1¿ lnx+2x=0denklemkökünümilyonda bir hatayla kirişler yöntemiyle bulun.

f (a ) . f (x1 )<0olmalıdır .

a=0.1 x1=1

f (0,1 )<0 f (1 )>0

xn+1=a+f (a )

f (a )−f (xn )( xn−a )

x2=0.1+f (0,1 )

f (0,1 )−f (1 )(1−0.1 )=0.1+ ln (0,1 )+2. (0,1 )

ln (0,1 )+2. (0,1 )−ln (1 )−2. (1 )(0,9 )=0.1+ −2,102585

−2,102585−2(0 ,9 )

x2=0,561252

x3=0,466315

x4=0,438952

x5=0,430382

x6=0,427626

x7=0,426442

x8=0,426348

x9=0,426307

x10=0,426304

x11=0,426303

x12=0,426302

x13=0,426302

x12=x13=0,426302denklemin kökümilyonda bir hataylahesaplanmıştır .

GrafikÇizimi

lnx=−2x

y=−2x

y=lnx

2¿e−x−2 x=0deklemin kökünümilyonda bir hatayla teğetler yöntemiylebulun .

xn+1=xn−f (xn)f ' (xn)

f (xn )=e−xn−2xn ab=lna. ab . b '

f ' (xn )= (e−xn )'+(−2 xn )'=−e− xn−2 e− x=lne. e− x .−1=−e− x

x1=1

x2=x1−f (x1 )f ' (x1 )

=1−

1e−2.1

−1e

−2=0.310724

x2=0.310724

x3=0.351511

x4=0.351733

x5=0.351733

f (0.351733 )=e−0.351733−2 (0.351733 )=0.000001 x4 köktürmilyondabir hatayla hesaplanmıştır .

GrafikÇizimi

e− x=2x

y=e−x

y=2x

3¿ y=(3 tanx )2cotanx

lny=2cotanx .tanx . ln 3 ab=lna.ab. b '

(lny) '=(2¿¿cotanx . tanx . ln 3) ' ¿ 2cotanx=ln 2.2cotanx .(cotanx) '

y 'y

=(2¿¿cotanx . tanx)' . ln3 ¿ (cotanx )'= −1sin 2x

(2¿¿cotanx . tanx)'=(ln 2.2cotanx . −1sin2x

. tanx)+ 1cos2x

(2¿¿cotanx)¿¿

y '=¿

4 ¿ limx→e

(ln3 x)1

1−lnx

limx→e

( ln3 x)1

1−lnx=1∞

lny=limx→e

11−lnx

.3. ln(lnx)

⟹ limx→e

11−lnx

.3. ln ( lnx )=00BelirsizliğiL ' Hopital

⟹ limx→e

(3. ln ( lnx ))'(1−lnx )'

=

3. ( lnx )'

lnxx '

x

=

3x .lnx1x

= 3lnx

lny=3

y=e3

5¿ f ( x , y )= y2−2 y x2+8 xy+5 fonksiyonunun x=1noktasındakiT .D ve N . D ?

y2−2 y+8 y+5= ( y+1 ) ( y+5 )=0 y=−1 , y=−5

y=−1i seçelim .nokta (1 ,−1 )noktasıdır .

Kapalı FonksiyonunTürevi=−f xf y

−f xf y

=−−4 xy+8 y2 y−2 x2+8x

= 4 xy−8 y2 y−2 x2+8 x

(1 ,−1)noktasındakiTeğet Eğimi

−4−(−8)−2−2+8

=44=1

mt=1mt .mn=−11.mn=−1mn=−1

Teğet Denklemi :y− y0x−x0

=mtmt=1 ( y− y0 )=mt (x−x0)

(1 ,−1 )noktası ( y−1 )=1. (x−(−1 ) )

y−1=x+1

Normal Denklemi :y− y0x−x0

=mnmn=1 ( y− y0 )=mn(x−x0)

(1 ,−1 )noktası ( y−1 )=−1. (x− (−1 ) )

y−1=−x−1

6¿ f ( x )=ln (16−x2 ) fonksiyonunun grafiğini çizim adımlarınagöre çiziniz .

