ˇ funkce ˚ eh Prub Robert Maˇr´ık ´ r´ı 2008 26. zaˇ
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
Obsah . . . . . . . . . . . . . . . . . . . . . . . .
3
y
. . . . . . . . . . . . . . . . . . . . . . . .
49
y y y
//
x . . . . . 1 + x2 3x + 1 . . . . . = x3 2(x 2 − x + 1) = . . (x − 1)2 3 x = . . . . . 3 − x2 2 x +1 = . . . . . x2 − 1
y=
/
.
..
. . . . . . . . . . . . . . . . . . . . . . . . 101 . . . . . . . . . . . . . . . . . . . . . . . . 149 . . . . . . . . . . . . . . . . . . . . . . . . 191
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
x 1 1 = lim = =0 2 x→±∞ x ±∞ 1+x
1(1 + x 2 ) − x(0 + 2x) y0 = (1 + )2 jmenovatele zlomku. • Omezen´ı na definiˇcn´ı obor vypl´ yvxa´2 ze
//
2 + x2 − • V´yraz x 2 + 1 nesm´ı =b´y1t nulov´ y. 2x (1 + x 2 )2 ˇ pro vˇsechna ´ a´ cˇ´ısla. • To je vˇsak zajiˇsteno realn 2 1−x = (1 + x 2 )2 / . ..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
x 1 1 = lim = =0 2 x→±∞ x ±∞ 1+x
y0 =
//
1(1 + x 2 ) − x(0 + 2x) (1 + x 2 )2
2 1+x − 2x 2 ˇ ´ • Citatel, x, je licha´ funkce, jmenovatel, (1 + x 2 ), je funkce suda. = 2 2 (1 + x ) ´ funkce. • Jako celek je tedy zlomek lich 2a 1−x = (1 + x 2 )2 c
Robert Maˇr´ık, 2008 × / . ..
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
x 1 1 = lim = =0 2 x→±∞ x ±∞ 1+x
y0 =
1(1 + x 2 ) − x(0 + 2x) (1 + x 2 )2
1 + x 2 − 2x 2 (1 + x 2 )2 ´2 ˚ c´ık s osou x a znamenko Urˇc´ıme pruseˇ funkce na jednotliv´ych 1−x intervalech. = (1 + x 2 )2 c
Robert Maˇr´ık, 2008 × // / . .. =
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
x 1 1 = lim = =0 2 x→±∞ x ±∞ 1+x
y0 =
1(1 + x 2 ) − x(0 + 2x) (1 + x 2 )2
1 + x 2 − 2x 2 (1 + x 2 )2 ´2 ˚ c´ık s osou x a znamenko Urˇc´ıme pruseˇ funkce na jednotliv´ych 1−x intervalech. = (1 + x 2 )2 c
Robert Maˇr´ık, 2008 × // / . .. =
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
x 1 1 = lim = =0 2 x→±∞ x ±∞ 1+x
y0 = =
1(1 + x 2 ) − x(0 + 2x) (1 + x 2 )2
1 + x 2 − 2x 2 (1 + x 2 )2 2
1 − x kdyˇz cˇ itatel je nulov´y. ´ =eˇ tehdy, Zlomek je roven nule prav (1 + x 2 )2 // / . ..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
x 1 1 = lim = =0 2 x→±∞ x ±∞ 1+x
y0 =
1(1 + x 2 ) − x(0 + 2x) (1 + x 2 )2
1 + x 2 − 2x 2 (1 + x 2 )2 ´ y bod ˚ c´ık x = 0 na osu 2x. Funkce nema´ zˇ adn´ Zakresl´ıme pruseˇ 1−x nespojitosti. = (1 + x 2 )2 c
Robert Maˇr´ık, 2008 × // / . .. =
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
//
x 1 1 = lim = =0 2 x→±∞ x ±∞ 1+x
2 ´ + xkladn´ • Jmenovatel (1 + x 2 ) je1(1 stale y. + 2x) ) − x(0 y0 = 2 ˇ (1 + e´x 2 )znam ´ • Citatel zlomku ma´ proto stejn enko jako cel´y zlomek x 2 2 1 + x − 2x . = 1 + x2 (1 + x 2 )2 ´ je-li x kladn ´ a naopak. • Funkce je kladna, 2e 1−x = (1 + x 2 )2 c
Robert Maˇr´ık, 2008 × / . ..
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
x 1 1 = lim = =0 2 x→±∞ x ±∞ 1+x
y0 = =
1(1 + x 2 ) − x(0 + 2x) (1 + x 2 )2
1 + x 2 − 2x 2 (1 + x 2 )2 2
1−x Urˇc´ıme limity v nekoneˇcnu. = (1 + x 2 )2 // / . ..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
//
x 1 1 = lim = =0 2 x→±∞ x ±∞ 1+x
• V´ıme, zˇ e o v´ysledku rozhoduj´2ı jenom vedouc´ı cˇ leny v cˇ itateli a ve 1(1 + x ) − x(0 + 2x) jmenovateli. y0 = (1 + x 2 )2 ˇ ´ • Zelenou cast lze vynechat. 2 1 + x − 2x 2 = 2 2 x 1 ´ ıme: • Zbytek zkrat´ = (1. + x ) 2 x 2 x 1−x = (1 + x 2 )2 c
Robert Maˇr´ık, 2008 × / . ..
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
x 1 1 = =0 = lim 2 x→±∞ x ±∞ 1+x
y0 =
1(1 + x 2 ) − x(0 + 2x) (1 + x 2 )2
=
1 + x 2 − 2x 2 (1 + x 2 )2
=
1−x (1 + x 2 )2
2
Dosad´ıme. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
y =0⇒
´ D(f ) = R; licha;
x =0⇒x=0 1 + x2 −
+ 0
lim
x→±∞
x 1 1 =0 = lim = 2 x→±∞ x ±∞ 1+x
y0 =
(1 + x 2 )2
1 1 1 jsou + x 2 nulov − 2x 2e. ´ i = ∞ −∞ 2 2 (1 + x ) • Funkce ma´ vodorovnou asymptotu y = 0 pro x jdouc´ı k ±∞. 2 1−x = (1 + x 2 )2 c
Robert Maˇr´ık, 2008 × / . .. • Obeˇ hodnoty
//
1(1 + x 2 ) − x(0 + 2x)
y=
x 1 + x2
´ D(f ) = R; licha;
−
+ 0
y0 =
1(1 + x 2 ) − x(0 + 2x) (1 + x 2 )2
2
=
1 + x − 2x (1 + x 2 )2
=
1−x (1 + x 2 )2
2
2
1 − x2 ; x1,2 = ±1 (1 + x 2 )2 • Vypoˇcteme derivaci.
y0 =
//
0 0 • Derivujeme pod´ıl podle vzorce ypro=derivaci pod´ılu. 2 1−x =0 / . .. 2 2
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
´ D(f ) = R; licha;
−
+ 0
y0 =
1(1 + x 2 ) − x(0 + 2x) (1 + x 2 )2
2
=
1 + x − 2x (1 + x 2 )2
=
1−x (1 + x 2 )2
2
2
y0 =
1 − x2 ; (1 + x 2 )2
x1,2 = ±1 y0 = 0
Uprav´ıme. //
/
.
..
1−x
2
2 2
=0
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
´ D(f ) = R; licha;
−
+ 0
y0 =
1(1 + x 2 ) − x(0 + 2x) (1 + x 2 )2
2
=
1 + x − 2x (1 + x 2 )2
=
1−x (1 + x 2 )2
2
2
y0 =
1 − x2 ; (1 + x 2 )2
x1,2 = ±1 y0 = 0
Uprav´ıme. //
/
.
..
1−x
2
2 2
=0
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
1 − x2 y0 = ; (1 + x 2 )2
+ 0
x1,2 = ±1 y0 = 0 2
1−x =0 (1 + x 2 )2 1 − x2 = 0 x2 = 1
x1 = 1 x2 = −1
´ ˇreˇsen´ı rovnice y 0 = 0. Hledame & min
//
/
.
..
%
MAX
&
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
´ D(f ) = R; licha;
1 − x2 y0 = ; (1 + x 2 )2
−
+ 0
x1,2 = ±1 y0 = 0 2
1−x =0 (1 + x 2 )2 1 − x2 = 0 x2 = 1
x1 = 1 x2 = −1
Dosad´ıme za derivaci. & //
/
.
..
min
%
MAX
&
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
´ D(f ) = R; licha;
1 − x2 y0 = ; (1 + x 2 )2
−
+ 0
x1,2 = ±1 y0 = 0 2
1−x =0 (1 + x 2 )2 1 − x2 = 0 x2 = 1
x1 = 1 x2 = −1
´ nulov´y cˇ itatel. Zlomek je nulov´y, ma-li % & min
//
/
.
..
MAX
&
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
´ D(f ) = R; licha;
1 − x2 y0 = ; (1 + x 2 )2
−
+ 0
x1,2 = ±1 y0 = 0 2
1−x =0 (1 + x 2 )2 1 − x2 = 0
x2 = 1
x1 = 1 x2 = −1
´ r´ıme x 2 . Vyjadˇ //
/
.
..
&
min
%
MAX
&
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
´ D(f ) = R; licha;
1 − x2 y0 = ; (1 + x 2 )2
−
+ 0
x1,2 = ±1 y0 = 0 2
1−x =0 (1 + x 2 )2 1 − x2 = 0 x2 = 1
x1 = 1 x2 = −1
´ ´ ame ´ Vypoˇc´ıtame x. Dostav dveˇ ˇreˇsen´ı. % & min
//
/
.
..
MAX
&
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1 &
MAX
%
min
1
−1 2
00
y =
1−x (1 + x 2 )2
!0
2 2
=
&
2
2
−2x(1 + x ) −(1 − x )2(1 + x )(0 + 2x)
(1 + x 2 )4 2 ´+ xı 2body. • Nakresl´ıme osu x a+ stacion −2x(1 x )[(1arn´ ) + (1 − x )2] = (1 + x 2 )4 ´ e´ body nespojitosti. • Nejsou zˇ adn 2 −2x[3 − x ] = / . .. 2
//
2 3
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1 &
MAX
%
min
1
−1 2
00
y =
&
1−x (1 + x 2 )2
!0
2 2
2
2
−2x(1 + x ) −(1 − x )2(1 + x )(0 + 2x) = (1 + x 2 )4 ´ ame ´ Testujeme x = −2. Dostav 2 2 2 −2x(1 + x )[(1 + x ) + (1 − x )2] 1 − 4 = y 0 (−2) = (1 + x 2 )4 < 0. kladna´ hodnota 2 −2x[3 − x ] = // / . .. 2 3
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1 &
MAX
%
min
1
−1 2
00
y =
1−x (1 + x 2 )2
!0
2 2
=
&
2
2
−2x(1 + x ) −(1 − x )2(1 + x )(0 + 2x) (1 + x 2 )4
2
2
2
Testujeme x = 0. −2x(1 + x )[(1 + x ) + (1 − x )2] = (1 +=x12 )4> 0 y 0 (0) 1 −2x[3 − x 2 ] = // / . .. 2 3
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1 &
MAX
%
min
1
−1 2
00
y =
&
1−x (1 + x 2 )2
!0
2 2
2
2
−2x(1 + x ) −(1 − x )2(1 + x )(0 + 2x) = ´ ´ eˇ x+=x 2−1. Funkce ma lokaln´ı minimum v bod(1 )4 Funkˇcn´ı hodnota je 2
=
//
/
.
..
=
2
2
−2x(1 + x )[(1 +−1 x ) + (1 −1x )2] y(−1) =(1 + x 2 )4 = − . 2 1 + (−1)2 2 −2x[3 − x ] 2 3
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1 &
MAX
%
min
1
−1 2
00
y =
&
1−x (1 + x 2 )2
!0
2 2
2
2
−2x(1 + x ) −(1 − x )2(1 + x )(0 + 2x) = (1 + x 2 )4 Testujeme x = 2. Plat´ı 2 2 2 −2x(1 + x )[(1 + x ) + (1 − x )2] 1 − 4 = y 0 (2) = (1 + x 2 )4 < 0. kladna´ hodnota 2 −2x[3 − x ] = // / . .. 2 3
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
´ D(f ) = R; licha;
−
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1 &
min
−1
%
MAX
&
1
!0 2 1−x y = (1 + x 2 )2 ´ ı maximum 2v 2bodeˇ x =2 1. Funkˇ 2 n´ı hodnota je Funkce ma´ lokaln´ −2x(1 + x ) −(1 − x )2(1 + x c)(0 + 2x) = 2 4 (1 + x ) 1 y(1)2 = −y(−1) = , 2 2 −2x(1 + x )[(1 + x ) + (12− x )2] = x 2 )a4´ a hodnota y(−1) jiˇz byla kde jsme vyuˇzili toho, zˇ e funkce(1je+lich 2 ´ vypoˇc´ıtana. −2x[3 − x ] = c
Robert Maˇr´ık, 2008 × // / . .. 00
2 3
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1
00
y =
=
1 − x2 (1 + x 2 )2
!0
−2x(1 + x 2 )2 −(1 − x 2 )2(1 + x 2 )(0 + 2x) (1 + x 2 )4
2
=
−2x(1 + x )[(1 + x 2 ) + (1 − x 2 )2] (1 + x 2 )4
2
=
−2x[3 − x ] (1 + x 2 )3 2
x(x − 3) =2 Vypoˇcteme druhou derivaci. (1 + x 2 )3 //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1
00
y =
=
1 − x2 (1 + x 2 )2
!0
−2x(1 + x 2 )2 −(1 − x 2 )2(1 + x 2 )(0 + 2x) (1 + x 2 )4
2
=
−2x(1 + x )[(1 + x 2 ) + (1 − x 2 )2] (1 + x 2 )4
2
−2x[3 − x ] • Derivuje pod´ = ıl podle vzorce pro derivaci pod´ılu. (1 + x 2 )3 • Jmenovatel derivujeme jako sloˇzenou funkci. T´ım se nezbav´ıme 2 x(x v−cˇ3) ´ zlomek. moˇznosti vytknout itateli a zkratit =2 (1 + x 2 )3 //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1 1 − x2 (1 + x 2 )2
00
y =
=
!0
−2x(1 + x 2 )2 −(1 − x 2 )2(1 + x 2 )(0 + 2x) (1 + x 2 )4
2
=
−2x(1 + x )[(1 + x 2 ) + (1 − x 2 )2] (1 + x 2 )4
2
=
−2x[3 − x ] (1 + x 2 )3 2
Vytkneme //
/
.
..
