = \begin{cases}1 - \sqrt{(1-\tfrac{x^2}{3})^3} & x\in (-\sqrt{3},\sqrt{3}),\\
1 & x\in (-\infty,-\sqrt{3}]\cup [\sqrt{3},\infty).\end{cases}$\\
\pr $y(x) = 2\arctg\Big(e^{1-\cos x}\Big)$, $x\in\er$
\pr $y(x) = \begin{cases}2\sin(\tfrac{x^2}{2} + \tfrac{\pi}{6}) & x\in (-\sqrt{\tfrac{2}{3}\pi},\sqrt{\tfrac{2}{3}\pi}),\\
2 & x\in (-\infty,-\sqrt{\tfrac{2}{3}\pi}]\cup [\sqrt{\tfrac{2}{3}\pi},\infty).\end{cases}$\\[5pt]
\textbf{2.}\resetpr \pr Pro ka\v{z}d\'e $K\in\er\setminus\{0\}$ jsou $y(x)=-\tfrac{x}{2} + \tfrac{K}{x}$, $x\in (-\infty,0)$ nebo $x\in (0,\infty)$ maxim\'aln\'i \v{r}e\v{s}en\'i. Dal\v{s}\'im \v{r}e\v{s}en\'im je $y(x)=-\tfrac{x}{2}$, $x\in\er$\\
\pr Pro ka\v{z}d\'e $c\in\er$ jsou maxim\'aln\'imi \v{r}e\v{s}en\'imi $y(x) = \pm x\sqrt{2c - \log x^2}$, $x\in (-e^c,0)$ nebo $x\in(0,e^c)$\\
\textbf{3.}\resetpr \pr $y(x) = -x\log|x| + cx$, $x\in (-\infty,0)$ nebo $x\in (0,\infty)$, $c\in\er$
\pr $y(x) = \log x + \tfrac{c}{x}$, $x\in (0,\infty)$, $c\in\er$\\
\pr $y(x) = (x^2 - \tfrac{2}{x})e^{-x}$, $x\in (0,\infty)$\\[5pt]
\textbf{4.}\resetpr 
\pr $y(x)=e^{-\tg \tfrac{x}{2}}$, $x\in (0,\pi)$
\pr $y(x)=2(\sin x - 1) + 5e^{-\sin x}$, $x\in \er$
\pr $y(x)=xe^{1+2x}$, $x\in (0,\infty)$\\
\pr $y(x) = -2x + x\exp(x\log 5)$, $x\in (0,\infty)$
\pr $y(x) = \begin{cases}x^4-4x^2 + 8(1-e^{-x^2/2}) & x\in [0,\infty),\\
-x^4-4x^2 + 8(e^{x^2/2}-1) & x\in (-\infty,0]\end{cases}$\\
\pr $y(x) = \begin{cases}-2 & x\in (-\infty, \log(1/3)],\\
-\sqrt[4]{16 + \big(\tfrac{8}{3}(e^x - \tfrac{1}{3})\big)^{3/2}} & x\in [ \log(1/3,\infty)\end{cases}$
\newpage
\centerline{\MakeUppercase{vii. oby\v{c}ejn\'e diferenci\'aln\'i rovnice druh\'eho \v{r}\'adu}}\ \\
\textbf{1. Uva\v{z}ujte rovnici $y'' - y = \frac{e^x - e^{-x}}{e^x + e^{-x}}$.}\resetpr \pr Doka\v{z}te, \v{z}e $\{e^x, e^{-x}\}$ tvo\v{r}\'i fundament\'aln\'i syst\'em \v{r}e\v{s}en\'i homogenn\'i rovnice na intervalu $(-\infty,\infty)$ \pr Nalezn\v{e}te obecn\'e \v{r}e\v{s}en\'i rovnice na intervalu $(-\infty,\infty)$ \pr Nalezn\v{e}te \v{r}e\v{s}en\'i spl\v{n}uj\'ic\'i po\v{c}\'ate\v{c}n\'i podm\'inku $y(0)=\frac{\pi}{2} + 3$, $y(1) = (e+\tfrac{1}{e})\arctg(e) + e + \tfrac{2}{e}$.\\[5pt]
\textbf{2. Uva\v{z}ujte rovnici $y'' - \tfrac{4}{x}y' + \tfrac{6}{x^2}y = 3x$ s po\v{c}\'ate\v{c}n\'i podm\'inkou $y(1)=3$, $y'(1)=10$.}\resetpr \pr Doka\v{z}te, \v{z}e $\{x^2, x^3\}$ tvo\v{r}\'i fundament\'aln\'i syst\'em \v{r}e\v{s}en\'i homogenn\'i rovnice na intervalu $(0,\infty)$ \pr Nalezn\v{e}te \v{r}e\v{s}en\'i rovnice.\\[5pt]