1)

T . A : (−4 ,+4 )

2)

limx→−4−¿ ln (16−x2)=Tanımsız lim

x→4−¿ln (16−x2)=−∞ ¿

¿ ¿¿

limx→−4+¿ ln (16−x2 )=−∞ lim

x →4+¿ ln (16−x2 )=Tanımsız ¿¿ ¿

¿

y=0 yatay asimptot

3)

f ( x )=ln (16−x2 )

y=0 x=0

16−x2=1 y= ln 16

x1=√15x2=−√15

4)

f ' ( x )= −2 x16−x2

= 2 xx2−16

2 x=0 x=0dır .

x2=16 x=−4 ,+4

5) 6)

7¿ f ( x )= x2+1x2−1

T . A=(−∞ ,−1 )U (−1,1 )U (1,+∞)

1)

limx→−∞

x2+1x2−1

=1 (L' Hopital ) limx→∞

x2+1x2−1

=1 (L'Hopital ) ( y=1Düşey Asimptot)

2)

limx→−1−¿ x2+1

x2−1=∞ lim

x →+1−¿ x 2+1x 2−1

=−∞ ¿

¿

¿

limx→−1+¿ x2+1

x2−1=−∞ lim

x→+1+¿ x2+1x2−1

=∞¿

¿¿

¿

y=−∞ ,∞ yatay asimptot

3)

f ( x )= x2+1x2−1

f ' ( x )=2 x .(x2−1 )−2x . (x2+1 )

(x2−1 )2=0

f ' ( x )=2x (−2 )(x2−1 )2

= −4 x(x2−1 )2

=0

x1=0 x2,3=1 x4,5=−1

4)

f ( x )= x2+1x2−1

x=0 için y=0için

y=−1dir . x= yoktur .(x2=−1)

5)

6)

7¿ f ( x )= x2−4x−1

T . A=(−∞ ,1 )U (1 ,+∞)

1)

limx→−∞

x2−4x−1

=−∞ (L'Hopital ) limx→∞

x2−4x−1

=∞ ( L'Hopital )

limx→1−¿ x2−4

x−1 =∞ (L' Hopital ) limx→1+ ¿ x2−4

x−1=−∞ (L ' Hopital) ¿

¿¿

¿

2)

f ' ( x )=2 x .( x−1 )−(x2−4 )

( x−1 )2=0

f ' ( x )=2x2−2 x−x2+4¿ ¿( x−1 )2

= x2−2 x+4(x−1 )2

=0

x1,2=1 x3=1.73205 x4=−1.73205

3)

f ( x )= x2−4x−1

x=0 için y=0 için

y=4 tür . x=2 ,−2dir .

4) 5)

8¿ f ( x , y , z )=4 x2−2 y2+6 z2−8 xy−10 yz+4 xz−12 x+10 y−20 z−33Fonksiyonunun ekstremumlarını bulup konveksliğini inceleyiniz.∇ f ( x )=0 Köklerini bulmaf x=8 x−8 y+4 z−12=0 f x2 =4 x−4 y+2 z−6=0

f y=−8 x−4 y−10 z+10=0 f x+¿f y=−12 y−6 z−2=0¿

f z=+4 x−10 y+12 z−20=0 − f z=−4 x+10 y−12 z+20=0

f x2

−f z=0+6 y−10 z+14=0

2¿

Yerine koyarsak x=13 , y=−23, z=1 x0=(1

3,−23,1)

f xx=8 , f xy=f yx=−8 , f xz=f zx=4 , f yz=f zy=−10 , f yy=−4 , f zz=12

H f ( x )=|a b cd e fg h i|H f ( x )=| 8 −8 4

−8 −4 −104 −10 12 |

|H 1|=|a|=|8|>0

|H 2|=|a bd e|=| 8 −8−8 −4|=−32−64=−96<0

Kural sağlanmadığından H 3 e gerek yoktur.

|H 1|=+ ,∨H 2|=− , (+,−,…) tanımsızdır .