=2
x(x − 3) (1 + x 2 )3
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1
00
y =
=
1 − x2 (1 + x 2 )2
!0
−2x(1 + x 2 )2 −(1 − x 2 )2(1 + x 2 )(0 + 2x) (1 + x 2 )4
2
=
−2x(1 + x )[(1 + x 2 ) + (1 − x 2 )2] (1 + x 2 )4
2
=
−2x[3 − x ] (1 + x 2 )3 2
x(x − 3) =2 ´ se zkrat´ ´ (1 ´ Zelene´ cˇ asti ı. Zjednoduˇ + x 2 )3 s´ıme v´yraz v hranate´ zavorce. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1 1 − x2 (1 + x 2 )2
00
y =
=
!0
−2x(1 + x 2 )2 −(1 − x 2 )2(1 + x 2 )(0 + 2x) (1 + x 2 )4
2
=
−2x(1 + x )[(1 + x 2 ) + (1 − x 2 )2] (1 + x 2 )4
2
=
−2x[3 − x ] (1 + x 2 )3 2
=2 //
/
.
..
x(x − 3) (1 + x 2 )3
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1
2
2
x(x − 3)
x(x − 3) 2 =0 2 3 2 3 (1 + x p ) p(1 + x ) x3 = 0, x4 = 3, x5 = − 3
y 00 = 2
⇒
∩
in.
√
− 3
∪
in.
0
⇒
∩
x(x 2 − 3) = 0
in.
√
∪
3
Vyˇreˇs´ıme y 00 = 0. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1
2
2
x(x − 3) 2 =0 2 3 2 3 (1 + x p ) p(1 + x ) x3 = 0, x4 = 3, x5 = − 3
y 00 = 2
x(x − 3)
⇒
∩
in.
√
− 3
∪
in.
0
⇒
∩
x(x 2 − 3) = 0
in.
√
∪
3
Zlomek je nulov´y, je-li nulov´y jeho cˇ itatel. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1
2
2
x(x − 3) 2 =0 2 3 2 3 (1 + x p ) p(1 + x ) x3 = 0, x4 = 3, x5 = − 3
y 00 = 2
x(x − 3)
⇒
∩
in.
√
− 3
∪
in.
0
⇒
∩
x(x 2 − 3) = 0
in.
√
∪
3
Jsou dveˇ moˇznosti: bud’ x = 0, nebo x 2 − 3 = 0. Druha´ z moˇznost´ı vede na rovnici x2 = 3 p x = ± 3. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
−
´ D(f ) = R; licha;
+ 0
1 − x2 y0 = ; (1 + x 2 )2
x1,2 = ±1
2
2
x(x − 3) 2 =0 2 3 2 3 (1 + x p ) p(1 + x ) x3 = 0, x4 = 3, x5 = − 3
y 00 = 2
x(x − 3)
⇒
∩
in.
√
− 3
∪
in.
0
⇒
∩
x(x 2 − 3) = 0
in.
√
∪
3
´ e´ body nespojitosti. Vyznaˇc´ıme body na osu x. Nejsou zde zˇ adn //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
x(x − 3)
⇒ 2 =0 2 3 (1 + x 2p )3 p(1 + x ) x3 = 0, x4 = 3, x5 = − 3 ∩
in.
√
∪
− 3
1
−1
2
x(x − 3)
min % MAX&
&
x1,2 = ±1
2
y =2
+ 0
1 − x2 y0 = ; (1 + x 2 )2 00
−
´ D(f ) = R; licha;
in.
0
⇒
∩
x(x 2 − 3) = 0
in.
√
∪
3
Testujeme x = −2. y 00 (−2) = 2 //
/
.
..
−2(4 − 3) < 0. kladna´ hodnota c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
&
x1,2 = ±1
2
y =2
+ 0
1 − x2 y0 = ; (1 + x 2 )2 00
−
´ D(f ) = R; licha;
2
x(x − 3)
x(x − 3)
min
% MAX&
−1
1
⇒ 2 =0 2 3 (1 + x 2p )3 p(1 + x ) x3 = 0, x4 = 3, x5 = − 3 ∩
in.
√
− 3
∪
in.
0
⇒
∩
x(x 2 − 3) = 0
in.
√
∪
3
Testujeme x = −1. Funkce je v tomto bodeˇ konvexn´ı, protoˇze je zde ´ ı minimum. lokaln´ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
x(x − 3)
⇒ 2 =0 2 3 (1 + x 2p )3 p(1 + x ) x3 = 0, x4 = 3, x5 = − 3 ∩
in.
√
− 3
∪
1
−1
2
x(x − 3)
min % MAX&
&
x1,2 = ±1
2
y =2
+ 0
1 − x2 y0 = ; (1 + x 2 )2 00
−
´ D(f ) = R; licha;
in.
0
⇒
∩
x(x 2 − 3) = 0
in.
√
∪
3
p V bodeˇ x = − 3 je inflexe. Funkˇcn´ı hodnota je √ p − 3 y(− 3) = ≈ −0.43. 1+3 //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
x(x − 3)
⇒ 2 =0 2 3 (1 + x 2p )3 p(1 + x ) x3 = 0, x4 = 3, x5 = − 3 ∩
in.
√
− 3
∪
1
−1
2
x(x − 3)
min % MAX &
&
x1,2 = ±1
2
y =2
+ 0
1 − x2 y0 = ; (1 + x 2 )2 00
−
´ D(f ) = R; licha;
in.
0
⇒
∩
x(x 2 − 3) = 0
in.
√
∪
3
´ ı, protoˇze je zde Testujeme x = 1. Funkce je v tomto bodeˇ konkavn´ ´ ı maximum. lokaln´ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
x(x − 3)
⇒ 2 =0 2 3 (1 + x 2p )3 p(1 + x ) x3 = 0, x4 = 3, x5 = − 3 ∩
in.
√
∪
− 3
1
−1
2
x(x − 3)
min % MAX&
&
x1,2 = ±1
2
y =2
+ 0
1 − x2 y0 = ; (1 + x 2 )2 00
−
´ D(f ) = R; licha;
in.
0
⇒
∩
x(x 2 − 3) = 0
in.
√
∪
3
´ ame ´ Testujeme x = 2. Dostav y 00 (2) = 2 //
/
.
..
2(4 − 3) > 0. ˇ kladneho ´ neco c
Robert Maˇr´ık, 2008 ×
y=
x 1 + x2
x(x − 3)
⇒ 2 =0 2 3 (1 + x 2p )3 p(1 + x ) x3 = 0, x4 = 3, x5 = − 3 ∩
in.
√
− 3 Inflexe v bodeˇ x =
//
/
.
..
∪
1
−1
2
x(x − 3)
min % MAX&
&
x1,2 = ±1
2
y =2
+ 0
1 − x2 y0 = ; (1 + x 2 )2 00
−
´ D(f ) = R; licha;
in.
⇒
∩
0
x(x 2 − 3) = 0
in.
√
∪
3
p 3. Funkˇcn´ı hodnota je √ p 3 ≈ 0.43. y( 3) = 1+3 c
Robert Maˇr´ık, 2008 ×
−
+ 0
f (0) = 0 f (±∞) = 0
&
min
−1
% MAX & 1
f (±1) = ±
∩
in.
√
− 3 1 2
∪
in.
0
∩
in.
√
∪
3
p f (± 3) ≈ ±0.433
ˇ s´ı v´ysledky. ˚ zitejˇ Vyp´ısˇ eme si nejduleˇ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
−
+
&
0
min
% MAX &
−1
f (0) = 0 f (±∞) = 0
in.
∩
1
√
− 3
f (±1) = ±
1 2
∪
in.
0
∩
in.
√
∪
3
p f (± 3) ≈ ±0.433
y
ments √ − 3
−1 1
√
3
x
´ Zakresl´ıme souˇradn´y system. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
−
+
&
0
min
% MAX &
−1
f (0) = 0 f (±∞) = 0
in.
∩
1
√
− 3
f (±1) = ±
1 2
∪
in.
0
∩
in.
√
∪
3
p f (± 3) ≈ ±0.433
y
ments √ − 3
−1 1
√
3
x
˚ c´ık s osou x. Funkˇcn´ı hodnoty se v tomto bodeˇ V bodeˇ x = 0 je pruseˇ ˇ ı z kladn´ych na zaporn ´ ´ men´ e. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
−
+
&
0
min
% MAX &
−1
f (0) = 0 f (±∞) = 0
in.
∩
1
√
− 3
f (±1) = ±
1 2
∪
in.
0
∩
in.
√
∪
3
p f (± 3) ≈ ±0.433
y
ments √ − 3
−1 1
√
3
x
´ ame ´ Zachyt´ıme informaci o vodorovne´ teˇcneˇ v ±∞. Dav si pozor na ´ ´ eˇ nakreslit nad nebo pod znamenko funkce, mus´ıme graf spravn asymptotu. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
−
+
&
0
min
% MAX &
−1
f (0) = 0 f (±∞) = 0
in.
∩
1
√
− 3
f (±1) = ±
1 2
∪
in.
0
∩
in.
√
∪
3
p f (± 3) ≈ ±0.433
y
ments √ − 3
−1 1
//
/
.
..
√
3
x
c
Robert Maˇr´ık, 2008 ×
−
+
&
0
min
% MAX &
−1
f (0) = 0 f (±∞) = 0
in.
∩
1
√
− 3
f (±1) = ±
1 2
∪
in.
0
∩
in.
√
∪
3
p f (± 3) ≈ ±0.433
y
ments √ − 3
−1 1
//
/
.
..
√
3
x
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; y =0 3x + 1 =0 x3 3x + 1 = 0 1 x=− 3 −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = +0 x3 3x + 1 1 lim− = −∞ = 3 x→0 −0 x 3 3 3x + 1 = lim = =0 lim 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; y =0 3x + 1 =0 x3 3x + 1 = 0 1 x=− 3 −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = +0 x3 • Urˇc´ıme definiˇcn´ı obor. 3x + 1 1 lim− = −∞ = 3 x→0 xnula. −0 • Ve jmenovateli nesm´ ı b´ y t 3 3 3x + 1 = lim = =0 lim 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; y =0 3x + 1 =0 x3 3x + 1 = 0 1 x=− 3 −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = +0 x3 3x + 1 1 lim− = −∞ = 3 x→0 −0 x 3 3 3x + 1 ˚ c´ık s osou =sen´ lim = y == lim x jako ˇreˇ Urˇc´ıme pruseˇ ı rovnice 0.0 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; y =0 3x + 1 =0 x3 3x + 1 = 0 1 x=− 3 −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = +0 x3 3x + 1 1 lim− = −∞ = 3 x→0 −0 x 3 3 3x + 1 = lim = =0 lim 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; y =0 3x + 1 =0 x3 3x + 1 = 0 1 x=− 3 −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = +0 x3 3x + 1 1 lim− = −∞ = 3 x→0 −0 x 3 3 3x + 1 = lim = =0 lim 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; y =0 3x + 1 =0 x3 3x + 1 = 0 1 x=− 3 −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = +0 x3 3x + 1 1 lim− = −∞ = 3 x→0 −0 1 x 3 3x + useˇ 1 ˚ c´ık x = 3 Funkce ma´ s osou x jedin´ y pr − = lim lim 3= ∞ = 0 3 2 x→±∞ x→±∞ x x lim
x→0+
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = lim = =0 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ ◦ 1c´ıme0znamenka ´ •− Urˇ funkce.
+
3
//
ˇ ıme osu x pomoc´ı pruseˇ ˚ c´ık˚u a bodu˚ nespojitosti • Rozdel´ na podin2 3 2 3x x − (3x + 1) ´ ´ ´ tervaly, kde se znam enko zachov av a. 3x − (3x + 1)3x = y0 = (x 3 )2 x6
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = lim = =0 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ ◦ Uvaˇzujme zcela vlevo. Zvolme x = −1 a vypocˇ teme 0 − 1 interval +
3
−3 + 1 y(−1) = = 22 >x0. 2 3x − (3x + 1) −1 3x − (3x + 1)3x = y0 = 3 2 (x ) x6 3
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = lim = =0 3 2 x→±∞ x→±∞ ∞ x x 1 +zujme−prostˇ+ Uvaˇ redn´ı interval, zvolme x = − a vypoˇcteme ◦ 4 0 − 31 1 − 34 + 1 1 4 −16 < 0. y(− ) = = = 2 1 34 2 − 13x x − (3x + 1) − 64 3x − (3x + 64 1)3x = y0 = 3 2 (x ) x6 lim
x→0+
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = lim = =0 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ ◦ V posledn´ − 1 ım 0intervalu zvolme x = 1 a vypoˇcteme +
3
3+1 2 0. y(1) = 4> 2 = 3x x − (3x + 1) 1 3x − (3x + 1)3x = y0 = 3 2 (x ) x6 3
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = lim = =0 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ − 31
+ ◦ 0
2 3x x − (3x + 1) 3x − (3x + 1)3x = nespojitosti. y0 = Najdeme jednostrann e´ limity v bodech (x 3 )2 x6 3
//
/
.
..
2
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+
+ ◦ 0
− 31
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = lim = =0 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ − 31
+ ◦ 0 3
3x k−v´y(3x + 1)3x Dosazen´ı x = 00 vede razu typu = y = 3 2 (x )
//
/
.
..
2 yraz 2nenulov´ 3x yxv´− (3x
nula
.
x6
+ 1)
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = lim = =0 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ ◦ ´ sky v´ıme, zˇ e jednostranne´ limity jsou nevlastn´ı. •− Z 1pˇredn0aˇ
+
3
//
´ ´ ˇ • Schema se znamenkem funkce umoˇznuje odhalit, zda se funkce 2 3 minus nekoneˇ 2 cnu. 3x x − (3x + 1) bl´ızˇ´ı k plus nebo 3x − (3x + 1)3x = y0 = (x 3 )2 x6
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+
+ ◦ 0
− 31
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = lim = =0 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ − 31
+ ◦ 0 3
2
3x − (3x + 1)3x = Urˇc´ıme limity yv0 nevlastn´ ıch bodech. = (x 3 )2
//
/
.
..
3x
2
x − (3x + 1) x6
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = lim = =0 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ − 31
+ ◦ 0
2 ˇ leny jsou podstatn ´xv−limit ˇ+tohoto 3 ıc 2 V´ıme, zˇ e pouze vedouc´ e e typu a 3x (3x 1) 3x − (3x + 1)3x ˚= = y 0uˇ ostatn´ı cˇ leny m zeme vynechat. (x 3 )2 x6
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+
+ ◦ 0
− 31
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = lim = =0 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ − 31
+ ◦ 0 3
0 ´ ıme x. y = Zkrat´ //
/
.
..
3x − (3x + 1)3x (x 3 )2
3x
2
=
2
x − (3x + 1) x6
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+
+ ◦ 0
− 31
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim = =0 = lim 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ − 31
+ ◦ 0 3
Dosad´ıme. //
/
.
