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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.
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