34
35
13–14
FUNKSIYA GRAFIGIGA O‘TKAZILGAN
URINMA VA NORMAL TENGLAMALARI
Urinma tenglamasi.
y = f
(
x
) funksiyaga grafigining (
x
0
;
f
(
x
0
))
nuqtasidan
o‘tuvchi urinma tenglamasini topamiz (19-rasm). Urinma to‘g‘ri chiziq
bo‘lgani uchun uning umumiy ko‘rinshi
y = k x
+
b
bo‘ladi. Hosila ning
geometrik ma’nosiga ko‘ra
k
=tg
a
=
f
′
(
x
0
), ya’ni urinma teglamasi
y=f
′
(
x
0
)
x+b
ko‘rinishini oladi. Bu urinma (
x
0
;
f
(
x
0
)) nuqtadan o‘tgani
uchun
f
(
x
0
)=
f
′
(
x
0
)
x
0
+ b
bo‘ladi, bundan
b = f
(
x
0
)
– f
′
(
x
0
)
x
0
. Topilgan
b
ni urinma tenglamasiga qo‘yib,
y=f
′
(
x
0
)
x+f
(
x
0
)
–f
′
(
x
0
)
x
0
yoki
y – f
(
x
0
)
= f
′
(
x
0
)(
x – x
0
) (1)
tenglamani hosil qilamiz.
y – f
(
x
0
)
= f
′
(
x
0
)(
x – x
0
) tenglama (
x
0
;
f
(
x
0
))
nuqtada
y = f
(
x
) funksiyaga
o‘tkazilgan urinma tenglamasi bo‘ladi.
19-rasm.
1-misol.
f
(
x
)
= x
2
–
5
x
funksiya grafigiga
x
0
=
2 abssissali nuqtada o‘tka-
zilgan urinma tenglamasini yozing.
Avval funksiyaning va funksiyadan olingan hosilaning
x
0
=
2 nuqtadagi
qiymatini topamiz:
f
(
x
0
) =
f
(2) = 2
2
– 5·2 = –6,
f
ʹ(
x
) = 2
x
– 5,
f
ʹ(2) = 2·2 – 5 = –1.
Topilganlarni (1) tenglamaga qo‘yib, urinma tenglamasini hosil qilamiz:
( )
(
)
6
1
2 yoki 4.
y
x
y
x
− − = − ⋅ −
= − −
Javob:
y =
–
x
– 4.
▲
34
35
2-misol.
f
(
x
)
=x
3
–
2
x
2
funksiya grafigiga
x
0
=
1
abssissali nuqtada
o‘tkazilgan urinma tenglamasini yozing.
Avval funksiyaning va funksiyadan olingan hosilaning
x
0
=
1 nuqtadagi
qiymatini topamiz:
f
(
x
0
)=
f
(1)=1
3
–2·1
2
=–1,
f
ʹ(
x
)=3
x
2
–4
x
,
f
ʹ(1)=3·1
2
–4·1= –1.
Topilganlarni (1) tenglamaga qo‘yib, urinma tenglamasini hosil
qilamiz:
y
–(–1) = – 1 (
x
– 1 ) yoki
y = –x.
Javob:
y = –x.
▲
Agar
y=f
(
x
) funksi
ya grafigining
x
0
abssissali nuqtasida o‘tkazilgan
urinma
y=kx+b
to‘g‘ri chiziqqa parallel bo‘lsa,
f
'(
x
0
)
= k
bo‘ladi. Bu
shart orqali funksiyaning berilgan to‘g‘ri chiziqqa parallel bo‘lgan
urinmasi
topiladi.
3-misol.
f
(
x
)
= x
2
–
3
x +
4
funksiya uchun
y =
2
x –
1 to‘g‘ri chiziqqa pa-
ral lel bo‘lgan urinma tenglamasini yozing.