..
y0 =
3x − (3x + 1)3x (x 3 )2
3x
2
=
2
x − (3x + 1) x6
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
D(f ) = R \ {0} ; −
+ − 31
+ ◦ 0
3x + 1 1 =∞ = 3 +0 x 3x + 1 1 lim = −∞ = x→0− −0 x3 3 3x + 1 3 lim =0 = lim = 3 2 x→±∞ x→±∞ ∞ x x lim
x→0+
−
+ − 31
+ ◦ 0
2 3 2 Limita je vypoˇctena. 3x x − (3x + 1) 3x − (3x + 1)3x y0 = Funkce ma´ vodorovnou asymptotu y = 0 v ±∞. (x 3 )2 x6
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
+ ◦ 0
− 31
3
y0 =
−
+
D(f ) = R \ {0} ;
3x − (3x + 1)3x
2
3x 2 x − (3x + 1)
= (x 3 )2 x6 −2x − 1 2x + 1 x − 3x − 1 =3 = −3 =3 4 4 x x x4
y 0 (x) = −3
2x + 1 1 ; x1 = − 2 x4 %
MAX
− 12 % MAX & − 21 //
/
.
◦ 0
&
◦ 0
&
&
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
+ ◦ 0
− 31
3
y0 =
−
+
D(f ) = R \ {0} ;
3x − (3x + 1)3x
3x 2 x − (3x + 1)
2
= (x 3 )2 x6 −2x − 1 2x + 1 x − 3x − 1 =3 = −3 =3 4 4 x x x4
y 0 (x) = −3
2x + 1 1 ; x1 = − 2 x4 %
/
.
&
− 12
Derivujeme pod´ıl. & % MAX & ◦ 0 − 21 //
MAX
u 0 v
0
=
..
◦ 0
u v − uv v2
&
0
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
+ ◦ 0
− 31
3
y0 =
−
+
D(f ) = R \ {0} ;
3x − (3x + 1)3x
2
3x 2 x − (3x + 1)
= (x 3 )2 x6 −2x − 1 2x + 1 x − 3x − 1 =3 = −3 =3 4 4 x x x4
y 0 (x) = −3
2x + 1 1 ; x1 = − 2 x4 %
MAX
− 12
&
◦ 0
&
% MAX &
& ◦ 0 z´ıme na souˇcin. −ı21m rozloˇ Vytknut´ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
+ ◦ 0
− 31
3
y0 =
−
+
D(f ) = R \ {0} ;
3x − (3x + 1)3x
2
3x 2 x − (3x + 1)
= (x 3 )2 x6 2x + 1 x − 3x − 1 −2x − 1 = −3 =3 =3 4 4 x x x4
y 0 (x) = −3
2x + 1 1 ; x1 = − 2 x4 %
MAX
− 12
&
◦ 0
&
´ ıme. • Zkrat´ & % MAX & ◦ ´ 0 ıme zavorku. ´ • Rozn − 1 asob´ 2
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
+ ◦ 0
− 31
3
y0 =
−
+
D(f ) = R \ {0} ;
3x − (3x + 1)3x
2
3x 2 x − (3x + 1)
= (x 3 )2 x6 x − 3x − 1 −2x − 1 2x + 1 =3 =3 = −3 4 4 x x x4
y 0 (x) = −3
2x + 1 1 ; x1 = − 2 x4 %
MAX
− 12 % MAX &
◦ 1 0 − Zjednoduˇ 2 s´ıme. //
/
.
&
◦ 0
&
&
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
+ ◦ 0
− 31
3
y0 =
−
+
D(f ) = R \ {0} ;
3x − (3x + 1)3x
2
3x 2 x − (3x + 1)
= (x 3 )2 x6 −2x − 1 x − 3x − 1 2x + 1 =3 =3 = −3 4 4 x x x4
y 0 (x) = −3
2x + 1 1 ; x1 = − 2 x4 %
MAX
− 12 % MAX &
◦ 1 0 − 2 ´ Mame derivaci. //
/
.
&
◦ 0
&
&
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
MAX
&
◦ 0
− 12 % MAX & − 21
◦ 0
+ ◦ 0
− 31
2x + 1 1 ; x1 = − 4 2 x %
−
+
D(f ) = R \ {0} ;
&
&
2x + 1 y = −3 x4 00
4
0 4
= −3
2x 4 − (2x + 1)4x 3 (x 4 )2
3
4
2x − 8x − 4x −6x − 4x = −3 x8 x8 3 4 3 (3x + 2)x 3x + 2x ´ Mame derivaci. =6 =6
3
= −3
//
/
.
..
8
8
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
MAX
&
◦ 0
− 12 % MAX & − 21
◦ 0
+ ◦ 0
− 31
2x + 1 1 ; x1 = − 4 2 x %
−
+
D(f ) = R \ {0} ;
&
&
2x + 1 y = −3 x4 00
4
0 4
= −3
2x 4 − (2x + 1)4x 3 (x 4 )2
3
4
2x − 8x − 4x −6x − 4x = −3 x8 x8 3 0 4 3 Rovnice y = 0 je ekvivalentn´ 2x++2)x 1 = 0. (3x 3x + 2xı rovnici =6 =6
3
= −3
//
/
.
..
8
8
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
MAX
&
◦ 0
− 12 % MAX & − 21
◦ 0
+ ◦ 0
− 31
2x + 1 1 ; x1 = − 4 2 x %
−
+
D(f ) = R \ {0} ;
&
&
2x + 1 y = −3 x4 00
4
0 4
= −3
2x 4 − (2x + 1)4x 3 (x 4 )2
3
4
3
2x − 8x − 4x −6x − 4x = −3 x8 x8 3 4 3 ´ 3x Vyznaˇc´ıme stacionarn´ ı bod a bod nespojitosti (3x + 2)x na osu x. + 2x =6 =6 = −3
//
/
.
..
8
8
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
MAX
&
◦ 0
− 12 % MAX & − 21
◦ 0
+ ◦ 0
− 31
2x + 1 1 ; x1 = − 4 2 x %
−
+
D(f ) = R \ {0} ;
&
&
2x + 1 y = −3 x4 00
4
0 4
= −3
2x 4 − (2x + 1)4x 3 (x 4 )2
3
4
2x − 8x − 4x −6x − 4x = −3 = −3 8 x x8 −2 + 1 y 0 (−1) = −3 = 3 4> 0 3 3 (3x + 2)x 3x + 2x 1 =6 =6
//
/
.
..
8
8
3
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
MAX
&
◦ 0
− 12 % MAX & − 21
◦ 0
+ ◦ 0
− 31
2x + 1 1 ; x1 = − 4 2 x %
−
+
D(f ) = R \ {0} ;
&
&
2x + 1 y = −3 x4 00
4
0 4
= −3
2x 4 − (2x + 1)4x 3 (x 4 )2
3
4
3
2x − 8x − 4x −6x − 4x = −3 8 x x8 1 ˇ ı znamenko ´ ´ ´ ´ y 0 (− ) < 0, protoˇze funkce men´ z3 kladneho na zaporn e. 4 3 (3x + 2)x 3x + 2x 3 =6 =6 = −3
//
/
.
..
8
8
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
D(f ) = R \ {0} ;
2x + 1 1 ; x1 = − 4 2 x MAX
%
−
+
&
◦ 0
− 12 % MAX & − 21
◦ 0
+ ◦ 0
− 31
&
&
0 2x 4 − (2x + 1)4x 3 2x + 1 y = −3 = −3 x4 1 (x 4 )2 ´ ı minimum ˇ v bod e Funkce ma´ lokaln´ x = − . Funkˇcn´ı hodnota je 4 4 3 2 −6x 4 − 4x 3 2x − 8x − 4x 3 1 = −3 = −3 −2 + 1 −2 1 8 x8 y(− ) = = = 4. x 1 1 4 3 3 2 −8 −3x (3x + 2)x + 2x 8 =6 =6 00
//
/
.
..
8
8
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
MAX
&
◦ 0
− 12 % MAX & − 21
◦ 0
+ ◦ 0
− 31
2x + 1 1 ; x1 = − 4 2 x %
−
+
D(f ) = R \ {0} ;
&
&
2x + 1 y = −3 x4 00
4
0 4
= −3
2x 4 − (2x + 1)4x 3 (x 4 )2
3
4
2x − 8x − 4x −6x − 4x = −3 8 x x8 3 y 0 (1) = −3 = −9 < 0 4 3 3 (3x + 2)x 3x + 2x 1 =6 =6
3
= −3
//
/
.
..
8
8
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
2x + 1 ; x4
% MAX & − 21
2x + 1 y = −3 x4 00
4
−
+
D(f ) = R \ {0} ;
0
− 31
◦ 0
+ ◦ 0
&
= −3
4
3
2x 4 − (2x + 1)4x 3 (x 4 )2 4
2x − 8x − 4x −6x − 4x = −3 8 x x8 3 4 3 (3x + 2)x 3x + 2x =6 =6 8 x x8 3x + 2 =6 x5
3
= −3
y 00 = 6 //
/
.
3x + 2 2 ; x2 = − 3 x5 ..
∪
in.
∩
∪
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
2x + 1 ; x4
% MAX & − 21
2x + 1 y = −3 x4 00
4
−
+
D(f ) = R \ {0} ;
0
− 31
◦ 0
+ ◦ 0
&
= −3
4
3
2x 4 − (2x + 1)4x 3 (x 4 )2 4
2x − 8x − 4x −6x − 4x = −3 8 x x8 3 4 3 (3x + 2)x 3x + 2x =6 =6 8 x x8 3x + 2 =6 x5
3
= −3
y 00 = 6 //
/
.
3x + 2 2 ; x2 = − 3 x5 ..
∪
in.
∩
∪
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
2x + 1 ; x4
% MAX & − 21
2x + 1 y = −3 x4 00
4
−
+
D(f ) = R \ {0} ;
0
− 31
◦ 0
+ ◦ 0
&
= −3
4
3
2x 4 − (2x + 1)4x 3 (x 4 )2 4
2x − 8x − 4x −6x − 4x = −3 8 x x8 3 4 3 (3x + 2)x 3x + 2x =6 =6 8 x x8 3x + 2 =6 x5
3
= −3
y 00 = 6 //
/
.
3x + 2 2 ; x2 = − 3 x5 ..
∪
in.
∩
∪
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
2x + 1 ; x4
% MAX & − 21
2x + 1 y = −3 x4 00
4
−
+
D(f ) = R \ {0} ;
0
− 31
◦ 0
+ ◦ 0
&
= −3
4
3
2x 4 − (2x + 1)4x 3 (x 4 )2 4
2x − 8x − 4x −6x − 4x = −3 8 x x8 3 4 3 (3x + 2)x 3x + 2x =6 =6 8 x x8 3x + 2 =6 x5
3
= −3
y 00 = 6 //
/
.
3x + 2 2 ; x2 = − 3 x5 ..
∪
in.
∩
∪
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
2x + 1 ; x4
% MAX & − 21
2x + 1 y = −3 x4 00
4
−
+
D(f ) = R \ {0} ;
0
− 31
◦ 0
+ ◦ 0
&
= −3
4
3
2x 4 − (2x + 1)4x 3 (x 4 )2 4
2x − 8x − 4x −6x − 4x = −3 8 x x8 3 4 3 (3x + 2)x 3x + 2x =6 =6 8 x x8 3x + 2 =6 x5
3
= −3
y 00 = 6 //
/
.
3x + 2 2 ; x2 = − 3 x5 ..
∪
in.
∩
∪
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
2x + 1 ; x4
% MAX & − 21
2x + 1 y = −3 x4 00
4
−
+
D(f ) = R \ {0} ;
0
− 31
◦ 0
+ ◦ 0
&
= −3
4
3
2x 4 − (2x + 1)4x 3 (x 4 )2 4
2x − 8x − 4x −6x − 4x = −3 8 x x8 3 4 3 (3x + 2)x 3x + 2x =6 =6 8 x x8 3x + 2 =6 x5
3
= −3
y 00 = 6 //
/
.
3x + 2 2 ; x2 = − 3 x5 ..
∪
in.
∩
∪
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3
2x + 1 ; x4
% MAX & − 21
2x + 1 y = −3 x4 00
4
−
+
D(f ) = R \ {0} ;
0
− 31
◦ 0
+ ◦ 0
&
= −3
4
3
2x 4 − (2x + 1)4x 3 (x 4 )2 4
2x − 8x − 4x −6x − 4x = −3 8 x x8 3 4 3 (3x + 2)x 3x + 2x =6 =6 8 x x8 3x + 2 =6 x5
3
= −3
y 00 = 6 //
/
.
3x + 2 2 ; x2 = − 3 x5 ..
∪
in.
∩
∪
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3 y 00 = 6
D(f ) = R \ {0} ;
2x + 1 ; x4
% MAX &
3x + 2 2 ; x2 = − 3 x5 ∪
− 21
in.
− 23
//
/
.
..
−
+
+ ◦ 0
− 31
◦ 0
&
∩
◦ 0
∪
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3 y 00 = 6
D(f ) = R \ {0} ;
2x + 1 ; x4
% MAX &
3x + 2 2 ; x2 = − 3 x5 ∪
− 21
−
+
+ ◦ 0
− 31
◦ 0
in.
− 23
&
∩
◦ 0
∪
2 y 00 = 0 pro 3x + 2 = 0, t.j. x = − . 3 //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3 y 00 = 6
D(f ) = R \ {0} ;
2x + 1 ; x4
% MAX &
3x + 2 2 ; x2 = − 3 x5 ∪
− 21
in.
− 23
//
/
.
..
−
+
+ ◦ 0
− 31
◦ 0
&
∩
◦ 0
∪
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3 y 00 = 6
D(f ) = R \ {0} ;
2x + 1 ; x4
% MAX &
3x + 2 2 ; x2 = − 3 x5 ∪
− 21
in.
− 23
y 00 (−1) = 6 //
/
.
..
−1 =6>0 −1
−
+
+ ◦ 0
− 31
◦ 0
&
∩
◦ 0
∪
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3 y 00 = 6
D(f ) = R \ {0} ; % MAX &
2x + 1 ; x4
3x + 2 2 ; x2 = − 3 x5 ∪
− 21
in.
− 23
−
+
+ ◦ 0
− 31
◦ 0
&
∩
◦ 0
∪
1 −1 + 2 <0 y 00 (− ) = 6 3 − 15 3
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3 y 00 = 6
D(f ) = R \ {0} ;
2x + 1 ; x4
% MAX &
3x + 2 2 ; x2 = − 3 x5 ∪
− 21
in.
