9|n3+ (n+1 )3+(n+2 )3

1. Indukcijska baza: (n=1 )

9|n3+ (n+1 )3+(n+2 )39|1+23+339|1+8+279|36 ,36 :9=4

2. Indukcijska hipoteza: (n=k )

9|k3+ (k+1 )3+(k+2 )3 , k3+(k+1 )3+ (k+2 )3=E ,k∈Z

3. Dokaz: (n=k+1 )

9|( k+1 )3+ (k+2 )3+(k+3 )39|( k+1 )3+ [ (k+1 )+1 ]3+[ (k+2 )+1 ]3

9|(k3+3k2+3k+1 )+[ (k+1 )3+3 (k+1 )2+3 (k+1 )+1 ]+[ (k+2 )3+3 (k+2 )2+3 (k+2 )+1 ]

9|k3+(k+1 )3+ (k+2 )3+3k2+3 (k+1 )2+3 (k+2 )2+3k+3 ( k+1 )+3 (k+2 )+3

9|k3+(k+1 )3+(k+2 )3⏟E

+3∙ [k 2+ (k+1 )2+ (k+2 )2+k+ (k+1 )+(k+2 )+1 ]

9|3 ∙ [k2+(k+1 )2+( k+2 )2+k+(k+1 )+ (k+2 )+1 ]+E9|3 ∙ [k2+(k2+2k+1 )+(k 2+4k+4 )+3k+4 ]+E9|3 ∙ [3k2+9k+9 ]+E9|9 ∙ (k 2+3k+3 )+E , tvrdnja jeispravna

Zapisati sistem u matričnom obliku i riješiti ga:

x+z=−2− y−2 z=2−2 x−z=3

___________________x+0+z=−20− y−2 z=2−2 x+0−z=3

( 1 0 10 −1 −2

−2 0 −1) ∙(xyz )=(−223 ) , A=( 1 0 1

0 −1 −2−2 0 −1) , X=( xyz ), B=(−223 ) ,

A ∙ X=BA ∙ X=B/→ ∙ ( A−1) , s lijeve strane ,det (A )≠0A−1 ∙ A ∙ X=A−1 ∙BX=A−1 ∙B

A−1= 1det ( A )

adj (A )

det (A )=| 1 0 10 −1 −2

−2 0 −1|1 00 −1

−2 0⏟Sarus

=+(1+0+0 )−(2−0−0 )=1−2=−1≠0

AT=( 1 0 10 −1 −2

−2 0 −1)T

=(1 0 −20 −1 01 −2 −1)=(1 0 −2

0 −1 01 −2 −1)

adj (A )=(|1 0 −20 −1 01 −2 −1| |1 0 −2

0 −1 01 −2 −1| |1 0 −2

0 −1 01 −2 −1|

O O O

|1 0 −20 −1 01 −2 −1| |1 0 −2

0 −1 01 −2 −1| |1 0 −2

0 −1 01 −2 −1|

O O O

|1 0 −20 −1 01 −2 −1| |1 0 −2

0 −1 01 −2 −1| |1 0 −2

0 −1 01 −2 −1|)←ovaj dio sene piše ,

tu je demonstrativno=¿

¿(+|−1 0−2 −1| −|0 0

1 −1| +|0 −11 −1|

−| 0 −2−2 −1| +|1 −2

1 −1| −|1 01 −2|

+| 0 −2−1 0 | −|1 −2

0 0 | +|1 00 −1|)=¿

¿(+[ (1 )− (−0 ) ] −[ (−0 )− (0 ) ] +[ (−0 )−(−1 ) ]−[ (−0 )− (4 ) ] +[ (−1 )− (−2 ) ] −[ (−2 )− (0 ) ]+[ (0 )−(2 ) ] −[ (0 )− (−0 ) ] +[ (−1 )− (0 ) ] )=¿

¿( 1 0 1−(−4 ) (−1+2 ) −(−2 )−2 0 −1 )=( 1 0 1

4 1 2−2 0 −1)

A−1= 1det ( A )

adj (A )= 1−1 ( 1 0 1

4 1 2−2 0 −1)=−1( 1 0 1

4 1 2−2 0 −1)

X=A−1 ∙B=−1( 1 0 14 1 2

−2 0 −1)3x 3 ∙(−223 )

3x 1

=−1( 1∙ (−2 )+0 ∙2+1 ∙34 ∙ (−2 )+1∙2+2∙3

(−2 ) (−2 )+0 ∙2+(−1 ) ∙3)3x 1=¿−1(−2+0+3−8+2+64+0−3 )

3x 1

=−1(101)3 x1=(−10−1)3x 1X=( xyz )=(−10−1) , x=−1, y=0 , z=−1

y=ln x−12 x−1

Domen:x−12x−1

>0 ,

Dom (f )={∀ x∈R :( x∈ ⟨−∞,12 ⟩∪ ⟨1 ,+∞ ⟩)}

Parnost i periodičnost:

f (−x )=ln −x−1−2 x−1

=ln−( x+1 )−(2 x+1 )

=ln x+12 x+1

, nije parnainijeneparna

nema trigonometr ijskeelemente panije periodična

Nule i znak:

lnx−12x−1

=0 , x−12 x−1

=1 , x−12x−1

−1=0 , x−1−2x+12 x+1

=0 , −x2x+1

=0 , x=0¿¿

y ≥0 za x∈ [0 , 12 ⟩

y<0 za x∈ ⟨−∞ ,0 ⟩∪ ⟨1 ,+∞ ⟩

Asimptote:vertikalne :¿

horizontalne :¿

nemakosu, jer imahorizontalne za ( x→−∞ ) i ( x→+∞ )