Urinmaning berilgan to‘g‘ri chiziqqa parallellik shartiga ko‘ra,
f
′(
x
0
)
=
2 yoki 2
x
0
–
3=2 tengla
mani hosil qilamiz. Bu tenglamada
x
0
=
2,5 bo‘lgani uchun urinma abssissasi
x
0
=
2,5 bo‘lgan
nuqtadan
o‘tadi. Hisoblashlarni bajaramiz:
f
(
x
0
)
= f
( 2,5)
=
2,5
2
– 3·2,5 + 4 = 6,25 – 7,5 + 4 = 2,75
f
′(
x
0
)
= f
(2,5)
=
2.
Endi urinma tenglamasini topamiz:
y –
2,75=2(
x–
2,5) yoki
y=
2
x–
2,25.
Javob:
y =
2
x
– 2,25.
▲
4-misol.
f
(
x
)
=x
3
– 2
x
2
+ 3
x –
2 funksiya grafigiga
x
0
=
4 abssissali nuqtada
o‘tkazilgan urinma tenglamasini tuzing va urinma bilan
Ox
o‘qining musbat
yo‘nalishi tashkil qilgan burchakning sinusini toping.
Avval funksiyaning va funksiyadan olingan hosilaning
x
0
=
4 nuqtadagi
qiymatini topamiz:
f
(
x
0
)=
f
(4)=3·4
3
–2·4
2
+3·4–2=170,
f
ʹ(
x
)=3
x
2
–4
x
+3,
f
ʹ(4)=3·4
2
–4·4+3= 35.
Topilganlarni (1) tenglamaga qo‘yib, urinma tenglamasini hosil qilamiz:
y –
170 = 35(
x –
4) yoki
y =
35
x
+ 30.
Hosilaning geometrik ma’nosiga ko‘ra tgα=35, bundan
36
37
2
2
2
2
1
1
tg
35
35
sin
.
1
1226
1 ctg
1
1 35
1
tg
tg
α
α =
=
=
=
=
+
α
+
α
+
+
α
tg
tg
.
Javob:
y=
35
x
+30;
35
sin
.
1226
α =
▲
5*-misol.
f
(
x
)
=x
2
parabolaga abssissasi
x
0
bo‘lgan
A
nuqtada o‘tka-
zilgan urinma
Ox
o‘qini
1
2
0
x
nuqtada kesib o‘tadi. Shu da’voni isbotlang.
f
ʹ(
x
) = 2
x
,
f
(
x
0
)=
x
0
2
,
f
ʹ(
x
0
) = 2
x
0
.
Urinma tenglamasi (1) ga ko‘ra
y
= 2
x
0
⋅
x
–
x
0
2
bo‘ladi.
Uning
Ox
o‘qi
bilan kesisish nuqtasi
x
0
2
0
;
ekani ravshan. Bundan
y
=
x
2
parabolaga
abssissasi
x
0
bo‘lgan
A
nuqtada o‘tkazilgan urinmani yasash usuli kelib
chiqadi:
A
nuqta va
x
0
2
0
;
nuqta orqali o‘tuvchi to‘g‘ri chiziq
y=x
2
parabolaga
A
nuqtada urinadi.
Normal tenglamasi.
y=f
(
x
) funksiya grafigiga
x
=
x
0
abssissali nuqtada
o‘tkazilgan urinmaga
x
=
x
0
nuqtada perpendikular bo‘lgan
y
–
f
(
x
0
)
0
0
1
( )
(
)
( )
y f x
x x
f x
−
= −
−
′
(2)
to‘g‘ri
chiziqqa
y = f
(
x
) funksiya grafigining
x
0
abssissali nuqtasida
o‘tkazilgan normal deyiladi (19- rasm).
6-misol.
f
(
x
) =
x
5
funksiya grafigiga
x
0
= 1 abssissali nuqtada o‘tkazilgan
normal tenglamasini tuzing.