− 23
−
+
+ ◦ 0
− 31
◦ 0
&
∩
◦ 0
∪
2 2 −2 + 1 ≈ 3.375 Inflexn´ı bod x = − . y(− ) = 5 3 3 − 235 //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
3x + 1 x3
y 0 (x) = −3 y 00 = 6
D(f ) = R \ {0} ;
2x + 1 ; x4
% MAX &
3x + 2 2 ; x2 = − 3 x5 ∪
− 21
in.
− 23
y 00 (1) = 6 //
/
.
..
−
+
+ ◦ 0
− 31
◦ 0
&
∩
◦ 0
∪
5 = 30 > 0 1 c
Robert Maˇr´ık, 2008 ×
−
+ − 31 1 f (− ) = 0 3 1 f (− ) = 4 2
+ ◦ 0
% MAX & − 12 2 f (− ) ≈ 3.4 3
◦ 0
&
∪
in.
∩
− 32
◦ 0
∪
f (0+) = ∞, f (0−) = −∞
f (±∞) = 0,
Shrneme dosaˇzene´ v´ysledky. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
−
+ − 31 1 3 1 f (− ) = 4x 2 y
f (− ) = 0 placements
% MAX &
+ ◦ 0
2 f (− ) ≈ 3.4 3
1
.
..
∪
in.
∩
− 32
◦ 0
∪
f (0+) = ∞, f (0−) = −∞
y
− 32 /
&
f (±∞) = 0,
−1 √ 3 √ − 3
//
◦ 0
− 12
− 21
− 31
x
c
Robert Maˇr´ık, 2008 ×
−
+ − 31 1 3 1 f (− ) = 4x 2 y
f (− ) = 0 placements
% MAX &
+ ◦ 0
2 f (− ) ≈ 3.4 3
1
.
..
∪
in.
∩
− 32
◦ 0
∪
f (0+) = ∞, f (0−) = −∞
y
− 32 /
&
f (±∞) = 0,
−1 √ 3 √ − 3
//
◦ 0
− 12
− 21
− 31
x
c
Robert Maˇr´ık, 2008 ×
−
+ − 31 1 3 1 f (− ) = 4x 2 y
f (− ) = 0 placements
% MAX &
+ ◦ 0
2 f (− ) ≈ 3.4 3
1
.
..
∪
in.
∩
− 32
◦ 0
∪
f (0+) = ∞, f (0−) = −∞
y
− 32 /
&
f (±∞) = 0,
−1 √ 3 √ − 3
//
◦ 0
− 12
− 21
− 31
x
c
Robert Maˇr´ık, 2008 ×
−
+ − 31 1 3 1 f (− ) = 4x 2 y
f (− ) = 0 placements
% MAX &
+ ◦ 0
2 f (− ) ≈ 3.4 3
1
.
..
∪
in.
∩
− 32
◦ 0
∪
f (0+) = ∞, f (0−) = −∞
y
− 32 /
&
f (±∞) = 0,
−1 √ 3 √ − 3
//
◦ 0
− 12
− 21
− 31
x
c
Robert Maˇr´ık, 2008 ×
−
+ − 31 1 3 1 f (− ) = 4x 2 y
f (− ) = 0 placements
% MAX &
+ ◦ 0
2 f (− ) ≈ 3.4 3
1
.
..
∪
in.
∩
− 32
◦ 0
∪
f (0+) = ∞, f (0−) = −∞
y
− 32 /
&
f (±∞) = 0,
−1 √ 3 √ − 3
//
◦ 0
− 12
− 21
− 31
x
c
Robert Maˇr´ık, 2008 ×
−
+ − 31 1 3 1 f (− ) = 4x 2 y
f (− ) = 0 placements
% MAX &
+ ◦ 0
2 f (− ) ≈ 3.4 3
1
.
..
∪
in.
∩
− 32
◦ 0
∪
f (0+) = ∞, f (0−) = −∞
y
− 32 /
&
f (±∞) = 0,
−1 √ 3 √ − 3
//
◦ 0
− 12
− 21
− 31
x
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1)
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
(x − 1)2
y(0) =
2(0 − 0 + 1) (0 − 1)2
=2 2(x 2 − x + 1) (x − 1)2
=0
x2 − x + 1 = 0 +
+ ◦ 1 2
lim
x→1+
//
/
.
..
lim−
2(x − x + 1)
=
2(x 2 − x + 1)
=
(x −
1)2 2
2 = +∞ +0 2
= +∞
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y(0) =
2(0 − 0 + 1) (0 − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ =2 2(x 2 − x + 1) (x − 1)2
=0
x2 − x + 1 = 0
+ Urˇc´ıme definiˇcn´ı obor z podm´ınky 1◦
Plat´ı //
/
.
..
x − 1 6= 0. 2 2(x − x + 1) 2 lim = +∞ = 2 x→1+ +0 (x − 1) x = 6 1. 2(x 2 − x + 1) 2 = +∞ = lim− 2
+
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1)
y(0) =
(x − 1)2
2(0 − 0 + 1) (0 − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ =2 2(x 2 − x + 1) (x − 1)2
=0
x2 − x + 1 = 0 +
+ ◦ 1
//
˚ 2(x • Urˇc´ıme pruseˇ c´ık2s−osou y. x + 1) 2 lim+ = +∞ = 2 x→1 +0 1) ame ´ • Dosad´ıme x = 0(xa−hled y(0). 2(x 2 − x + 1) 2 = +∞ = lim / . .. − 2
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ 2(x 2 − x + 1) (x − 1)2
=0
x2 − x + 1 = 0
+
+ ◦ 1
2
2(x − x + 1)
2 = = +∞ +0 (x − 1)2 ˚ c´ık2s osou x. • Urˇc´ıme pruseˇ 2(x − x + 1) 2 limy− = 0 a ˇreˇs´ıme rovnici = +∞ = • Dosad´ıme x→1 +0 (x − 1)2 lim+
x→1
//
/
.
..
2(x 2 − x + 1)
2x
2
2
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ 2(x 2 − x + 1) (x − 1)2
=0
x2 − x + 1 = 0
+
+ ◦ 1
2
lim+
x→1
2(x − x + 1) (x − 1)2
=
2 = +∞ +0
2
2(x − x + 1) 2 lim− = +∞ = 2 x→1 +0 (x − 1) ˇ Citatel mus´ı b´yt nula. 2 2(x 2 − x + 1) 2x // / . ..
2
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ 2(x 2 − x + 1) (x − 1)2
=0
x2 − x + 1 = 0
+ + ◦ ı, protoˇze ze vzorce Tato kvadraticka´ rovnice nema´ ˇreˇsen´ 1 p −b ± b2 − 4ac 2 x1,2 = 2(x − x + 1) 2 2a = lim+ = +∞ x→1 +0 (x − 1)2 ´ Obdrˇz´ıme zaporn´ y diskriminant. 2 2(x − x + 1) 2 lim− = +∞ = 2 2 = 1+0 − 4.1.1 = −3 < 0 x→1 D =(xb− − 1)4ac //
/
.
..
2(x 2 − x + 1)
2x
2
2
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ +
+ ◦ 1
2
lim+
x→1
2(x − x + 1)
=
2 = +∞ +0
2(x − x + 1)
=
2 = +∞ +0
2(x 2 − x + 1)
= lim
(x −
1)2
(x −
1)2
2
lim−
x→1
lim
x→±∞
(x − 1)2
2
2x 2 =2 = lim x→±∞ x 2 x→±∞ 1
!0 x2 − x + 1 y osu = 2 x a bod nespojitosti x = 1. Nakresl´ıme (x − 1)2 0
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ +
+ ◦ 1
2
lim+
x→1
2(x − x + 1)
=
2 = +∞ +0
2(x − x + 1)
=
2 = +∞ +0
2(x 2 − x + 1)
= lim
(x −
1)2
(x −
1)2
2
lim−
x→1
lim
x→±∞
(x − 1)2
2
2x 2 =2 = lim x→±∞ x 2 x→±∞ 1
!0 x2 − x + 1 y ==2 2 > 0. Funkce je kladna´ na (−∞, 1). V´ıme, zˇ e y(0) (x − 1)2 0
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ +
+ ◦ 1
2
lim+
x→1
2(x − x + 1)
=
2 = +∞ +0
2(x − x + 1)
=
2 = +∞ +0
2(x 2 − x + 1)
= lim
(x −
1)2
(x −
1)2
2
lim−
x→1
lim
x→±∞
(x − 1)2
2
2x 2 =2 = lim x→±∞ x 2 x→±∞ 1
! 2(4 2 − 2 + 1) 0 Vypoˇcteme 0 y(2) = x − x + 1 > 0. Funkce je kladna´ na (1, ∞). y =2 (2 − 1)2 (x − 1)2
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ +
+ ◦ 1 2
lim+
x→1
2(x − x + 1)
=
2 = +∞ +0
2(x − x + 1)
=
2 = +∞ +0
2(x 2 − x + 1)
= lim
(x −
1)2
(x −
1)2
2
lim−
x→1
lim
x→±∞
(x − 1)2
2
2x 2 =2 = lim x→±∞ x 2 x→±∞ 1
!0 x2 − x + 1 y =2 Urˇc´ıme jednostrann e´ limity 2v bodeˇ nespojitosti (x − 1) 0
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ +
+ ◦ 1 2
lim+
x→1
2(x − x + 1)
=
2 = +∞ +0
2(x − x + 1)
=
2 = +∞ +0
2(x 2 − x + 1)
= lim
(x −
1)2
(x −
1)2
2
lim−
x→1
lim
x→±∞
(x − 1)2
x2 − x + 1 y = 2 Dosad´ıme x = 1. (x − 1)2 0
//
/
.
..
!0
2
2x 2 =2 = lim x→±∞ x 2 x→±∞ 1
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ +
+ ◦ 1 2
lim+
x→1
2(x − x + 1)
=
2 = +∞ +0
2(x − x + 1)
=
2 = +∞ +0
2(x 2 − x + 1)
= lim
(x −
1)2
(x −
1)2
2
lim−
x→1
lim
x→±∞
(x − 1)2
x2 − x + 1 y = 2 Odvod´ıme v´ysledek. (x − 1)2 0
//
/
.
..
!0
2
2x 2 =2 = lim x→±∞ x 2 x→±∞ 1
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ +
+ ◦ 1 2
lim+
x→1
2(x − x + 1)
=
2 = +∞ +0
2(x − x + 1)
=
2 = +∞ +0
2(x 2 − x + 1)
= lim
(x −
1)2
(x −
1)2
2
lim−
x→1
lim
x→±∞
(x − 1)2
x2 − x + 1 y = 2 Urˇc´ıme limity v ±∞. (x − 1)2 0
//
/
.
..
!0
2
2x 2 =2 = lim x→±∞ x 2 x→±∞ 1
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ +
+ ◦ 1 2
lim+
x→1
2(x − x + 1)
=
2 = +∞ +0
2(x − x + 1)
=
2 = +∞ +0
2(x 2 − x + 1)
= lim
(x −
1)2
(x −
1)2
2
lim−
x→1
lim
x→±∞
(x − 1)2
2
!0 x2 − x + 1 y jenom = 2 vedouc´ı cˇ leny. Uvaˇzujeme (x − 1)2
2x 2 =2 = lim x→±∞ x 2 x→±∞ 1
0
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ +
+ ◦ 1 2
lim+
x→1
2(x − x + 1)
=
2 = +∞ +0
2(x − x + 1)
=
2 = +∞ +0
2(x 2 − x + 1)
= lim
(x −
1)2
(x −
1)2
2
lim−
x→1
lim
x→±∞
(x − 1)2
2
2 2x =2 = lim x→±∞ 1 x→±∞ x 2
!0 Funkce ma´ kladnoux 2limitu − x +v 1±∞. Vodorovna´ pˇr´ımka y = 2 je 0 y ke = 2grafu v bodech ±∞. asymptotou (x − 1)2
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1)
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
(x − 1)2
2
y0 = 2 =2
x −x+1 (x − 1)2
(2x − 1)(x − 1)2 − (x 2 − x + 1)2(x − 1)(1 − 0) ((x − 1)2 )2
= 2(x − 1) 2
=2
!0
(2x − 1)(x − 1) − (x 2 − x + 1)2 (x − 1)4
2x − 2x − x + 1 − (2x 2 − 2x + 2)
(x − 1)3 −x − 1 x+1 =2 = −2 (x − 1)3 (x − 1)3 Vypoˇcteme derivaci / . ..x + 1
//0
3
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ 2
y0 = 2 =2
x −x+1 (x − 1)2
!0
(2x − 1)(x − 1)2 − (x 2 − x + 1)2(x − 1)(1 − 0) ((x − 1)2 )2
(2x − 1)(x − 1) − (x 2 − x + 1)2 = 2(x − 1) (x −ılu. 1)4 • Uˇzijeme vzorec pro derivaci pod´ 2 2 2x − 2x − x+ 1− (2x0 − 2x 0+ 2) =2 u 0 3 u v − uv (x − 1) = . v v2 −x − 1 x+1 =2 = −2 3 (x − 1) (x −sloˇ 1)z3 ene´ funkce pˇri derivovan´ ´ ı v´yrazu • Uˇzijeme vzorec pro derivaci 2 (x − 1) . //0
/
.
x+1
..