x0bükümnoktası , ne iç bükey ,nedış bükey

9) ln (sin ( x3 ))−cos (ecosy )−e5x .4siny−ln (sin (cos (2x2 )))−sin (cos ( ln (3lny ) ))

−ln (6x 2y )−sin (cos (exy ) )−ln( 2cosx

3√ y2 )=0 y’=?a=ln (sin (x3 ) ) e=−sin (cos ( ln (3lny ) ) )

b=−cos (ecosy ) f=−ln (6x 2y )

c=−e5 x .4siny g=−sin (cos (exy ))

d=−ln (sin (cos (2x2 ))) h=−ln( 2

cosx

3√ y2 )

y '=−f xf y

a=ln (sin (x3 ) )

a '=¿¿

f x=(3x¿¿2)cos (x3)

sin (x3 )f y=0¿

b=−cos (ecosy )

b '=−1.(cos (ecosy ))'=−1.(e¿¿ cosy) ' .(−sin (ecosy))=( y ) ' . ecosy . (−siny ) .(−sin(ecosy))¿

au=lna.au . u'

(e¿¿cosy)'=lne . ecosy . (cosy )'=ecosy . (−siny ) . y ' ¿

f x=0 f y=ecosy . (−siny ) .(−sin(ecosy))

c=−e5 x .4siny

c '=−1. (e5x .4siny )'=−1.¿¿

(e¿¿5 x) '=lne . e5x .(5x )' ¿

(4¿¿ siny)'=ln 4.(4¿¿ siny ).(siny) '=ln 4.(4¿¿ siny) .(cosy) . y ' ¿¿¿

c '=−¿

f x=5e5x .5 .4siny f y=ln 4.(4¿¿siny ). (cosy ) . e5x¿

-------------------------------------------------------------------------------------------------------------d=−ln (sin (cos (2x

2 )))

d '=−1. (ln (sin (cos (2x2 )) ))

'

=−1.(lnd)'=−1. d 'd

=−1.( sin (cos (2x

2 ))) 'sin (cos (2x

2 ))

sin (cos (2x2 ))=dcos (2x2 )=v 2x2=w

(sin (cos (2x2 )) )'=cosv . v '

v'=(cos (2x2 ))'=−sinw .w '

w '=ln 2.2x2

. (x¿¿2) '=ln 2.2x2

.2x ¿

d '=−cosv . (−sinw ) . ln 2. 2x2

.2 x

sin (cos (2x2 ))

f x=−cosv . (−sinw ) . ln 2.2x

2

.2 x

sin (cos (2x2 ))

f y=0

----------------------------------------------------------------------------------------------------------------

e=−sin (cos ( ln (3lny )) )

v=cos ( ln (3lny ))w=ln (3lny ) t=3lny e '=−1. (cos ( v ) ). v '

v'=−sin (w ) .w '

w '=(lnt )'= t 't=

(3¿¿lny) '3 lny

=ln 3. 3lny . ln ( y )'

3lny=ln 3.3lny . y '

y3lny

= ln 3.3lny . y '

3lny . y¿

e '=−1. cos (v ) .−sin (w ) . ln3. 3lny

3lny . y=cos ¿¿

f x=0 f y=cos ¿¿

------------------------------------------------------------------------------------------------------f=−ln (6x 2y )

f '=−u 'u

=−(6¿¿x 2y) '6x2y

=−(6¿¿x )' .2y+(2¿¿ y )' .6x

6x2 y=

−ln 6.6x . (x )' .2y+ ln2. 2y . ( y )' .6x

6x2y¿¿¿

f x=− ln6. 6x .2y

6x 2yf y=

−ln 2.2y .6x

6x2 y

--------------------------------------------------------------------------------------------------------g=−sin (cos (exy ))

u=cos (exy )

v=exy

g'=−1. sin (u) . u '

u'=−sinv . v '

v'=lne. exy . ( xy )'=exy .(xy )'

g'=−sin (cos (exy )) .−sin (exy ) . exy . x=sin (cos (exy )) .sin (exy ) .e xy .(xy ) '

f x=sin (cos (exy )) . sin (exy ) . exy . y f y=sin (cos (exy )) . sin (exy ). exy . x