\textbf{3. Uva\v{z}ujte rovnici $y'' - \tfrac{2x+2}{x^2+2x}y' + \tfrac{2}{x^2+2x}y = x^2+2x$ s po\v{c}\'ate\v{c}n\'i podm\'inkou $y(-1)=0$, $y'(-1)=4/3$.}\resetpr \pr Doka\v{z}te, \v{z}e $\{x^2, x+1\}$ tvo\v{r}\'i fundament\'aln\'i syst\'em \v{r}e\v{s}en\'i homogenn\'i rovnice na intervalu $(-2,0)$ \pr Nalezn\v{e}te \v{r}e\v{s}en\'i rovnice.\\[5pt]
\textbf{4. Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i diferenci\'aln\'ich rovnic s po\v{c}\'ate\v{c}n\'i podm\'inkou.}\\ (Hint: nejprve dokažte, že $\{\cos x,\sin x\}$ tvo\v{r}\'i fundament\'aln\'i syst\'em \v{r}e\v{s}en\'i homogenn\'i rovnice)\\ \resetpr 
\pr $y'' + y = \tfrac{1}{\sin x}$, $y(\tfrac{\pi}{2}) = 1$, $y'(\tfrac{\pi}{2}) = 0$
\pr $y'' + y = \sin^2 x$, $y(0) = 1$, $y'(0) = 1$
\pr $y'' + y = |x|$, $y(0) = 1$, $y'(0) = 2$
\\[5pt]
%%% OBECNE SE DAJI NEJAKE PRIKLADY GENEROVAT TAKTO %%%%%%%%
\iffalse
obecn\v{e}ji:\\
- $\{x^2+cx+d,x+a\}$ je F.S. \v{r}e\v{s}en\'i rovnice $y''+p(x)y'+q(x)y=0$ pokud $p(x)=-q(x)(x+a)$ a $q(x) = \frac{2}{x^2+2ax+ac-d}$\\
- partikul\'arn\'i \v{r}e\v{s}en\'i rovnice $y''+p(x)y'+q(x)y=r(x)$ je pak $y_p(x) = c_1(x)(x^2+cx+d) + c_2(x)(x+a)$, kde $c_1(x)$ je primitivn\'i funkce k $\frac{r(x)(x+a)}{x^2+2ax+cd-d}$ a $c_2(x)$ je primitivn\'i funkce k $-\frac{r(x)(x^2+cx+d)}{x^2+2ax+cd-d}$\\
- speci\'aln\v{e} partikul\'arn\'i \v{r}e\v{s}en\'i rovnice $y''+p(x)y'+q(x)y=x^2+2ax+cd-d$ je $y_p(x) = (\tfrac{x^2}{2} + ax)(x^2+cx+d) - (\tfrac{x^3}{3} + c\tfrac{x^2}{2} + dx)(x+a) = \tfrac{1}{6}x^4 + \tfrac{2}{3}ax^3 + \tfrac{1}{2}(ac-d)x^2$\\
- tak\v{z}e obecn\'e \v{r}e\v{s}en\'i rovnice
\[y'' - \frac{2(x+a)}{x^2+2ax+ac-d}y' + \frac{2}{x^2+2ax+ac-d} = x^2+2ax+cd-d\]
je
\[y(x) = \tfrac{1}{6}x^4 + \tfrac{2}{3}ax^3 + \tfrac{1}{2}(ac-d)x^2 + c_1(x^2+cx+d) + c_2(x+a)\quad (c_1,c_2\in\er)\]

\fi
%%%%%%%%%%%%%%%%%%
\textbf{5. Nalezn\v{e}te obecn\'e \v{r}e\v{s}en\'i rovnic}\\ \resetpr 
\pr $y''+6y'+10y=0$
\pr $y''-8y'+7y=0$
\pr $y''-6y'+9y=8e^x$
\pr $y'' - 2y' + 5y = \cos x$\\
% a na dalsi cviko ...
\pr $y'' -2y' + 10y = \tfrac{9e^x}{\cos 3x}$
\pr $y'' + 3y = \sin(\sqrt{3}x)  + 3\cos(\sqrt{3}x)$
\\[5pt]
\textbf{6. Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i diferenci\'aln\'ich rovnic s po\v{c}\'ate\v{c}n\'i podm\'inkou}\\ \resetpr
\pr $y'' - 2y' = 8x+4$, $y(0)=1$, $y'(0)=4$
% a na dalsi cviko ...