3
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1)
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
(x − 1)2
2
x −x+1 (x − 1)2
y0 = 2 =2
(2x − 1)(x − 1)2 − (x 2 − x + 1)2(x − 1)(1 − 0) ((x − 1)2 )2
= 2(x − 1) 2
=2
!0
(2x − 1)(x − 1) − (x 2 − x + 1)2 (x − 1)4
2x − 2x − x + 1 − (2x 2 − 2x + 2)
(x − 1)3 −x − 1 x+1 =2 = −2 (x − 1)3 (x − 1)3 Vytkneme (x − 1). / . ..x + 1
//0
3
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1)
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
(x − 1)2
2
y0 = 2 =2
x −x+1 (x − 1)2
(2x − 1)(x − 1)2 − (x 2 − x + 1)2(x − 1)(1 − 0) ((x − 1)2 )2
= 2(x − 1) 2
=2
!0
(2x − 1)(x − 1) − (x 2 − x + 1)2 (x − 1)4
2x − 2x − x + 1 − (2x 2 − 2x + 2)
(x − 1)3 −x − 1 x+1 =2 = −2 (x − 1)3 (x − 1)3
´ ´ ´ ıme (x − 1). Roznasob´ ıme zavorky a zkrat´ / . ..x + 1
//0
3
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1)
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
(x − 1)2
2
x −x+1 (x − 1)2
y0 = 2 =2
(2x − 1)(x − 1)2 − (x 2 − x + 1)2(x − 1)(1 − 0) ((x − 1)2 )2
= 2(x − 1) 2
=2
!0
(2x − 1)(x − 1) − (x 2 − x + 1)2 (x − 1)4
2x − 2x − x + 1 − (2x 2 − 2x + 2)
(x − 1)3 −x − 1 x+1 =2 = −2 (x − 1)3 (x − 1)3
Uprav´ıme cˇ itatel. / . ..x + 1
//0
3
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1)
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
(x − 1)2
2
y0 = 2 =2
x −x+1 (x − 1)2
(2x − 1)(x − 1)2 − (x 2 − x + 1)2(x − 1)(1 − 0) ((x − 1)2 )2
= 2(x − 1) 2
=2
!0
(2x − 1)(x − 1) − (x 2 − x + 1)2 (x − 1)4
2x − 2x − x + 1 − (2x 2 − 2x + 2)
(x − 1)3 −x − 1 x+1 =2 = −2 (x − 1)3 (x − 1)3 Derivace je nalezena. / . ..x + 1
//0
3
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1)
y 0 = −2
(x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) −2
&
min
−1
x+1 =0 (x − 1)3 x+1=0 x = −1 %
◦ 1
&
0 x+1 y 00 = −2 ˇ s´ıme rovnici y 0 = 0. (x − 1)3 Reˇ // / . .. 1(x − 1)3 − (x + 1)3(x − 1)2 (1 − 0)
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1)
y 0 = −2
(x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) −2
&
min
−1
x+1 =0 (x − 1)3 x+1=0 x = −1 %
◦ 1
&
0 x+1 y 00 = −2 ˇ (x − arn´ ´1)3 ım bodem je tedy x = −1. Citatel mus´ı b´yt nula. Stacion // / . .. 1(x − 1)3 − (x + 1)3(x − 1)2 (1 − 0)
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y 0 = −2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) &
%
min
−1 y 00 = −2 = −2
x+1 (x − 1)3 3
&
◦ 1
0
2
1(x − 1) − (x + 1)3(x − 1) (1 − 0)
= −2(x − 1)
((x − 1)3 )2 2 (x − 1) − (x + 1)3
(x − 1)6 −2x − 4 x+2 = −2 =4 4 ´ ı (x ´ Zakresl´ıme stacionarn´ bod a bod nespojitosti osu. − 1) (x − 1)4 na realnou
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y 0 = −2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) &
%
min
−1 y 00 = −2
Urˇc´ıme y 0 (−2).
= −2
x+1 (x − 1)3 3
&
◦ 1
0
2
1(x − 1) − (x + 1)3(x − 1) (1 − 0) ((x − 1)3 )2 2 (x − 1) − (x + 1)3
= −2(x − 1) ´6 hodnota zap. −2 + 1 (x − 1) y (−2) = −2 <0 = −2 −2x − 4 3 x + z2ap. ´ hodnota = −2 (−2 − 1) =4 (x − 1)4 (x − 1)4 0
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y 0 = −2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) &
%
min
−1 y 00 = −2 = −2
Urˇc´ıme y 0 (0).
x+1 (x − 1)3 3
&
◦ 1
0
2
1(x − 1) − (x + 1)3(x − 1) (1 − 0) ((x − 1)3 )2 2 (x − 1) − (x + 1)3
= −2(x − 1) (x kladn − 1)6 a´ hodnota 0+1 y (0) = −2−2x − 4 = −2 x + 2 >0 3 ´ a´ hodnota = −2 (0 − 1) = 4 zaporn (x − 1)4 (x − 1)4 0
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y 0 = −2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) &
%
min
−1 y 00 = −2
x+1 (x − 1)3 3
0
&
◦ 1
2
1(x − 1) − (x + 1)3(x − 1) (1 − 0) = −2 ((x − 1)3 )2 ´ ı minimum je v bodeˇ x = −1. Funkˇcn´ı hodnota je Lokaln´ (x − 1) − (x + 1)3 = −2(x − 1)2 2 (x −+1)1)6 2((−1) − (−1) 3 2.3 y(−1) =−2x − 4 x2 + 2 = 4 = 2 . (−1 − 1) = −2 =4 (x − 1)4 (x − 1)4
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y 0 = −2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) &
%
min
−1 y 00 = −2 = −2
x+1 (x − 1)3 3
&
◦ 1
0
2
1(x − 1) − (x + 1)3(x − 1) (1 − 0)
= −2(x − 1)
((x − 1)3 )2 2 (x − 1) − (x + 1)3
(x − 1)6 3 2+1 y (2) =− −24 −2x x +=2−2 < 0 1 = −2 = −4 1)3 (2 (x − 1)4 (x − 1)4 0
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y 0 = −2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) 0 x+1 00 y = −2 (x − 1)3 3
= −2
2
1(x − 1) − (x + 1)3(x − 1) (1 − 0)
= −2(x − 1)
((x − 1)3 )2 2 (x − 1) − (x + 1)3
(x − 1)6 x+2 −2x − 4 =4 = −2 4 (x − 1) (x − 1)4
x+2 ; x2 = −2 y 00 = 4 (x − 1)4 Vypoˇcteme druhou derivaci. //
/
.
..
x+2
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y 0 = −2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) 0 x+1 00 y = −2 (x − 1)3 3
= −2
2
1(x − 1) − (x + 1)3(x − 1) (1 − 0)
= −2(x − 1)
((x − 1)3 )2 2 (x − 1) − (x + 1)3
(x − 1)6 x+2 −2x − 4 =4 = −2 4 (x − 1) (x − 1)4 • Pouˇzijeme pravidlo pro derivaci pod´ılu. x+2 ; x2budeme = −2 derivovat jako sloˇzenou funkci. y 00 =• 4Jmenovatel (x − 1)4 //
/
.
..
x+2
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y 0 = −2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) 0 x+1 00 y = −2 (x − 1)3 3
= −2
2
1(x − 1) − (x + 1)3(x − 1) (1 − 0)
= −2(x − 1)
((x − 1)3 )2 2 (x − 1) − (x + 1)3
(x − 1)6 x+2 −2x − 4 =4 = −2 4 (x − 1) (x − 1)4 x+2 ; x2 = −2 y 00 = 4 (x − 1)4 2 Vytkneme (x − 1) v cˇ itateli. //
/
.
..
x+2
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y 0 = −2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) 0 x+1 00 y = −2 (x − 1)3 3
= −2
2
1(x − 1) − (x + 1)3(x − 1) (1 − 0)
= −2(x − 1)
((x − 1)3 )2 2 (x − 1) − (x + 1)3
(x − 1)6 −2x − 4 x+2 = −2 =4 4 (x − 1) (x − 1)4 x+2 ; x2 = −2 y 00 = 4 (x − 1)4 Uprav´ıme. //
/
.
..
x+2
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
y 0 = −2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) 0 x+1 00 y = −2 (x − 1)3 3
= −2
2
1(x − 1) − (x + 1)3(x − 1) (1 − 0)
= −2(x − 1)
((x − 1)3 )2 2 (x − 1) − (x + 1)3
(x − 1)6 x+2 −2x − 4 =4 = −2 4 (x − 1) (x − 1)4
x+2 ; x2 = −2 y 00 = 4 (x − 1)4 Obdrˇzeli jsme druhou derivaci. //
/
.
..
x+2
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) x+2 00 y =4 ; x2 = −2 (x − 1)4 y 0 = −2
4
x+2 =0 (x − 1)4 x+2=0 x = −2
∩
in.
−2
∪
◦ 1
∪
ˇ s´ıme y 00 = 0. Reˇ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) x+2 00 y =4 ; x2 = −2 (x − 1)4 y 0 = −2
4
x+2 =0 (x − 1)4 x+2=0 x = −2
∩
in.
−2
∪
◦ 1
∪
Jedine´ ˇreˇsen´ı je x = −2. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) x+2 00 y =4 ; x2 = −2 (x − 1)4 y 0 = −2
∩
in.
−2
∪
◦ 1
∪
Budeme urˇcovat intervaly konvexnosti a konkavity. Zakresl´ıme bod, ´ kde je druha´ derivace nulova´ a bod nespojitosti na realnou osu. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) x+2 00 y =4 ; x2 = −2 (x − 1)4 y 0 = −2
∩
in.
−2
y 00 (−3) = 4 //
/
.
..
∪
◦ 1
∪
−3 + 2 <0 kladna´ hodnota c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) x+2 00 y =4 ; x2 = −2 (x − 1)4 y 0 = −2
∩
in.
−2
y 00 (0) = 4 //
/
.
..
∪
◦ 1
∪
0+2 >0 kladna´ hodnota c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) x+2 00 y =4 ; x2 = −2 (x − 1)4 y 0 = −2
∩
in.
∪
◦ 1
−2
∪
Inflexn´ı bod je v bodeˇ x = −2. Funkˇcn´ı hodnota je y(−2) =
14 . 9
ˇ si sami.) (Vypoˇctete //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
2(x 2 − x + 1) (x − 1)2
˚ c´ık s osou x D(f ) = R \ {1}; y(0) = 2; nen´ı pruseˇ
x+1 3 ; x1 = −1. . . lok. minimum, y(−1) = 3 2 (x − 1) x+2 00 y =4 ; x2 = −2 (x − 1)4 y 0 = −2
∩
in.
−2
y 00 (2) = 4 //
/
.
..
∪
◦ 1
∪
2+1 >0 kladna´ hodnota c
Robert Maˇr´ık, 2008 ×
+
+ ◦ 1
&
min
−1
f (0) = 2 f (±∞) = 2
%
◦ 1
f (1±) = +∞ 3 f (−1) = 2
&
∩
in.
∪
−2 f (−2) =
◦ 1
∪
14 9
Shrneme dosavadn´ı znalosti. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y + 1
&
+ ◦ 1
−1 √ 3 √ − 3
min
%
−1
f (0) = 2 x y =2 f (±∞) − 23
◦ 1
&
∩
in.
∪
−2
f (1±) = +∞ 3 f (−1) = 2
f (−2) =
∪
◦ 1
14 9
y
− 13 − 12
2
−2
−1
1
x
Nakresl´ıme souˇradnou soustavu. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y + 1
&
+ ◦ 1
−1 √ 3 √ − 3
min
−1
f (0) = 2 x y =2 f (±∞) − 23
%
◦ 1
&
∩
in.
∪
−2
f (1±) = +∞ 3 f (−1) = 2
f (−2) =
∪
◦ 1
14 9
y
− 13 − 12
2
−2
−1
1
x
˚ c´ık s osou y. Funkce v tomto bodeˇ roste. Vyznaˇc´ıme pruseˇ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y + 1
&
+ ◦ 1
−1 √ 3 √ − 3
min
−1
f (0) = 2 x y =2 f (±∞) − 23
%
◦ 1
&
∩
in.
∪
−2
f (1±) = +∞ 3 f (−1) = 2
f (−2) =
∪
◦ 1
14 9
y
− 13 − 12
2
−2
−1
1
x
Nakresl´ıme asymptoty. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y + 1
&
+ ◦ 1
−1 √ 3 √ − 3
min
−1
%
◦ 1
&
in.
∪
−2
f (1±) = +∞ 3 f (−1) = 2
f (0) = 2 x y =2 f (±∞) − 23
∩
f (−2) =
∪
◦ 1
14 9
y
− 13 − 12
2
−2
−1
1
x
Nakresl´ıme funkci v okol´ı svisle´ asymptoty. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y + 1
&
+ ◦ 1
−1 √ 3 √ − 3
min
−1
%
◦ 1
&
in.
∪
−2
f (1±) = +∞ 3 f (−1) = 2
f (0) = 2 x y =2 f (±∞) − 23
∩
f (−2) =
∪
◦ 1
14 9
y
− 13 − 12
2
−2
−1
1
x
Nakresl´ıme funkci v okol´ı vodorovne´ asymptoty. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y + 1
&
+ ◦ 1
−1 √ 3 √ − 3
min
−1
f (0) = 2 x y =2 f (±∞) − 23
%
◦ 1
&
∩
in.
∪
−2
f (1±) = +∞ 3 f (−1) = 2
f (−2) =
∪
◦ 1
14 9
y
− 13 − 12
2
−2
−1
1
x
´ ı minimum funkce. Nakresl´ıme lokaln´ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y + 1
&
+ ◦ 1
−1 √ 3 √ − 3
min
−1
f (0) = 2 x y =2 f (±∞) − 23
%
◦ 1
&
∩
in.
∪
−2
f (1±) = +∞ 3 f (−1) = 2
f (−2) =
∪
◦ 1
14 9
y
− 13 − 12
2
−2
−1
1
x
Hotovo! //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
lim√ − x→− 3 lim√ + x→− 3
√◦ 3
0
− 3 √ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ lim
y0 = //
/
.
..
3x 2 · (3 − x 2 ) − x 3 · (0 − 2x) 2
(3 − x 2 )2 2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ lim
2
2
3
´ ame ´ 3x · ı(3 − 3x −) x −2x6= ·0. (0Dost − 2x) nky Definiˇcn´ı oborpurˇc´ıme0 z podm´ av dva body y = 2 2 nespojitosti ± 3. (3 − x )
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x lim = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ ˚ c´ık s osou y ma´ druhou souˇradnici Pruseˇ 0 2 =30. 3xy(0) · (3=− x 2 ) − x · (0 − 2x) 3−0 y = 2 2 (3 − x ) 0
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ lim
3
ˇ sen´ım rovnice Reˇ x = 0. //
/
.