-------------------------------------------------------------------------------------------------------------h=−ln( 2

cosx

3√ y2 )

h'=−u'

u=

−( 2cosx

3√ y2 )'

( 2cosx

3√ y2 )u=2

cosx

3√ y2( 3√ y2 )'= y

23−1= 2

3 3√ y

u'=( ln 2.2cosx . (cosx )' . 3√ y2)−( 2

3 3√ y.2cosx)

3√ y4=

ln 2.2cosx . (sinx ) . x ' . 3√ y2−( 23 3√ y

.2cosx)3√ y4

h'=−u'

u =

−ln 2.2cosx . (si nx ) . x ' . 3√ y2−( 23 3√ y

.2cosx)( 2cosx

3√ y2 )

f x=− ln2. 2cosx . (sinx ) .1 . 3√ y2−( 2

3 3√ y.2cosx)

( 2cosx

3√ y2 )f y=

3√ y2.2 .2cosx2cosx 3√ y4 .3 3√ y

= 23√ y3

=-¿

ecosy . (−siny ) . (−sin (ecosy ))+ ln 4.(4¿¿ siny ). (cosy ) . e5 x+cos ¿¿¿

11¿ x=√ t ve y=cos (sin ( log2 (2cotan ( sint ) ))) dydx=?

x2=t

y=cos(sin ( log2 (2cotan (sin x2 )) )) log2 (2cotan ( sin x2) )=cotan ( sin x2 ) log22

y=cos (sin (cotan (sin x2 )))

u=sin (cotan ( sin x2 ))

v=cotan ( sin x2 )

w=sin x2

cotan (u )= cosusinu ( cosusinu )

'

= (cosu )' . sinu−( sinu )' . cosu( sinu )2

=u ' (− (sinu )2−( cosu )2)(sinu )2

= −u '( sinu )2

y '=−sinu.u '

u'=cos (v ) . v '

v'= (cotan (w ) )'= −w '(sinw )2

w '=2x . cos x2

y '=−sinu.cos v .(−2x .cos x2( sinw )2 )y '=2x .cos x

2 . sinu .cosv( sinw )2

12¿ f ( x )=2 x3−6 x2+12 x+9 fonksiyonuna [1,3 ] aralığındaODT uygulayınız .

f∈C[1,3]

f∈D(1,3)

f (a )=f (b )olsun. f ' (c )=0olacak şekildec∈ [a ,b ] vardır .

f (3 )− f (1)3−1 =

54−54+36+9−(2−6+12+9)2 =

62=3

f ' ( x )=6 x2−12 x+12=3

6 x2−12 x+9=0

∆=b2−4.a . c=144−4.6.9=−72

x1,2=−b±√∆2a

x1=−(−12 )−(√−72)

12=12−√−72

12

x2=−(−12 )+(√−72)

12=12+√−72

12

13¿ f ( x )=2 x3−6 x2+9 fonksiyonuna [0,3 ] aralığında Rolle teoremiuygulayınız .

f∈C[0,3]

f∈D(0,3)

f (0 )=9

f (3 )=54−54+9=9

f ' ( x )=6 x−12 x=0

x=0 x=2

Z∈ [0,3 ]Rolle TeoremiGerçekleştirildi .

14¿ f (x )=lnx fonksiyonunu x=1civarındaTaylor serisine açıp ln 2 sayısını

7 terimiçin yaklaşık hesapl ayınız .

x=2 , a=1

x=∑n=0

∞ (x−a)n . f n(a)n!

f (a )=lna

f ' (a )=1a

f ' ' (a )=−1a2

f ' ' ' (a )= 2a3

f ' v(a)=−6a4

lnx=(x−1). ln 11

+(x−1)1

− ( x−1 )2

2+ ( x−1 )2

3+ ( x−1 )2

4+¿

ln 2=ln 1+1−12+ 13−14….

lnx=∑n=1

∞ (−1)n+1

n

ln 2≅ 1−12+ 13−14+ 15−16+ 17=0,1428571

15¿cos1 sayısını onbindebir hataylabulunuz .