\pr $y'' + y' = \frac{e^{x}}{1+e^{2x}}$, $y(0)=\tfrac{\pi}{4} - \tfrac{1}{2}\log 2$, $y'(0)=0$\\
\pr $y'' -4y'+4y = x^2 + 2e^{2x}$, $y(0)=0$, $y'(0)=0$
\\[20pt]

  \noindent Dal\v{s}\'i p\v{r}\'iklady k procvi\v{c}en\'i lze nal\'ezt nap\v{r}\'iklad:
\begin{itemize}
	\item v materiálech ke cvičení prof. Kalendy, viz. příklady ze sekce VI zde: \url{https://www.karlin.mff.cuni.cz/~kalenda/archiv/s21314l.pdf/}
        \item na webu kolegyně Kuncové zde: \url{https://www.karlin.mff.cuni.cz/~kuncova/historie28.php}
        \item v online sbírce Tomáše Bárty zde: \url{http://www.karlin.mff.cuni.cz/~barta/pcODR/}
\end{itemize}
\vysledky
\textbf{1.}\resetpr \pr \pr $y(x) = (e^x + e^{-x})\arctg e^x + c_1e^x + c_2e^{-x}$, $x\in\er$ \pr $y(x) = (e^x + e^{-x})\arctg e^x + e^x + 2e^{-x}$, $x\in\er$\\
\textbf{2.} $y(x) = 3x^3\log x + 2x^2 + x^3$, $x\in (0,\infty)$\\
\textbf{3.} $y(x) = \tfrac{1}{6}x^4 + \tfrac{2}{3}x^3 + \tfrac{1}{2}x^2 + x + 1$, $x\in (-2,0)$\\
\textbf{4.}\resetpr \pr $y(x) = (\tfrac{\pi}{2}-x)\cos x + (1+\log(\sin x))\sin x$, $x\in (0,\pi)$\\
\pr $y(x) = \tfrac{1}{2} + \tfrac{1}{6}\cos(2x) + \tfrac{1}{3}\cos x + \sin x$, $x\in \er$
\pr $y(x) = \begin{cases}|x| + \cos x + \sin x & x\geq 0,\\ |x| + \cos x + 3\sin x & x\leq 0.\end{cases}$\\
\textbf{5.}\resetpr \pr $y(x) = c_1e^{-3x}\sin x + c_2e^{-3x}\cos x$, $x\in\er$ \pr $y(x) = c_1e^x + c_2e^{7x}$, $x\in\er$\\ \pr $y(x) = 2e^x + c_1e^{3x} + c_2xe^{3x}$, $x\in\er$ \pr $y(x) = \tfrac{1}{5}\cos x - \tfrac{1}{10}\sin x+ c_1e^{x}\cos 2x + c_2e^{x}\sin 2x$, $x\in\er$\\ \pr $y(x) = 3xe^x\sin 3x + \log|\cos 3x|e^x\cos 3x + c_1e^x\sin 3x + c_2e^x\cos 3x$, $x\in (\tfrac{\pi}{6} + k\tfrac{\pi}{3}, \tfrac{\pi}{2} + k\tfrac{\pi}{3})$, $k\in\zet$\\ \pr $y(x) = x(\tfrac{\sqrt{3}}{2}\sin(\sqrt{3}x) - \tfrac{1}{6}\sqrt{3}\cos(\sqrt{3}x)) + c_1\sin(\sqrt{3}x) + c_2\cos(\sqrt{3}x)$, $x\in\er$
\\
\textbf{6.}\resetpr \pr $y(x) = -3+4e^{2x}-2x^2-4x$, $x\in\er$ \pr $y(x) = \arctg(e^x) - \tfrac{1}{2}\log(1+e^{2x})e^{-x} -\tfrac{\log 2}{2} + \tfrac{\log 2}{2}e^{-x}$, $x\in\er$\\
\pr $y(x) = x^2e^{2x} + \tfrac{x^2}{4} + \tfrac{x}{2} + \tfrac{3}{8} - \tfrac{3}{8}e^{2x} + \tfrac{1}{4}xe^{2x}$, $x\in\er$
	\newpage
\centerline{\MakeUppercase{viii. Funkce více proměnných}}\ \\
	\textbf{1. Spočtěte parciální derivace funkcí všude, kde existují}\\[3pt] \resetpr
\pr $e^{xy}$
\pr $xy+yz+zx$
\pr $|x|\cdot|y|$
\pr $|y+\cos x|$
\pr $f(x,y)=\begin{cases}e^{-\frac{\pi}{x^2+3xy+3y^2}} & (x,y)\neq 0,\\
0 & (x,y)=(0,0)\end{cases}$\\[5pt]
\textbf{2. Ur\v{c}ete a nakreslete defini\v{c}n\'i obor funkce $f$ a vy\v{s}et\v{r}ete její parci\'aln\'i derivace}\\ \resetpr