..
x 2 2 3 3x (3 · (0y −pr2x) ´ xame ´ ) − xjedin´ ˚ c´ık s osou x, bod ısk−av useˇ = 0 ·z´ 0 3 y− x=2 2 2 (3 − x ) 2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ lim
3x 2 · (3 − x 2 ) − x 3 · (0 − 2x) y0 = ´ Nulov´y bod a body nespojitosti vyneseme osu. (3 − x 2 )2na realnou
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ Plat´ı −8 y(−2) = =8>0 2 3x−2 )4− x 3 · (0 − 2x) 3x · (3 − p y0 = a graf funkce je nad osou x na intervalu − 3). (3 − x 2(−∞, )2 lim
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ Plat´ı −1 1 y(−1) = =− <0 2 3−−x 21) − x 32· (0 − 2x) 3x · (3 p y0 = a graf funkce je pod osou x na intervalu (3 − x 2(− )2 3, 0). lim
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ Plat´ı 1 1 y(1) = = >0 2 3−−x 21) − x23 · (0 − 2x) 3x · (3 p y0 = a graf funkce je nad osou x na intervalu (3 − x 2(0, )2 3). lim
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ Plat´ı 8 y(2) = = −8 < 0 2 3−−x42 ) − x 3 · (0 − 2x) 3x · (3 p y0 = a graf funkce je pod osou x na intervalu (3 − x 2()2 3, ∞). lim
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 = =∞ 0 3 − x2 √ 3 − 27 x = = −∞ 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim = =∞ √ − 2 0 x→ 3 3 − x √ 3 27 x lim = = −∞ √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ lim
3x 2 · (3 − x 2 ) − x 3 · (0 − 2x) y0 = 2 )2 Budeme zkoumat jednostranne´ limity nespojitosti. (3 −vxbodech
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 = =∞ 0 3 − x2 √ 3 − 27 x = = −∞ 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim = =∞ √ − 2 0 x→ 3 3 − x √ 3 27 x lim = = −∞ √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ nenulov´y v´yraz Vˇsechny limity jsou typu a jednostranne´ limity jsou 0 2 2 3 3x · (3 − x ) − x · (0 − ´ e´ znam ´ nevlastn´ı. Spravn snadno zjist´ıme ze2x) schematu y 0 = enko 2 2 ´ uvedeneho v´ysˇ e. (3 − x ) lim
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ 3 − x 2 x→±∞ −x 2 x→±∞ lim
3x 2 · (3 − x 2 ) − x 3 · (0 − 2x) y0 = Vypoˇcteme limity v nevlastn´ıch bodech. (3 − x 2 )2
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0 −
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
0
− 3 lim√ − x→− 3 lim√ + x→− 3
√ 3 x − 27 =∞ = 0 3 − x2 √ 3 − 27 x = −∞ = 2 0 3−x
+ √◦ 3
−
√ 3 27 x lim =∞ = √ − 2 0 x→ 3 3 − x √ 3 27 x lim = −∞ = √ + 2 0 x→ 3 3 − x
3
3
x x = lim = lim −x = ∓∞ x→±∞ −x 2 x→±∞ x→±∞ 3 − x 2 lim
3x˚ 2 V· (3 − x 2 ) −ıch x 3bodech · (0 − 2x) Jedna´ se o pod´ıl polynom nevlastn´ je podstatna´ pouze y 0 = u. 2 2 ´ zavislost na vedouc´ıch cˇ lenech polynom (3 − x u˚) v cˇ itateli a ve jmenovateli.
//
/
.
..
2
2
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
−
− 3 y0 =
=
+ √◦ 3
0
3x 2 · (3 − x 2 ) − x 3 · (0 − 2x)
(3 − x 2 )2
x 2 3(3 − x 2 ) + 2x 2 x
2
(3 − x 2 )2 2 9−x
= Budeme hledat derivaci funkce. pod´ıl (3 − x 2Derivujeme )2 x y0 = //
/
2
9−x
2
(3 ..− x 2 )2
.
;
&
−
u 0 v
min
−3
0
= %
0
3
x podle vzorce 3 − x2
u ·v −u·v . v2 % MAX & % % ◦ ◦ √ √ c Maˇr´ık, 2008 × 0 3 Robert
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
−
=
2 2 x 9 − 2x 0 x . ; yVytkneme = (3 ..− x 2 )2 // / .
&
−
3x 2 · (3 − x 2 ) − x 3 · (0 − 2x)
(3 − x 2 )2
x 2 3(3 − x 2 ) + 2x 2 x
=
√◦ 3
0
− 3 y0 =
+
2
(3 − x 2 )2 2 9−x
(3 − x 2 )2
min
−3
%
◦ √
%
% 0
√◦
%
MAX
&
c Maˇr´ık, 2008 × 3 Robert
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0
+
˚ c´ık s x: x = 0 pruseˇ
◦ √
−
=
2 2 x 9−x 0 ´ ; yUprav´ = ıme zavorku. (3 ..− x 2 )2 // / .
&
−
3x 2 · (3 − x 2 ) − x 3 · (0 − 2x)
(3 − x 2 )2
x 2 3(3 − x 2 ) + 2x 2 x
=
√◦ 3
0
− 3 y0 =
+
2
(3 − x 2 )2 2 9−x
(3 − x 2 )2
min
−3
%
◦ √
%
% 0
√◦
%
MAX
&
c Maˇr´ık, 2008 × 3 Robert
y=
x3 3 − x2
p D(f ) = R \ {± 3}; +
˚ c´ık s x: x = 0 pruseˇ x y0 =
2
9−x
2
(3 − x 2 )2
◦ √
=
−
;
&
min
−3
%
◦ √
+ √◦ 3
0
− 3
3
y 00 =
y(0) = 0
%
% 0
− 3 2 2
2
4
√◦ 3
−
%
MAX
&
3 2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) (3 − x 2 )4
h
i 2 2 4 2 4 2x · 27−9x − 6x +2x +18x −2x ˇ sen´ım rovnice x 2 (9 − x 2 ) = 0 jsou body x = 0 a x = ±3. Tyto Reˇ = ´ ı body vyneseme(3 ´ stacionarn´ spolu osu. − x 2s)3body nespojitosti na realnou h i
//
/
.
..
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; +
˚ c´ık s x: x = 0 pruseˇ x y0 =
2
9−x
2
(3 − x 2 )2
◦ √
−
;
&
min
−3
%
◦ √
+
%
% 0
− 3 2 2
√◦ 3
0
− 3
3
y 00 =
y(0) = 0
2
4
√◦ 3
−
%
MAX
&
3 2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x)
= cene´ v´yrazy v derivaci jsou kladne´ a neovlivn´ı v´ysledne´ ˇ Cerven eˇ oznaˇ (3 − x 2 )4 h Staˇc´ı tedy zjiˇst’ovat znamenko ´ ´ i v´yrazu (9 − x 2 ). Pro znamenko derivace. 2 2 4 2 4 x = −4 plat´ı 2x · 27−9x − 6x +2x +18x −2x 2 2 = 9 − x = 92 − (−4) < 0. (3 − x )3 h i
//
/
.
..
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; +
˚ c´ık s x: x = 0 pruseˇ x y0 =
2
9−x
2
(3 − x 2 )2
◦ √
=
−
;
&
min
−3
%
◦ √
+ √◦ 3
0
− 3
3
y 00 =
y(0) = 0
%
% 0
− 3 2 2
2
4
(3 − x 2 )4
2xı · 27−9x 2 − 6x 2 +2x 4 +18x 2 −2x 4 Pro x = −2 plat´ = 9 − x 2 = 92 − (−2)2 > 0. (3 − x )3 h i /
.
..
%
MAX
&
3 2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) h
//
√◦ 3
−
2
i c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; +
˚ c´ık s x: x = 0 pruseˇ x y0 =
2
9−x
2
(3 − x 2 )2
y(0) = 0 ◦ √
−
&
min
−3 3
%
◦ √
%
% 0
− 3 2 2
√◦ 3
0
− 3 ;
+
2
4
√◦ 3
−
%
MAX
&
3 2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) ´ ı minimum. Funkˇ V bodeˇ x ==−3 je lokaln´ cn´ı hodnota je (3 − x 2 )4 h i 2 2−27 4 −272 94 2x · 27−9x − 6x +2x +18x −2x y(−3) = = = 3−9 −6 2 = 2 3 (3 − x ) h i y 00 =
//
/
.
..
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; +
˚ c´ık s x: x = 0 pruseˇ x y0 =
2
9−x
2
(3 − x 2 )2
◦ √
=
−
;
&
min
−3
%
◦ √
+ √◦ 3
0
− 3
3
y 00 =
y(0) = 0
%
% 0
− 3 2 2
2
4
(3 − x 2 )4
2xı · 27−9x 2 − 6x 2 +2x 4 +18x 2 −2x 4 Pro x = −1 plat´ = 9 − x 2 = 92 − (−1)2 > 0. (3 − x )3 h i /
.
..
%
MAX
&
3 2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) h
//
√◦ 3
−
2
i c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; +
˚ c´ık s x: x = 0 pruseˇ x y0 =
2
9−x
2
(3 − x 2 )2
◦ √
=
−
;
&
min
−3
%
◦ √
+ √◦ 3
0
− 3
3
y 00 =
y(0) = 0
%
% 0
− 3 2 2
2
4
(3 − x 2 )4
2 2 4 2 4 2x ı · 27−9x − 6x +2x +18x −2x Pro x = 1 plat´ 2 2 = 9− (3x− x=2 )93 − 1 > 0. h i /
.
..
%
MAX
&
3 2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) h
//
√◦ 3
−
2
i c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; +
˚ c´ık s x: x = 0 pruseˇ x y0 =
2
9−x
2
(3 − x 2 )2
◦ √
=
−
;
&
min
−3
%
◦ √
+ √◦ 3
0
− 3
3
y 00 =
y(0) = 0
%
% 0
− 3 2 2
2
4
(3 − x 2 )4
2 2 4 2 4 2x ı · 27−9x − 6x +2x +18x −2x Pro x = 2 plat´ 2 2 = 9− (3x− x=2 )93 − 2 > 0. h i /
.
..
%
MAX
&
3 2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) h
//
√◦ 3
−
2
i c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; +
˚ c´ık s x: x = 0 pruseˇ x y0 =
2
9−x
2
(3 − x 2 )2
◦ √
=
−
;
&
min
−3
%
◦ √
+ √◦ 3
0
− 3
3
y 00 =
y(0) = 0
%
% 0
− 3 2 2
2
4
(3 − x 2 )4
2 2 4 2 4 2x ı · 27−9x − 6x +2x +18x −2x Pro x = 4 plat´ 2 2 = 9− (3x− x=2 )93 − 4 < 0. h i /
.
..
%
MAX
&
3 2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) h
//
√◦ 3
−
2
i c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; +
˚ c´ık s x: x = 0 pruseˇ x y0 =
2
9−x
2
(3 − x 2 )2
y(0) = 0 ◦ √
−
&
min
−3 3
%
◦ √
%
% 0
− 3 2 2
√◦ 3
0
− 3 ;
+
2
4
√◦ 3
−
%
MAX
&
3 2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) ´ ı maximum. Funkˇ V bodeˇ x ==3 je lokaln´ cn´ı hodnota je (3 − x 2 )4 h i 2 227 4 27 2 94 2x · 27−9x − 6x +2x +18x −2x y(3) = = =− 3−9 −6 2 = 2 3 (3 − x ) h i y 00 =
//
/
.
..
2
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; 3
y 00 =
=
y(0) = 0 2 2
2
4
2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) h
(3 − x 2 )4
2x · 27−9x 2 − 6x 2 +2x 4 +18x 2 −2x 4 Derivujeme=funkci 2 (3 − x22 )3 x i(9 − x ) 9x 2 − x 4 h = 2 2x · 27+3x (3 − x 2 )2 (3 − x 2 )2 = podle vzorce (3 − x 2 )3 u 0 u0 · v − u · v 0 h i = . 2 v 2x · 27 + 3x v2 ∩ ∪ ;x=0 y 00 = ◦ // / . .. √ (3 − x 2 )3
i
∪
√◦
∩ c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; 3
y 00 =
=
=
y(0) = 0 2 2
2
4
2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) (3 − x 2 )4
h
2x · 27−9x 2 − 6x 2 +2x 4 +18x 2 −2x 4 h
2
i
(3 − x 2 )3
i
2x · 27+3x ´ • Protoˇze jsme ve jmenovateli neroznasobovali, ale derivovali jako = 2 3 sloˇzenou funkci, nezbavili jsme se moˇ z nosti vytknout. (3 − x ) i • Nyn´hı tedy vytkneme cˇ leny, ktere´ se v cˇ itateli opakuj´ı. 2x · 27 + 3x 2 ∩ ∪ ∪ ;x=0 y 00 = ◦ // / . .. √◦ √ (3 − x 2 )3
∩ c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; 3
y 00 =
=
=
=
y(0) = 0 2 2
2
4
2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) (3 − x 2 )4
h
2x · 27−9x 2 − 6x 2 +2x 4 +18x 2 −2x 4 h
2x · 27+3x
2
i
(3 − x 2 )3
(3 − x 2 )3 h i 2 2x · 27 + 3x ´ ıme a roznasob´ ´ ´ Zkrat´ ıme zavorky. ;x=0 y 00 = 2 3 // / . .. (3 − x )
∪
◦ √
∩
i
∪
√◦
∩ c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3}; 3
y 00 =
=
=
=
y(0) = 0 2 2
2
4
2
(18x − 4x ) · (3 − x ) − (9x − x ) · 2(3 − x )(−2x) 2 (3 − x 2 )2 h i 2x(3 − x 2 ) · (9 − 2x 2 )(3 − x 2 ) + (9x − x 3 )(2x) (3 − x 2 )4
h
2x · 27−9x 2 − 6x 2 +2x 4 +18x 2 −2x 4 h
2x · 27+3x
2
i
(3 − x 2 )3
i
(3 − x 2 )3 h i 2 2x · 27 + 3x ´ V´yrazy v hranate´ zavorce se seˇctou resp. odeˇ ∩ctou. ∪ ∪ y 00 = ;x=0 ◦ 2 3 // / . .. √◦ √ (3 − x )
∩ c
Robert Maˇr´ık, 2008 ×
p x3 D(f ) = R \ {± 3}; 3 − x2 h i 2x · 27 + 3x 2 y 00 = ;x=0 (3 − x 2 )3 y=
y(0) = 0
ˇ ım se zbytkem zjist´ıme, zˇ e plat´ı Delen´
∪
◦ √
− 3
∩
∪ 0
√◦ 3
∩
3x x3 = −x + 2 3−x 3 − x2 ´ je pˇr´ımka, druha´ cˇ ast ´ se bl´ızˇ´ı k nule pro x bl´ızˇ´ıc´ı se do plus Prvn´ı cˇ ast nebo minus nekoneˇcna. Funkce ma´ proto v nevlastn´ıch bodech asymptotu y = −x. ´ ˇreˇsen´ı rovnice y 00 = 0. Druha´ derivace je vypoˇctena. Nyn´ı hledame 2 ´ kladn´y, je jedin´ym ˇreˇsen´ım teto ´ Protoˇze v´yraz (27 + 3x ) je stale rovnice bod x = 0. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
p x3 D(f ) = R \ {± 3}; 3 − x2 h i 2x · 27 + 3x 2 y 00 = ;x=0 (3 − x 2 )3 y=
y(0) = 0
ˇ ım se zbytkem zjist´ıme, zˇ e plat´ı Delen´
∪
◦ √
− 3
∩
∪ 0
√◦ 3
∩
3x x3 = −x + 2 3−x 3 − x2 ´ je pˇr´ımka, druha´ cˇ ast ´ se bl´ızˇ´ı k nule pro x bl´ızˇ´ıc´ı se do plus Prvn´ı cˇ ast nebo minus nekoneˇcna. Funkce ma´ proto v nevlastn´ıch bodech asymptotu y = −x. ´ osu vyneseme bod x = 0 (y 00 (0) = 0) a body, kde je druha´ Na realnou ´ derivace nespojita. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
p x3 D(f ) = R \ {± 3}; 3 − x2 h i 2x · 27 + 3x 2 y 00 = ;x=0 (3 − x 2 )3 y=
y(0) = 0
ˇ ım se zbytkem zjist´ıme, zˇ e plat´ı Delen´
∪
◦ √
− 3
∩
∪ 0
√◦ 3
∩
3x x3 = −x + 2 3−x 3 − x2 ´ je pˇr´ımka, druha´ cˇ ast ´ se bl´ızˇ´ı k nule pro x bl´ızˇ´ıc´ı se do plus Prvn´ı cˇ ast nebo minus nekoneˇcna. Plat´ı Funkce ma´ proto v nevlastn´ıch bodech asymptotu y = −x. ´ 2 · (−2) · [kladn´y v´yraz] zaporn´ y v´yraz y 00 (−2) = >0 = 2 3 ´ z aporn´ y v´yraz (3 − (−2) ) a funkce je konvexn´ı na intervalu obsahuj´ıc´ım cˇ´ıslo −2. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
p x3 D(f ) = R \ {± 3}; 3 − x2 h i 2x · 27 + 3x 2 y 00 = ;x=0 (3 − x 2 )3 y=
y(0) = 0
ˇ ım se zbytkem zjist´ıme, zˇ e plat´ı Delen´
∪
◦ √
− 3
∩
∪ 0
√◦ 3
∩
3x x3 = −x + 2 3−x 3 − x2 ´ je pˇr´ımka, druha´ cˇ ast ´ se bl´ızˇ´ı k nule pro x bl´ızˇ´ıc´ı se do plus Prvn´ı cˇ ast nebo minus nekoneˇcna. Plat´ı Funkce ma´ proto v nevlastn´ıch bodech asymptotu y = −x. ´ 2 · (−1) · [kladn´y v´yraz] zaporn´ y v´yraz y 00 (−1) = <0 = 2 3 kladn´ y v´yraz (3 − (−1) ) ´ ı na intervalu obsahuj´ıc´ım cˇ´ıslo −1. a funkce je konkavn´ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
p x3 D(f ) = R \ {± 3}; 3 − x2 h i 2x · 27 + 3x 2 y 00 = ;x=0 (3 − x 2 )3 y=
y(0) = 0
ˇ ım se zbytkem zjist´ıme, zˇ e plat´ı Delen´
∪
◦ √
− 3
∩
∪ 0
∩
√◦ 3
3x x3 = −x + 2 3−x 3 − x2 ´ je pˇr´ımka, druha´ cˇ ast ´ se bl´ızˇ´ı k nule pro x bl´ızˇ´ıc´ı se do plus Prvn´ı cˇ ast nebo minus nekoneˇcna. Funkce Plat´ı ma´ proto v nevlastn´ıch bodech asymptotu y = −x. 2 · 1 · [kladn´y v´yraz] kladn´y v´yraz y 00 (1) = = >0 2 3 kladn´ y v´yraz (3 − 1 ) a funkce je konvexn´ı na intervalu obsahuj´ıc´ım cˇ´ıslo 1.