x=1 , a=0(Maclaurin)

x=∑n=0

∞ (x )n . f n(a)n!

f (a )=cos (a )=1

f ' (a )=−sin (a)=0

f ' ' (a )=−cos (a)=−1

f ' ' ' (a )=sin (a)=0

f ıv (a )=cos (a )=1

cosx=1−0− x2

2+0+ x

4

24….

x=∑n=0

∞ (−1)n+2. x2n

(2n )!

cos1≅ 1−12+ 124

− 1120

+ 1720

=0,00138

16¿2x5−3x4+3x3−23 x2+31x−10=0denklemini Horner metodunu

kullanarak yaklaşık hesaplayınız .

|r1 . r2 . r3 . r4 . r5|=|a0an|=|−102 |→∓1 ,∓2 ,∓5∓1,∓2 Kök Adayları :∓1 ,∓ 1

2,∓2 ,∓ 5

2,∓5

( x−1 ) (2 x4−x3+2x2−21 x+10 )=0

( x−1 ) ( x−2 ) (2x3+3 x2+8x−5 )=0

( x−1 ) ( x−2 )(x−12 ) (2 x2+4 x+10 )=0

Diğer Kökler Horner Metoduyla bulunamadığından klasik kök bulma yöntemi kullanılmalıdır.

f ( x )=(2x2+4 x+10)

∆=b2−4 ac=16−80=−64

x1,2=−b∓√∆2a

x1=−4+√−64

4=−1+2 i

x2=−4−√−64

4=−1−2 i

Kökler={1 ,2 , 12 ,(−1+2i),(−1−2 i)}

17¿ (− j−√3 )her biçimde yazın

3.bölge (π+ϑ )

r=√(−1)2+(−√3)2=2 sinϑ= yr=−12ϑ=30

tgϑ= yx= 1

−√3 cosϑ= yr=−√3

2

ϑ=arctan ( −1−√3 )=180+30=210

z=−3− j=2/210=2( cos 210↓+ j .sin 210

↓ )=2e j7π6

c os (180+30 )−12

sin (180+30)−√32

¿−√3− j

18¿ j sayısının 4 kökünedir ?

zn=x+iy=(x+iy)1n

z4=0+ j

z=(0+1 j)14 ϑ=ϑ+2kπ (k∈R)

z ¿(r . e j (ϑ ))14 r=√02+12=1

zk+1 ¿(r . ej (ϑ +2 kπ ))

14 tgϑ=1

0=∞,ϑ=90°

zk+1=r14 .(cjs (ϑ +2kπ4 ))

z1=cjsπ8z2=cjs

5 π8z3=cjs

9π8z4=cjs

13π8z5=cjs

17 π8

=cjs π8

19¿(√3− j)2000=?

r=√(√3)2+(−1)2=2

tgϑ=−1√3→ϑ=arctan(−1√3 )=4.Bölge(2 π−ϑ )

z=2/330=2cjs330=2ej 11π6

(2e j 11 π6 )1453=2e j .1453.11π6 =2ej 5π6

20¿ log (0.001 )−2 ln e0.675−6 log2 (0.5 )+6( 1log 56 )+4. log0.5 (0.25 )− log2 (3 log 100 )=?

0.001=10−3

0.5=2−10.25=4−1=2−2

1log56

=log65→6log65=x→ log6 x=log65→x=5 tir .

¿ log10 (10−3 )−2.(0.675)lne−6 log2 (2−1 )+6( 1log56 )+4. log2−1 (2−2 )− log2 (3 log10 (102 ) )

¿−3−1,35+5+8−log26

¿8,65−log26

21¿minz=5 x+10 y

Kısıtlar

x+ y≥12

x+2 y ≥14

2 x+ y ≥14

x≥0 y ≥0

𝐵 𝑛𝑜𝑘𝑡𝑎𝑠𝚤: Cnoktası :

x+ y=12 x+ y=12

2 x+ y=14 x+2 y=14

x=2 y=10 y=2 x=10

minz=5 x+10 y

A (0,14 )=140

B (2,10 )=110

C (10,2 )=70

D (14,0 )=70 C,D Noktaları min noktalarıdır.