Ekstremne tačke i tok funkcije:

y '=(ln x−12x−1 )

'

=2 x−1x−1 ( x−1

2 x−1 )'

=2 x−1x−1

( x−1 )' (2 x−1 )−( x−1 ) (2 x−1 )'

(2 x−1 )2=¿ 1

x−1(2 x−1 )−( x−1 ) (2 )

2x−1= 1x−1

2x−1−2x+22 x−1

= 1( x−1 ) (2x−1 )

y '= 1( x−1 ) (2x−1 )

≠0 , nema ekstremne

−∞121+∞

x−1 −¿ −¿ +¿2 x−1 −¿ +¿ +¿Dom>0 +¿ −¿ +¿

−∞ 0121+∞

−x +¿ −¿ −¿2 x+1 −¿ −¿ +¿znak −¿ +¿ −¿

Prevojne tačke, konveksnost/konkavnost:

y ' '=( 1( x−1 ) (2x−1 ) )

'

=0−[ ( x−1 ) (2 x−1 ) ]'

[ ( x−1 ) (2 x−1 ) ]2=

−( x−1 )' (2x−1 )+( x−1 ) (2 x−1 )'

[ (x−1 ) (2 x−1 ) ]2=¿−

(2x−1 )+2 ( x−1 )

[ ( x−1 ) (2x−1 ) ]2= −4 x−3

[ ( x−1 ) (2x−1 ) ]2= −4 x+3

[ ( x−1 ) (2x−1 ) ]2I¿

y ' '= −4 x+3

[ ( x−1 ) (2 x−1 ) ]2=0 ,−4 x+3=0 , x≠ 3

4, nema prevojne

Graf:

−∞121+∞

1 +¿ +¿( x−1 ) −¿ +¿

(2 x−1 ) −¿ +¿y ' +¿ +¿tok ↗ ↗

−∞12341+∞

−4 x+3 +¿ −¿[ ( x−1 ) (2x−1 ) ]2 +¿ +¿

y ' ' +¿ −¿ykonv /konk ∪ ∩

I=∫exsin 2x dx=|u=sin 2 x du=2cos2x dx

dv=exdx v=ex |=exsin 2 x−2∫ e

xcos2x dx⏟

I1

I 1=∫ excos2x dx=|u=cos 2x du=−2sin 2 xdx

dv=ex dx v=e x |=excos2 x−∫ e

x (−2sin 2 x )dx=¿excos 2x−(−2∫e

xsin 2x dx⏟

I)=e

xcos2 x+2 I

I=ex sin 2x−2 I 1=ex sin 2x−2 (excos2 x+2 I )=ex sin2 x−2ex cos2 x−4 I

I=ex (sin 2x−2cos 2x )−4 I I+4 I=ex (sin 2 x−2cos2 x )5 I=ex (sin 2x−2cos2 x )

I= ex

5(sin 2 x−2cos2x )+C

Izračunati površinu zatvorenu grafovima funkcijama: y1=√ x+1+2y2=x+√ x+1x=0x=2

površinanije isječenai čitavase nalazina području y>0pa integralne trebamo razdvajati na

više dijelova zbog negativnogdijela ( y<0 )

I 1=∫0

2

(√ x+1+2 )dx=¿∫0

2

√x+1dx+∫0

2

2dx=¿∫0

2

( x+1 )12dx+2 x 2

¿¿¿¿

I 2=∫0

2

( x+√x+1 )dx=∫0

2

xdx+∫0

2

√ x+1dx=¿ x2

22

¿¿¿¿

P=I 1−I 2=6√3+103

−12√3+86

=12√3+20−(12√3+8 )

6=20−8

6=126

=2

y=sin (2 x+1 )

x+1+ ex−1sin x

y '=(sin (2 x+1 )x+1

+ ex−1sin x )

'

y '=[sin (2x+1 ) ]' (x+1 )−sin (2x+1 ) ( x+1 )'

( x+1 )2+

(ex−1 )' (sin x )−(ex−1 ) (sin x )'

(sin x )2

y '=cos (2 x+1 ) (2 x+1 )' ( x+1 )−sin (2 x+1 )

( x+1 )2+exsin x−(ex−1 )cos x

(sin x )2

y '=2cos (2x+1 ) ( x+1 )−sin (2 x+1 )

( x+1 )2+exsin x−(ex−1 )cos x

(sin x )2

-------------------------------------------------------------------------------------------------------------------------------------- y=ln tg x

y '=( ln tg x )'

y '= 1tg x

( tg x )'

y '= 1sin xcos x

( sin xcos x )'

y '= cos xsin x

(sin x )' (cos x )−(sin x ) (cos x )'

(cos x )2

y '= 1sin x

(cos x ) (cos x )− (sin x ) (−sin x )cos x

= 1sin x

(cos x )2+(sin x )2

cos x= 1sin x

1cos x

= 1sin x cos x

y ' '=( 1sin x cos x )

'

=0−(sin x cos x )'

(sin x cos x )2=

−(sin xcos x )'

(sin xcos x )2

y ' '=0−(sin x cos x )'

(sin x cos x )2=

−(sin xcos x )'

(sin xcos x )2

y ' '=− (sin x )' (cos x )+ (sin x ) (cos x )'

(sin xcos x )2

y ' '=− (cos x ) (cos x )+(sin x ) (−sin x )

(sin x cos x )2=

−(cos x )2−(sin x )2

(sin x cos x )2