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
p x3 D(f ) = R \ {± 3}; 3 − x2 h i 2x · 27 + 3x 2 y 00 = ;x=0 (3 − x 2 )3 y=
y(0) = 0
ˇ ım se zbytkem zjist´ıme, zˇ e plat´ı Delen´
∪
◦ √
− 3
∩
∪ 0
√◦ 3
∩
3x x3 = −x + 2 3−x 3 − x2 ´ je pˇr´ımka, druha´ cˇ ast ´ se bl´ızˇ´ı k nule pro x bl´ızˇ´ıc´ı se do plus Prvn´ı cˇ ast nebo minus nekoneˇcna. Funkce Plat´ı ma´ proto v nevlastn´ıch bodech asymptotu y = −x. 2 · 2 · [kladn´y v´yraz] kladn´y v´yraz y 00 (2) = <0 = 2 3 ´ (3 − 2 ) zaporn´ y v´yraz ´ ı na intervalu obsahuj´ıc´ım cˇ´ıslo 1. a funkce je konkavn´
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x3 3 − x2
p D(f ) = R \ {± 3};
y(0) = 0
ˇ ım se zbytkem zjist´ıme, zˇ e plat´ı Delen´ x3 3x = −x + 2 3−x 3 − x2 ´ je pˇr´ımka, druha´ cˇ ast ´ se bl´ızˇ´ı k nule pro x bl´ızˇ´ıc´ı se do plus Prvn´ı cˇ ast nebo minus nekoneˇcna. Funkce ma´ proto v nevlastn´ıch bodech asymptotu y = −x.
//
/
.
..
c
Robert Maˇr´ık, 2008 ×
+ − √◦ − 3 0 3
+
◦ √
−
f (0) = 0; f (±∞) = ∓∞;
% %MAX& ∪ ∩ ∪ ∩ ◦ √◦ √ √◦ −3 − 3 0 3 3 3 − 3 0 9 f (±3) = ∓ 2 p p f (− 3±) = ∓∞; f ( 3±) = ∓∞ &min%
◦ √
%
ˇ s´ı v´ysledky. ˚ zitejˇ Shrneme nejduleˇ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
− 32 + − + − ◦ √◦ − 31 √ − 3 0 3 − 21 f (0) = 0; x f (±∞) = ∓∞; y 1
% %MAX& ∪ ∩ ∪ ∩ ◦ √◦ √ √◦ −3 − 3 0 3 3 3 − 3 0 9 f (±3) = ∓ 2 p p f (− 3±) = ∓∞; f ( 3±) = ∓∞ &min%
◦ √
%
y
−1 −2 2
−3
√ − 3
√
3
x
3
ˇ Zakresl´ıme svisle´ asymptoty a funkci v okol´ı techto asymptot. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
− 32 + − + − ◦ √◦ − 31 √ − 3 0 3 − 21 f (0) = 0; x f (±∞) = ∓∞; y 1
% %MAX& ∪ ∩ ∪ ∩ ◦ √◦ √ √◦ −3 − 3 0 3 3 3 − 3 0 9 f (±3) = ∓ 2 p p f (− 3±) = ∓∞; f ( 3±) = ∓∞ &min%
◦ √
%
y
−1 −2 2
−3
√ − 3
√
3
3
x
´ asymptoty. Podobneˇ zakresl´ıme sˇ ikmou asymptotu a funkci v okol´ı teto ´ ame ´ Dav pozor na konkavitu/konvexitu. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
− 32 + − + − ◦ √◦ − 31 √ − 3 0 3 − 21 f (0) = 0; x f (±∞) = ∓∞; y 1
% %MAX& ∪ ∩ ∪ ∩ ◦ √◦ √ √◦ −3 − 3 0 3 3 3 − 3 0 9 f (±3) = ∓ 2 p p f (− 3±) = ∓∞; f ( 3±) = ∓∞ &min%
◦ √
%
y
−1 −2 2
−3
√ − 3
√
3
x
3
´ ıho bodu, kter´y nen´ı lokaln´ ´ ım Zakresl´ıme funkci v okol´ı stacionarn´ ´ extremem. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
− 32 + − + − ◦ √◦ − 31 √ − 3 0 3 − 21 f (0) = 0; x f (±∞) = ∓∞; y 1
% %MAX& ∪ ∩ ∪ ∩ ◦ √◦ √ √◦ −3 − 3 0 3 3 3 − 3 0 9 f (±3) = ∓ 2 p p f (− 3±) = ∓∞; f ( 3±) = ∓∞ &min%
◦ √
%
y
−1 −2 2
−3
√ − 3
√
3
3
x
´ ı extremy. ´ Zakresl´ıme lokaln´ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
− 32 + − + − ◦ √◦ − 31 √ − 3 0 3 − 21 f (0) = 0; x f (±∞) = ∓∞; y 1
% %MAX& ∪ ∩ ∪ ∩ ◦ √◦ √ √◦ −3 − 3 0 3 3 3 − 3 0 9 f (±3) = ∓ 2 p p f (− 3±) = ∓∞; f ( 3±) = ∓∞ &min%
◦ √
%
y
−1 −2 2
−3
√ − 3
√
3
3
x
Dokresl´ıme cel´y graf. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ −
+
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
+ ◦ 1 2
2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0 lim−
lim −
2
2
x +1 x = lim =1 x→∞ x 2 − 1 x→∞ x 2 lim
2
y0 =
//
/
.
..
=
0
2
2
2
(x + 1) · (x − 1) − (x + 1) · (x − 1)
0
(x 2 − 1)2
2x · (x 2 − 1) − (x 2 + 1) · 2x
c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0 lim −
2
−
+ ◦ 1 2
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0 lim−
2
x +1 x lim = lim =1 x→∞ x 2 − 1 x→∞ x 2 ˇ sen´ım Urˇc´ıme definiˇcn´ı obor – ve jmenovateli nesm´ı b´yt nula. Reˇ rovnice 2 0 22 − 1 = 0 2 2 0 (x + 1) · (xx − 1) − (x + 1) · (x − 1) 0 y = 2 je x = ±1. Tyto body je nutno vylouˇ definiˇ cn´ıho oboru a jedna´ se (xc2it −z 1) o body nespojitosti. 2 2 2x · (x − 1) − (x + 1) · 2x = c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ −
+
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
+ ◦ 1 2
2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0 lim−
lim −
2
2
x +1 x = lim =1 x→∞ x 2 − 1 x→∞ x 2 lim
2
0
2
2
2
0
(x + 1) · (x − 1) − (x + 1) · (x − 1) y0 = 0(x +21− 1)2 ˚ c´ık s osou y. Dosazen´ım x = 0 urˇc´ıme y(0) = = −1, coˇz je pruseˇ −21+ 1) · 2x 2x · (x 2 − 1) −0(x = c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0 lim −
2
−
+ ◦ 1 2
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0 lim−
2
x +1 x = lim =1 x→∞ x 2 − 1 x→∞ x 2 lim
Rovnice
2 0 22 + 1 = 0 2 2 0 (x + 1) · (xx − 1) − (x + 1) · (x − 1) y = ´ ych cˇ´ısel ˇreˇsen´ nema´ v oboru realn´ ı a2 − funkce (x 1)2 tedy nen´ı nikdy rovna ˚ c´ık 2s osou x. 2 nule. Graf nema´ pruseˇ 2x · (x − 1) − (x + 1) · 2x = c
Robert Maˇr´ık, 2008 × // / . .. 0
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ −
+
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
2
+ ◦ 1 2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0
lim −
lim−
2
2
x +1 x = lim =1 x→∞ x 2 − 1 x→∞ x 2 lim
2
0
2
2
2
0
(x +˚ z1)e zm · (xenit − (x ys+ · (xeˇ nespojitosti − 1) ´ ˇ − 1) ˇ v1)bod funkce nanejv´ Znamenko y 0 = se muˇ 2 2 (x − 1) ˚ c´ık s osou x). Vyneseme (protoˇze nen´ı pruseˇ tedy body nespojitosti na 2 2 ´ realnou osu. 2x · (x − 1) − (x + 1) · 2x = c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ −
+
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
2
+ ◦ 1 2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0
lim −
lim−
2
2
x +1 x = lim =1 x→∞ x 2 − 1 x→∞ x 2 lim
2
0
2
2
2
2
0
(−2) (x + 1) · (x − 1) − (x + 1)+· 1(x −5 1) = > 0 a funkce je Dosazen´ım x = y 0 −2 = zjist´ıme, zˇ e y(−2) = 2 2 (x − 1) (−2)2 − 1 3 kladna´ na intervalu obsahuj´ ıc´ım cˇ´ıslo −2. 2x · (x 2 − 1) − (x 2 + 1) · 2x = c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ −
+
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
2
+ ◦ 1 2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0
lim −
lim−
2
2
x +1 x = lim =1 x→∞ x 2 − 1 x→∞ x 2 lim
2
0
2
2
2
0
(x + 1) · (x − 1) − (x + 1) · (x − 1) ˚ c´ık Dosazen´ım x = y 0 0=jsme jiˇz dˇr´ıve zjistili (kdyˇz jsme poˇc´ıtali pruseˇ (xz2 aporn ´− 1)2 a´ na intervalu obsahuj´ıc´ım s osou y), zˇ e y(0) = −1 a funkce je cˇ´ıslo 0. 2x · (x 2 − 1) − (x 2 + 1) · 2x = c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
2
−
+ ◦ 1 2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0
lim −
lim−
2
2
x +1 x = lim =1 x→∞ x 2 − 1 x→∞ x 2 lim
2
0
2
2 2
2
0
1 1) ·5(x − 1) (x + 1) · (x − 1) (2) − (x+ + = > 0 a funkce je Dosazen´ım x = y 0 2=zjist´ıme, zˇ e y(2) = 2 2 2 3 (x (2) − 1)− 1 ˇ´ıslo2 2. kladna´ na intervalu obsahuj´ 2 ıc´ım c 2x · (x − 1) − (x + 1) · 2x = c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0 lim −
2
−
+ ◦ 1 2
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0 lim−
2
x +1 x = lim =1 x→∞ x 2 − 1 x→∞ x 2 lim
Urˇc´ıme jednostranne´ 2limity 0v bodech nespojitosti. Vˇ 2 2 2sechny 0 (x + 1) · (x − 1) − (x + 1) · (x − 1) 2 0 y = jednostranne´ limity jsou typu a v´ysledkem budou nevlastn´ı limity, tj. 0 (x 2 − 1)2 ´ ym znamenkem”. ´ “nekoneˇcno, opatˇrene´ spr2avn´ 2x · (x − 1) − (x 2 + 1) · 2x = c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ −
+
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
+ ◦ 1 2
2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0 lim−
lim −
2
2
x +1 x = lim =1 x→∞ x 2 − 1 x→∞ x 2 lim
2
0
2
2
2
0
(x + 1) · (x − 1) − (x + 1) · (x − 1) y0 = 2 ´ Podle znamenek funkce na jednotliv´ podintervalech snadno (x 2ych − 1) ´ e´ v´ysledky. odvod´ıme spravn 2 2 2x · (x − 1) − (x + 1) · 2x = c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ −
+
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
+ ◦ 1 2
2
2 x +1 = = +∞ x→−1 x 2 − 1 0 2 x +1 2 lim = = −∞ x→−1+ x 2 − 1 0
2 x +1 = = −∞ x→1 x 2 − 1 0 2 x +1 2 lim+ = = +∞ 2 x→1 x − 1 0 lim−
lim −
2
2
x +1 x = lim =1 x→∞ x 2 − 1 x→∞ x 2 lim
2
0
2
2
2
0
(x +ıch 1) bodech. · (x − 1)Protoˇ − (xze+se 1) jedn · (x a´−o1)racionaln´ ´ ı Urˇc´ıme limity v nevlastn´ y0 = 2 2 funkci, jsou pro limitu v nevlastn´ım(xbod −eˇ1)rozhoduj´ıc´ı pouze vedouc´ı cˇ leny cˇ itatele a jmenovatele. 2x · (x 2 − 1) − (x 2 + 1) · 2x = c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ
◦ −1
2
y0 =
0
Derivujeme pod´ıl=
podle vzorce y0 = //
/
−4x 1)2
2 .. (x . −
2
+ ◦ 1 2
(x + 1) · (x − 1) − (x + 1) · (x − 1) 2
=
2
−
0
(x 2 − 1)2
2x · (x − 1) − (x 2 + 1) · 2x
2
(x 2 − 1)2 2
2x · x − 1 − (x + 1)
(x 2 − 1)2x 2 + 1 y= x2 − 1 2x −2 −4x = = (x 2 − 1)u2 0 (xu20 −· v1)−2 u · v 0 = . v% v2 & & % MAX ◦ ◦
c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ ◦ −1
2
y0 =
2
0
=
=
+ ◦ 1
2
2
(x + 1) · (x − 1) − (x + 1) · (x − 1)
0
(x 2 − 1)2
2
=
−
+
˚ c´ık s osou x nen´ı pruseˇ
2x · (x − 1) − (x 2 + 1) · 2x
2
(x 2 − 1)2 2
2x · x − 1 − (x + 1)
2x −2 (x 2
´ Dopoˇc´ı−4x tame derivace. y0 = 2 .. // / (x . − 1)2
−
(x 2 − 1)2
1)2
=
%
−4x − 1)2
(x 2
◦
%
MAX
&
◦
& c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ ◦ −1
2
y0 =
2
0
=
=
+ ◦ 1
2
2
(x + 1) · (x − 1) − (x + 1) · (x − 1)
0
(x 2 − 1)2
2
=
−
+
˚ c´ık s osou x nen´ı pruseˇ
2x · (x − 1) − (x 2 + 1) · 2x
2
(x 2 − 1)2 2
2x · x − 1 − (x + 1)
2x −2 (x 2
−
(x 2 − 1)2
1)2
=
−4x − 1)2
(x 2
−4xv´yraz 2x v cˇ itateli. % % Vytkneme y0 = ◦ 2 // / (x . .. − 1)2
MAX
&
◦
& c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ ◦ −1
2
y0 =
2
0
=
= −4xzavorku. ´ Uprav´ıme y0 = 2 // / (x . .. − 1)2
+ ◦ 1
2
2
(x + 1) · (x − 1) − (x + 1) · (x − 1)
0
(x 2 − 1)2
2
=
−
+
˚ c´ık s osou x nen´ı pruseˇ
2x · (x − 1) − (x 2 + 1) · 2x
2
(x 2 − 1)2 2
2x · x − 1 − (x + 1)
2x −2 (x 2
−
(x 2 − 1)2
1)2
=
%
−4x − 1)2
(x 2
◦
%
MAX
&
◦
& c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ ◦ −1
2
y0 =
2
0
=
= ´ıme upravy ´ Dokonˇc−4x y0 = 2 .. // / (x . − 1)2
+ ◦ 1
2
2
(x + 1) · (x − 1) − (x + 1) · (x − 1)
0
(x 2 − 1)2
2
=
−
+
˚ c´ık s osou x nen´ı pruseˇ
2x · (x − 1) − (x 2 + 1) · 2x
2
(x 2 − 1)2 2
2x · x − 1 − (x + 1)
2x −2 (x 2
−
(x 2 − 1)2
1)2
=
%
−4x − 1)2
(x 2
◦
%
MAX
&
◦
& c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2
%
◦ −1 2
y 00 = −4 ·
&
0 2
+ ◦ 1 &
◦ 1 2
(x 2 − 1)4 2
(x − 1) · x − 1 − 4x 2
= −4 ·
MAX
−
1 · (x − 1) − x · 2(x − 1) · 2x 2
= −4 ·
%
◦ −1
(x 2 − 1)4
2
2
−3x − 1 3x + 1 =4· 2 3 (x − 1) (x 2 − 1)3
´ ı bod. Vyneseme Derivace3x je2nula jedin´y stacionarn´ + 1 pro x = 0, coˇz je 00 ∪ ∩na realnou ∪ ytento = 4stacion · ´ ´ arn´ ı bod a body nespojitosti osu. ◦ ◦ (x 2 − 1)3 1 −1 //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2
%
◦ −1 2
y 00 = −4 ·
MAX
− &
0 2
+ ◦ 1 &
◦ 1 2
1 · (x − 1) − x · 2(x − 1) · 2x 2
= −4 ·
%
◦ −1
(x 2 − 1)4 2
(x − 1) · x − 1 − 4x
2
(x 2 − 1)4 ´ nezaporn´ ´ ´ se o sudou moc• Jmenovatel zlomku je poˇ 2 rad 2y (jedna −3x rozhoduje −1 3x + 1 ´ ˇ ninu). O znam enku tedy pouze c itatel zlomku. = −4 · =4· (x 2 − 1)3 (x 2 − 1)3 ´ • Protoˇze v cˇ itateli je (−4x), ma´ derivace pˇresneˇ opaˇcne´ znamenko 2 ˇ ´ jako prom enn a x. 3x + 1 ∪ ∩ ∪ y 00 = 4 · ◦ ◦ (x 2 − 1)3 1 −1 //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2
%
◦ −1 2
y 00 = −4 ·
&
0 2
+ ◦ 1 &
◦ 1 2
(x 2 − 1)4 2
(x − 1) · x − 1 − 4x 2
= −4 ·
MAX
−
1 · (x − 1) − x · 2(x − 1) · 2x 2
= −4 ·
%
◦ −1
(x 2 − 1)4
2
2
−3x − 1 3x + 1 =4· 2 3 (x − 1) (x 2 − 1)3
´ ı maximum. Funkˇcn´ı hodnota v tomto V bodeˇ x3x =20+m1a´ funkce lokaln´ 00 ∩˚ c´ık s∪osou y). ybod =eˇ 4je· y(0) = −1 (bylo poˇc´ıtano ´ ∪jako ◦ ◦ pruseˇ (x 2 − 1)3 1 −1 c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2
%
◦ −1 2
y 00 = −4 ·
%
◦ −1
MAX
0 2
− &
+ ◦ 1 ◦ 1
&
2
1 · (x − 1) − x · 2(x − 1) · 2x
(x 2 − 1)4 2 2 2 Budeme hledat druhou derivaci. (x − 1)Derivujeme · x − 1 − pod´ 4x ıl = −4 · x4 2 y 0 = −4(x· − 1) 2 22 2 −3x − 1 (x − 1) 3x + 1 = −4 · =4· (x 2 − 1)3 (x 2 − 1)3 podle vzorce u 0 u0 · v − u · v 0 2 = . 3x + 1 v ∪ ∪ v 2∩ y 00 = 4 · ◦ ◦ (x 2 − 1)3 1 −1 // / . ..
c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2
%
◦ −1 2
y 00 = −4 ·
&
0 2
+ ◦ 1 &
◦ 1 2
(x 2 − 1)4 2
(x − 1) · x − 1 − 4x 2
= −4 ·
MAX
−
1 · (x − 1) − x · 2(x − 1) · 2x 2
= −4 ·
%
◦ −1
(x 2 − 1)4
2
2
−3x − 1 3x + 1 =4· 2 3 (x − 1) (x 2 − 1)3
2 2 2 Protoˇze jsme jako sloˇzenou funkci, nezbavili 3x +v´1yraz (x − 1) derivovali 00 ∪ ∩e´ zkratit. ∪ a pot yjsme = 4se· moˇznosti vytknout v cˇ itateli ◦ ´ ◦ (x 2 − 1)3 1 −1 c
Robert Maˇr´ık, 2008 × // / . ..
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2
%
◦ −1 2
y 00 = −4 ·
&
0 2
+ ◦ 1 &
◦ 1 2
(x 2 − 1)4 2
(x − 1) · x − 1 − 4x 2
= −4 ·
MAX
−
1 · (x − 1) − x · 2(x − 1) · 2x 2
= −4 ·
%
◦ −1
(x 2 − 1)4
2
2
3x + 1 −3x − 1 =4· 2 3 (x − 1) (x 2 − 1)3
2
3x + 1 00 ∪yraz v z∩avorce. yProvedeme =4· ´ ´ ◦ ∪ kracen´ ı a uprav´ıme v´ ◦ (x 2 − 1)3 1 −1 //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2
%
%
◦ −1
2
&
◦ 1 2
(x 2 − 1)4 2
(x − 1) · x − 1 − 4x 2
= −4 ·
&
+ ◦ 1
1 · (x − 1) − x · 2(x − 1) · 2x 2
= −4 ·
MAX
−
0
2
y 00 = −4 ·
◦ −1
(x 2 − 1)4
2
2
−3x − 1 3x + 1 =4· 2 3 (x − 1) (x 2 − 1)3
2
3x + 1 00 yDokonˇ = 4 c· ´ıme upravy. ´ (x 2 − 1)3 //
/
.
..
∪
◦ −1
∩
◦ 1
∪ c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2 2
3x + 1 y =4· (x 2 − 1)3 00
%
◦ −1
∪
%
◦ −1
MAX
− &
0 ◦ −1
∩
◦ 1
+ ◦ 1 ◦ 1
&
∪
´ protoˇze rovnice (3x 2 + 1) = 0 • Druha´ derivace nen´ı nikdy nulova, ´ ych cˇ´ısel. nema´ ˇreˇsen´ı v oboru realn´ ´ ˇ nejv´ysˇ e skokem v bodeˇ ne˚ ze zmenit • Znamenko derivace se muˇ ´ spojitosti. Vyneseme na realnou osu body nespojitosti. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2 2
3x + 1 y =4· (x 2 − 1)3 00
%
◦ −1
%
◦ −1
MAX
− &
0
∪
◦ −1
∩
◦ 1
+ ◦ 1 ◦ 1
&
∪
Funkce je konvexn´ı na intervalu (−∞, −1), protoˇze cˇ ´ıslo (−2) leˇz´ı v tomto intervalu a y 00 (−2) = 4 · //
/
.
..
kladn´y v´yraz [(−2)2 − 1]3
>0
c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2 2
3x + 1 y =4· (x 2 − 1)3 00
%
◦ −1
∪
%
◦ −1
MAX
− &
0 ◦ −1
∩
◦ 1
+ ◦ 1 ◦ 1
&
∪
´ ı na intervalu (−1, 1), protoˇze cˇ´ıslo 0 leˇz´ı v tomto Funkce je konkavn´ ´ ı (je zde stacionarn´ ´ ı intervalu a funkce je v tomto bodeˇ nutneˇ konkavn´ ´ ı maximum – funkce je pod teˇcnou). bod a lokaln´ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
y=
x2 + 1 x2 − 1
˚ c´ık s osou y: [0, −1]; D(f ) = R \ {−1, 1}; pruseˇ +
˚ c´ık s osou x nen´ı pruseˇ y0 =
−4x (x 2 − 1)2 2
3x + 1 y =4· (x 2 − 1)3 00
%
◦ −1
∪
%
◦ −1
MAX
− &
0 ◦ −1
∩
◦ 1
+ ◦ 1 ◦ 1
&
∪
Funkce je konvexn´ı na intervalu (1, ∞), protoˇze cˇ´ıslo 2 leˇz´ı v tomto intervalu a kladn´y v´yraz >0 y 00 (2) = 4 · (22 − 1)3 //
/
.
..
c
Robert Maˇr´ık, 2008 ×
+ − % MAX & % & ◦ ◦ ◦ ◦ −1 1 −1 0 1 f (0) = −1; f (±∞) = 1; f (−1±) = ∓∞; +
∪
∩ ∪ ◦ ◦ −1 1 f (1±) = ±∞
ˇ s´ı v´ysledky. ˚ zitejˇ Shrneme nejduleˇ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
− 21 +
+ − % MAX & % & ◦ ◦ ◦ ◦ x −1 1 −1 0 1 y = −1; f (0) f (±∞) = 1; f (−1±) = ∓∞; 1
−1
∪
∩ ∪ ◦ ◦ −1 1 f (1±) = ±∞
y
−2 2
x y 1
3 −3 √ 3 √ − 3
−1
1 −1
x
Zakresl´ıme soustavu souˇradnic a asymptoty. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
− 21 +
+ − % MAX & % & ◦ ◦ ◦ ◦ x −1 1 −1 0 1 y = −1; f (0) f (±∞) = 1; f (−1±) = ∓∞; 1
−1
∪
∩ ∪ ◦ ◦ −1 1 f (1±) = ±∞
y
−2 2
x y 1
3 −3 √ 3 √ − 3
−1
1 −1
x
Naˇcrtneme funkci v okol´ı svisl´ych asymptot. Vyuˇzijeme monotonie. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
− 21 +
+ − % MAX & % & ◦ ◦ ◦ ◦ x −1 1 −1 0 1 y = −1; f (0) f (±∞) = 1; f (−1±) = ∓∞; 1
−1
∪
∩ ∪ ◦ ◦ −1 1 f (1±) = ±∞
y
−2 2
x y 1
3 −3 √ 3 √ − 3
−1
1
x
−1
ˇ vyuˇzijeme Naˇcrtneme funkci v okol´ı vodorovne´ asymptoty. Opet schema s monotoni´ı. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
− 21 +
+ − % MAX & % & ◦ ◦ ◦ ◦ x −1 1 −1 0 1 y = −1; f (0) f (±∞) = 1; f (−1±) = ∓∞; 1
−1
∪
∩ ∪ ◦ ◦ −1 1 f (1±) = ±∞
y
−2 2
x y 1
3 −3 √ 3 √ − 3
−1
1 −1
x
´ ı maximum. Zakresl´ıme lokaln´ //
/
.
..
c
Robert Maˇr´ık, 2008 ×
− 21 +
+ − % MAX & % & ◦ ◦ ◦ ◦ x −1 1 −1 0 1 y = −1; f (0) f (±∞) = 1; f (−1±) = ∓∞; 1
−1
∪
∩ ∪ ◦ ◦ −1 1 f (1±) = ±∞
y
−2 2
x y 1
3 −3 √ 3 √ − 3
−1
1 −1
x
Dokresl´ıme cel´y graf. //
/
.
..
c
Robert Maˇr´ık, 2008 ×
K ONEC
//
/
.
..
c
Robert Maˇr´ık, 2008 ×