\pr $f(x,y) = \arcsin\frac{|y|}{|x|+1}$
\pr $f(x,y) = \sqrt{y^6-x^3}$\\[5pt]
\textbf{3. Pomoc\'i v\v{e}t z p\v{r}edn\'a\v{s}ky doka\v{z}te, \v{z}e funkce $f(x,y)=(xy + \sin (e^{x+y}))^2$ je spojit\'a.}\\[5pt]
\textbf{4. Rozhodn\v{e}te, zda n\'asleduj\'ic\'i mno\v{z}iny jsou otev\v{r}en\'e ev. uzav\v{r}en\'e a ur\v{c}ete vnit\v{r}ek.}\\ \resetpr	
	\pr $\{\tfrac{1}{n}\setsep n\in\en\}\cup\{0\}$
	\pr  $\{[x,y]\in\er^2\setsep x \leq 0,\,y > 0\}$	
	\pr $\{[x,y]\in\er^2\setsep x^2+e^y>17\}$\\
	\pr $\{[x,y]\in\er^2\setsep |x-y|=x-y\}$
	\pr $\{[x,y,z]\in\er^3\setsep x\geq 0,y>0,x+y=2,z\leq 0\}$
\\[20pt]
Dal\v{s}\'i p\v{r}\'iklady k procvi\v{c}en\'i lze nal\'ezt nap\v{r}\'iklad:
 \begin{itemize}
     \item v sekci II zde: \url{https://www.karlin.mff.cuni.cz/~cuth/doc/FSV/m2-2015-16.pdf}
     \item v sekci IX zde:
\url{https://www.karlin.mff.cuni.cz/~cuth/doc/MFF/MA/ma2-2016-17.pdf}
    \item na webu kolegyně Kuncové zde: \url{https://www.karlin.mff.cuni.cz/~kuncova/historie30.php}
 \end{itemize}
P\v{r}\'iklady zkou\v{s}kov\'e obt\'i\v{z}nosti lze \v{c}erpat ze zkou\v{s}kov\'ych p\'isemek z Matematiky II na FSV zde: \url{http://www.karlin.mff.cuni.cz/~kalenda/edu.php?edutype=archpis}\\[20pt]
\vysledky
\textbf{1.}\resetpr \pr $\pdx=y e^{xy}$, $\pdy=x e^{xy}$ pro $(x,y)\in\er^2$.
\pr $\pdx=y+z$, $\pdy=x+y$, $\pdz=x+y$ pro $(x,y,z)\in\er^3$.\\
\pr $\pdx(x,y)= |y|\sgn x$ pro $x\ne 0$. $\pdy(x,y)=|x|\sgn y$ pro $y\ne 0$. $\pdx(0,0)=\pdy(0,0)=0$. $\pdx(0,y)$ pro $y\ne 0$ a $\pdy(x,0)$ pro $x\ne0$ neexistuj\'{\i}.
\pr $\pdx(x,y)=-\sgn(y+\cos x)\cdot\sin x$, $\pdy(x,y)=\sgn(y+\cos x)$, pokud $y\neq -\cos x$. $\pdy(x,-\cos x)$ neexistuje pro $x\in\er$. $\pdx(k\pi,(-1)^{k+1})=0$ pro $k\in\zet$. $\pdx(x,-\cos x)$ neexistuje pro $x\neq k\pi$.
\pr $\pdx=e^{-\frac{\pi}{x^2+3xy+3y^2}} \cdot \frac{\pi(2x+3y)}{(x^2+3xy+3y^2)^2}$, $\pdy=e^{-\frac{\pi}{x^2+3xy+3y^2}} \cdot \frac{\pi(3x+6y)}{(x^2+3xy+3y^2)^2}$ pro $(x,y)\ne(0,0)$; v bod\v{e} $(0,0)$ jsou ob\v{e} parci\'aln\'{\i} derivace nulov\'e.\\
\textbf{2.} viz. v\'ysledky zkou\v{s}kov\'ych p\'isemek z Matematiky II na FSV zde: \url{http://www.karlin.mff.cuni.cz/~kalenda/edu.php?edutype=archpis}. Konkr\'etn\v{e}ji, viz. varianta A z roku 2004/2005 a varianta D z roku 2010/2011.\\[3pt]
\textbf{4.}\resetpr 
\pr Mno\v{z}ina je uzav\v{r}en\'a s prázdným vnitřkem
\pr Mno\v{z}ina nen\'{\i} uzav\v{r}en\'a ani otev\v{r}en\'a. Vnit\v{r}ek je $\{[x,y]\in\er^2\setsep x<0,\,y> 0\}$
\pr Mno\v{z}ina nen\'{\i} uzav\v{r}en\'a, je otevřená
\pr Mno\v{z}ina je uzav\v{r}en\'a, není otevřená, vnit\v{r}ek $\{[x,y]\setsep x > y\}$
\pr Ani uzav\v{r}en\'a ani otev\v{r}en\'a, vnit\v{r}ek pr\'azdn\'y
\end{document}
