\input /home/pasky/school/fastex/lib.tex

\chapter{Pr�b�h funkce}{





\exercise{
	Vy�et�ete pr�b�h funkce
	$$ f(x) = \arcsin\left(2x\over1+x^2\right) $$

	\list{
	\nlistItem{ $D(f)$

		$$ \eqalign{D\left(2x\over1+x^2\right) &= \real \cr
		            D(\arcsin)                 &= [-1,1] } $$
		$$ \Rightarrow D(f) = \{x\in\real,\ -1 \le {2x\over1+x^2} \le 1\} $$

		$$ \sesac{
			\eqalign{ -1     &\le {2x\over1+x^2}\cr
			          -1-x^2 &\le 2x\cr
			          0      &\le 1+2x+x^2 = (x+1)^2
			}
		}{ \Rightarrow x \in \real} $$

		$$ \sesac{
			\eqalign{ {2x\over1+x^2} &\le 1\cr
			          2x             &\le 1+x^2\cr
			          0              &\le 1-2x+x^2 = (x-1)^2
			}
		}{ \Rightarrow x \in \real} $$

		$$ \Rightarrow D(f) = \real $$


		\shortdef{Obor spojitosti}{
			$2x$, $1+x^2$ --- polynomy $\Rightarrow$ spojit� na $\real$.
			$\sin$ spojit� a ryze monotonn� na $\left[-{\pi\over2},{\pi\over2}\right]$
			$\Rightarrow$ dle \href{v�ta o inverzn� funkci}{v�ty o inverzn� funkci}
			je $\arcsin$ spojit� na $[-1,1]$ $\Rightarrow$ dle \ref{VOLSF} (P1)
			je $f$ spojit� na $\real$.
		}
	}
	\nlistItem{
		Pr�se��k s osou $y$: $f(0) = 0$, $[0,0]$
		Pr�se��k(y) s osou $x$: $f(x) = 0 \Leftrightarrow {2x\over1+x^2}=0 \Rightarrow x=0$, $[0,0]$
	}
	\nlistItem{
		Symetrie:

		Funkce je lich�, nebo�
		$$ f(-x) = \arcsin\left(-2x\over1+(-x)^2\right) = -\arcsin\left(2x\over1+x^2\right) = -f(x) $$
		($\arcsin$ je lich�).

		Funkce nen� sud� (nebo� $f \not\equiv 0$) nebo $f(1) = {\pi\over2}$, ale $f(-1) = -{\pi\over2}$.

		Funkce nen� periodick�, nebo� $f(0) = 0$, ale $f(x) \ne 0\ \forall x \in \real \setminus \{0\}$.
	}
	\nlistItem{
		Limity v $\pm\infty$:

		\scpart{$$\lim_{x\to -\infty} f(x)$$}{
			$$\lim_{x\to -\infty} {2x\over1+x^2} = 0$$
			$$\lim_{x\to 0} {\arcsin(y)} = 0$$

			$\arcsin$ je spojit� v 0 $\Rightarrow$ (P1)

			$$ \Rightarrow \lim_{x\to -\infty} f(x) = 0$$
		}

		Z lichosti (nebo analogick�m v�po�tem)
		$$\lim_{x\to +\infty} f(x) = 0$$
	}
	\nlistItem{
		Prvn� derivace:

		$$f'(x) = {1\over\sqrt{1-\left(2x\over1+x^2\right)^2}}\cdot{2(1+x^2)-2x\cdot2x\over(1+x^2)^2}$$
		$$= {2(1-x^2)\over(1+x^2)\sqrt{(1+x^2)^2-4x^2}} = {2(1-x^2)\over(1+x^2)\sqrt{(1+x^2)^2}}$$

		$$\Rightarrow f'(x) = {2\over1+x^2}\cdot{1-x^2\over\sqrt{(1-x^2)^2}}$$
		pro $x \ne \pm1$.

		\shortdef{Pozn.}{$\sqrt{y^2} = |y|$}

		$$\Rightarrow f'(x) = {2\over1+x^2}\cdot{1-x^2\over|1-x^2|},\ x \in \real \setminus \{\pm1\}$$

		\shortdef{Pozn.}{${y\over|y|} = \sign(y)$ jen pro $y\ne0$!}

		\startRuledTable
		            & $(-\infty,-1)$   & $(-1,1)$        & $(1,\infty)$     \cr
		$f'(x)$     & $-{2\over1+x^2}$ & ${2\over1+x^2}$ & $-{2\over1+x^2}$ \cr
		\endRuledTable

		Existuje $f'(\pm1)$? Nev�me, ale spo��t�me $f'_\pm(\pm1)$.
		Dv� mo�nosti:
		\list{
			\olistItem{Z definice: $f'_+(1) = \lim_{x\to 1+} {f(x)-f(1)\over x-1}$}
			\olistItem{Z v�ty o limit� derivac�: $f'_+(1) = \lim_{x\to 1+} f'(x)$, $f$ je spojit� na $\cP^+(1)$.}
		}

		Z v�ty o limit� derivac�:
		$$ f'_+(1) = \lim_{x\to 1_+} f'(x) = \lim_{x\to 1_+} {-2\over1+x^2} \becauseof{dosazen� + spojitost}{=} {-2\over1+x} = -1 $$

		$$ f'_-(-1) = \lim_{x\to 1_-} {f(x)-f(1)\over x-1} = \lim_{x\to 1_-} {\arcsin\left(2x\over1+x^2\right)-{\pi\over2}\over x-1} $$
		$$ \becauseof{L'H ``${0\over0}$''}{=} \lim_{x\to 1_-} {2\over1+x^2} \becauseof{dosazen�}{=} {2\over1+1} = 1 $$

		$$ \Rightarrow f'(1) \textbox{neexistuje} $$

		(Analogicky $f'(-1)$ neexistuje. Bu� z lichosti nebo analogick�m v�po�tem.)
		}

	\nlistItem{
		Intervaly monotonie:

		\startRuledTable
		 $(-\infty,-1)$   & $(-1,1)$        & $(1,\infty)$     \cr
		 $f'<0 \Rightarrow \nearrow$
			 & $f'>0 \Rightarrow \searrow$
			 & $f'<0 \Rightarrow \nearrow$ \cr
		\endRuledTable
	}
	
	\nlistItem{
		Extr�my:

		Mohou b�t jen v bodech $x$, kde $f'(x)$ neexistuje nebo $f'(x) = 0$.

		$$\eqalign{f'(x) = 0         &\ \ldots\ \textbox{neexistuj�}\cr
		           f'(x) \not\exists &\ \ldots\ x = \pm1,\ \textbox{kandid�ti na extr�m}}$$

		\list{
		\rlistItem{$x=-1$:

			$$\sesac{\eqalign{
					f \searrow &\ \ldots\ \cP^-(-1,\delta)\cr
					f \nearrow &\ \ldots\ \cP^+(-1,\delta)}
				}{\Rightarrow -1 \textbox{je bod lok�ln�ho minima}}$$
		}
		\rlistItem{$x=1$:

			Z lichosti: $x=1$ je bod lok�ln�ho maxima. Dal�� extr�my $f$ nem�.
		}
		}

		Jsou extr�my glob�ln�?
		\list{
		\rlistItem{$x=-1$:

			$$\eqmalign{
				f \searrow&\ \ldots&\ (-\infty,-1) &\Rightarrow f(y) \ge f(-1)&\hfill\qquad \forall y \in (-\infty,-1)\cr
				f \nearrow&\ \ldots&\ (-1,1)       &\Rightarrow f(y) \ge f(-1)&\hfill\qquad \forall y \in (-1,1)\crcr
				}$$
			$$f(-1) = -{\pi\over2},\qquad f(y) \le 0\quad \forall x \in (1,\infty) \Rightarrow f(y) \ge f(-1)$$

			Tedy -1 je glob�ln� minimum.
		}
		\rlistItem{$x=1$:

			Z lichosti: $x=1$ je bod glob�ln�ho maxima.
		}
		}
	}

	\nlistItem{
		Druh� derivace, konvexita:

		$$ \left(2\over1+x^2\right)' = {-2\over(1+x^2)^2}\cdot2x = {-4x\over(1+x^2)^2} $$

		\startRuledTable
		            & $(-\infty,-1)$   & $(-1,1)$        & $(1,\infty)$     \cr
		$f'(x)$     & $-{2\over1+x^2}$ & ${2\over1+x^2}$ & $-{2\over1+x^2}$ \cr
		$f''(x)$    & ${4x\over(1+x^2)^2}$ & $-{4x\over(1+x^2)^2}$ & ${4x\over(1+x^2)^2}$ \cr
		            & $f''<0$ $\Rightarrow$ $\cup$ (konk.)
		              & \vbox{\everycr{}\halign{#&#\cr
			                $f''>0$ na $(-1,0)$ & $\Rightarrow$ $\cap$\cr
		                        $f''<0$ na $(0,1)$  & $\Rightarrow$ $\cup$\cr}}
		              & $f''>0$ $\Rightarrow$ $\cap$ (konv.)\cr
		\endRuledTable

		\startRuledTable
		$(-\infty,-1)$ & $(-1,0)$   & $(0,1)$        & $(1,\infty)$     \cr
		$\cap$         & $\cup$     & $\cap$         & $\cup$ \cr
		\endRuledTable

		$f''(\pm1)$ neexistuje, nebo� $f'(\pm1)$ neexistuje.
	}

	\nlistItem{
		Inflexn� body:

		Bu� $f''$ $\not\exists$ nebo $f''(x) = 0$.

		$f'' \not\exists$: $x=\pm1$ --- bod inflexe nen�, nebo� $\not\exists f'$.
		$f''(x) = 0$: $x=0$ --- kandid�t na inflexi.

		\medskip
		$$\sesac{
			\eqalign{
				f'' > 0&\ \ldots\ \cP^-(0,\delta)\cr
				f'' < 0&\ \ldots\ \cP^+(0,\delta)
			}
			}{ \Rightarrow x=0 \textbox{je inflexn�m bodem.} } $$
		\medskip

		Jin� inflexn� body $f$ nem� (nutn� podm�nka jinde nespln�na).
	}

	\nlistItem{
		Asymptoty:

		$$\lim_{x\to \infty}{f(x)\over x} = \lim_{x\to \infty}{\arcsin({2x\over1+x^2})\over x} = 0$$
		$$\lim_{x\to \infty}\left(f(x)-0x\right) = 0$$
		$$ a = b = 0 $$

		Nulov� p��mka, tedy osa x.
	}

	\nlistItem{
		Obor hodnot:

		$$H(f) = \left[-{\pi\over2},{\pi\over2}\right]$$
		$$\eqalign{-{\pi\over2} &\ \ldots\ \textbox{glob�ln� minimum,} -{\pi\over2} = f(-1)\cr
		           {\pi\over2}  &\ \ldots\ \textbox{glob�ln� maximum,} {\pi\over2} = f(1)}$$

		$$\becauseof{\ref{Darboux}}{\Rightarrow} f \textbox{nab�v� ka�d� mezihodnoty} $$
	}
	}
}

}

\bend