Hosila formulasiga ko‘ra
f
ʹ(
x
) = 5
x
4
bo‘ladi. Funksiya va uning
hosilasining
x
0
=1 nuqtadagi qiymatlarini hisoblaymiz:
f
(1)=1
5
=1 va
f
′(1) = 5·1
4
= 5. Bu qiymatlarni normalning tenglamasiga qo‘yamiz va
1
1
( 1)
5
y
x
− = −
−
yoki
1
6
5
5
y
x
= −
+
tenglamani hosil qilamiz.
Javob:
1
6
5
5
y
x
= −
+
.
▲
36
37
Eslatma:
f
(
x
)=
x
5
funksiya grafigiga
x
0
=1 abssissali nuqtada
o‘tkazilgan urinma tenglamasi
y
=5
x
–4 bo‘ladi (isbotlang!). Urinma
va normalning burchak koeffitsiyenti ko‘paytmasi 5·(
1
1
( 1)
5
y
x
− = −
−
)= –1
ekaniga
e’tibor bering.
?
Savol va topshiriqlar
1.
y
=
f
(
x
) funksiya grafigiga
x
0
abssissali nuqtada o‘tkazilgan urinma
tenglamasini yozing.
2.
y
=
f
(
x
) funksiya grafigiga
x
0
abssissali nuqtada o‘tkazilgan normal
tenglamasini yozing.
3. Berilgan funksiyaning biror to‘g‘ri chiziqqa parallel bo‘lgan
urinmasi qanday topiladi? Misolda tushuntiring.
Mashqlar
45.
Funksiya grafigiga abssissasi
x
0
=
1;
x
0
=–2;
x
0
=0 bo‘lgan nuq-
tada o‘tkazilgan urinma tenglamasini yozing:
1)
f
(
x
)
=
2
x
2
–
5
x
+1;
2)
f
(
x
)
=
3
x–
4;
3)
f
(
x
)
=
6;
4)
f
(
x
)
=x
3
–
4
x
;
5)
f
(
x
)
=e
x
;
6)
f
(
x
)=2
x
;
7)
f
(
x
)
=
2
x
+
ln2;
8)
f
(
x
)
=
sin
x
; 9)
f
(
x
)
=
cos
x
;
10)
f
(
x
)
=
cos
x–
sin
x
; 11)
f
(
x
)
=e
x
x
;
12)
f
(
x
)
=x
·sin
x.
46.
Funksiya uchun
y=
7
x–
1 to‘g‘ri chiziqqa parallel bo‘lgan urinma
tenglamasini yozing:
1)
f
(
x
)
=x
3
–
2
x
2
+
6;
2)
f
(
x
)
=
4
x
2
–
5
x+
3; 3)
f
(
x
)
=
8
x–
4
.
47.
Berilgan
f
(
x
) va
g
(
x
) funksiyalarning urinmalari parallel bo‘ladigan
nuqtalarni toping:
1)
( )
( )
2
3
5
4,
4
5;
f x
x
x
g x
x
=
−
+
=
−
( )
( )
2
3
5
4,
4
5;
f x
x
x
g x
x
=
−
+
=
−
2)
( )
( )
8
9,
5
8;
f x
x
g x
x
=
+
= − +
( )
( )
8
9,
5
8;
f x
x
g x
x
=
+
= − +
3)
( )
( )
7 11,
7 9;
f x
x
g x
x
=
+
=
−
( )
( )
7 11,
7 9;
f x
x
g x
x
=
+
=
−
4)
( )
( )
3
2
,
;
f x
x g x
x
=
=
–8,
( )
( )
3
2
,
;
f x
x g x
x
=
=
+5;
5)
( )
( )
3
2
,
5
7;
f x
x
x g x
x
=
+
=
−
( )
( )
3
2
,
5
7;
f x
x
x g x
x
=
+
=
−
6)
( )
( )
4
3
,
;
f x
x g x
x
=
=
+11,
( )
( )
4
3
,
;
f x
x g x
x
=
=
+10.