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


\subchapter{Konvexn� a konk�vn� funkce}{


\definition{
	Nech� $f$ m� v bod� $a \in \real$ vlastn� derivaci.
	Ozna��me
	$$T_a = \{ [x,y] \in \real^2,\ y = f(a) + f'(a)(x-a) \}.$$

	�ekneme, �e bod $[x,f(x)] \in \real^2$ le�� nad (pod)
	{\bf te�nou} $T_a$, jestli�e $f(x) > f(a) + f'(a)(x-a)$ (resp.
	$f(x) < \ldots$).
}


\definition{
	Nech� $f$ m� v bod� $a \in \real$ vlastn� derivaci.
	�ekneme, �e $f$ m� v bod� $a$ {\bf inflexi}, jestli�e
	$\existss \delta$ tak, �e:

	\indentLevel{
		\item{Bu�:}{$\forall x \in (a-\delta,a): [x,f(x)]$ le�� nad $T_a$
		          a $\forall x \in (a,a+\delta): [x,f(x)]$ le�� pod $T_a$}

		\item{nebo:}{$\forall x \in (a-\delta,a): [x,f(x)]$ le�� pod $T_a$
		           a $\forall x \in (a,a+\delta): [x,f(x)]$ le�� nad $T_a$.}
	}
}


\theorem{27}{nutn� podm�nka existence inflexe}{
	Jestli�e $f''(a) \ne 0$, pak $f$ nem� v bod� $a$ inflexi.

	\proof{
		Bez �jmy na obecnosti nech� $f''(a) > 0$.
		Potom $$\lim_{x\to a} {f'(x) - f'(a) \over x - a} > 0$$

		Tedy $\existss \delta > 0$ takov�, �e
		$$ \forall x \in (a,a+\delta): f'(x) > f'(a) \,\et\, \forall x \in (a-\delta,a): f'(x) < f'(a) $$

		Zvolme $y \in (a,a+\delta)$. Pak $f$ je spojit� na $[a,y]$ a $\existss f'$ na $(a,y)$.
		Tedy dle pana Lagrange $\existss \xi \in (a,y)$:
		$$ {f(x) - f(a) \over y - a} = f'(\xi) > f'(a) $$
		$$ f(y) > f(a) + f'(a)(y - a)\qquad \forall y \in (a,a+\delta) $$
		Tedy jsme nad $T_a$.

		Analogicky:

		Zvolme $z \in (a-\delta,a)$. Pak $f$ je spojit� na $[z,a]$ a $\existss f'$ na $(z,a)$.
		Tedy dle pana Lagrange $\existss \psi \in (z,a)$:
		$$ {f(a) - f(z) \over a - z} = f'(\psi) < f'(a) $$
		$$ f(z) < f(a) + f'(a)(z - a)\qquad \forall z \in (a-\delta,a) $$

		Tedy nen� inflexe v $a$!
	}

	\penalty-1000
	\notes{
	\list{
		\listItem{
			$f''(a) = 0$ $\not\Rightarrow$ $f$ m� v bod� $a$ inflexi!

			Viz $f(x) = x^4,\ a = 0$. $f'(x) = 4x^3$, $f''(x) = 12x^2$,
			$f''(0) = 0$.
		}

		\listItem{
			Jestli�e $f''(a)$ neexistuje, pak $f$ m��e a nemus� m�t
			inflexi v bod� $a$.

			Viz $f(x) = x|x|$. $f''(0)$ neexistuje, ale $f$ m� v bod�
			nula inflexn� bod.
		}
	}
	}
}


\theorem{28}{posta�uj�c� podm�nka pro existenci inflexe}{

	Nech� $f$ m� spojitou prvn� derivaci na intervalu $(a,b)$.
	Nech� $z \in (a,b)$. Nech� pro $\forall x \in (a,z)$ plat� $f''(x) > 0$
	a pro $\forall x \in (z,b)$ plat� $f''(x) < 0$ (nebo naopak).
	Pak $z$ je bod inflexe funkce $f$.

	\proof{
		Nech� $f'' > 0$ na $(a,z)$, $f'' < 0$ na $(z,b)$.
		Potom dle \ias{}{26}{} je $f'$ rostouc� na $(a,z)$ a klesaj�c� na $(z,b)$.

		$$ \existss \xi \in (z,b):\ {f(b) - f(z) \over b-z} = f'(\xi) < f'(z) $$
		$$ \Rightarrow f(b) < f(z) + f'(z)(b-z) $$
		(pod te�nou)

		$$ \existss \eta \in (a,z):\ {f(z) - f(a) \over z-a} = f'(\eta) < f'(z) $$
		$$ \Rightarrow f(a) > f(z) + f'(z)(a-z) $$
		(nad te�nou)

		Tot� plat� pro ka�d� bod $x \in (z,b)$ a $y \in (a,z)$,
		tedy
		$$\sesac{\eqalign{f(x) &< f(z) + f'(z)(x-z)\cr
		                  f(y) &> f(z) + f'(z)(y-z)}}
			{\Rightarrow \textbox{$f$ m� v bod� $z$ inflexi.}}$$

		\qed
	}

	\example{
		$$f(x) = \arctg x$$
		$$f'(x) = {1 \over 1 + x^2} $$
		$$f''(x) = {-1 \over (1+x^2)^2} 2x $$
		$$f''(x) = {-2x \over (1+x^2)^2} \cases{>0 & pro $x < 0$\cr
		                                        <0 & pro $x > 0$} \Rightarrow \textbox{$f$ m� v 0 inflexi} $$
		$$f''(0) \textbox{existuje} \Rightarrow f''(0) = 0$$
	}

}



\penalty-1000
\TODO{TADY TOHO {\bf HODN�} CHYB�! (DEFINICE KONK/KONV, T�I V�TY, ASYMPTOTY, \dots)}

\scpart{Asymptoty}{
	(Zhruba.)

	$$\lim_{x\to\infty} f(x)/x = a$$
	$$\lim_{x\to\infty} f(x)-ax = b$$

	Pokud jsou ob� limity kone�n�, asymptota je
	$$ y = ax + b $$
	(jinak asymptota nen�).

	(Ekviv. druh� asymptota pro $-\infty$.)
}

}


\bend
