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GI-103B
NANG
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DVL.013442
m Nha
xuat
ban Dai
hoc
Quoc
gia Ha Noi
NGUYEN oCfC CHI
^UJOI
bai tdfL
I AI TICH 12
XAXG CAO
THU VlifJ l\m BINH TH'JAN
NHA XUAT BAN DAI HOC QU6c GIA HA
LCJl
NOI DAU
GIAI
BAI TAP DAI SO 12,
duac
bien
soan
v6i muc dich
giiip
hoc
sinh doi chieu va kiem tra lai cac ket qua khi thtfc hien giai cac bai
tap trong

KHAO
SAT VA VE
D6 THI
CUA HAM SO
§1.
TiNH
DdN
DIEU
CUA HAM SO
ivdl
DUI\
cAi^ I^IOf
1,
Dinh
li: Gia sil hkm so f c6 dao Mm
tren
khoang I.
• Neu f '(x) > 0, Vx e I thi h^m so f dong bien
tren
khoang I
• N§'u f '(x) < 0, Vx e I thi h^m so' f nghich bien
tren
khoang I
• Neu f '(x) = 0, Vx e I thi h^m so f khong d6i
tren
khoang I
Chii
y: Khoang I neu
duoc
thay bkng mot doan

R
Ta c6; y' = Gx^ + 6x
y' = 0 o 6x^ + 6x = 0 o 6x(x + 1) = 0 o
Bang bien
thien:
X
= 0
x = -l
X
—00
-1
0
+0C
y'
+
0 0
+
y
—'—* — , —
Vay ham so' dong bien
tren
moi khoang (-oo; -1) va (0; +oo), nghich bien
tren
khoang (-1; 0)
b) y,= x^ - 2x^ + X + 1
Ham
so' xdc
dinh tren
K
Ta c6: y' = 3x^ - 4x + 1

-73
0
• J3
+00
y'
+
0
II -
0
+
y
-— ' ^
~—-————"
—' ——>
Vay ham so'
dong
bien
tren
m6i khoang
(-00;
- 73 ) va (0; 73 ), nghich
bien
tren
moi khoang (- 73 ; 0) va (73 ;
+00)
2
d) y = X
X
Ham
so \±c

y' = 0 <=> 4x'' - 4x = 0 o 4x(x'' - 1) = 0 o
Bdng bien
thien
x = ±1
Vay ham so
dong
bien
tren
m6i khoang (-1; 0) va (1;
+00),
nghich bien
tren
moi khoang
(-00;
-1) va (0; 1).
Oy = 74-x'
Ham
so' xac
dinh tren
[-2; 2]
Tac.
y-^^
y' = 0 » X = 0
Bang bien
thign
X
-2
0
+2
y'

m6i khoang
(-00;
-2) va (-2;
+00)
-x'-2x + 3
b) y =
X +1
Ham
so xac
dinh tren
M\{-11
(-2x - 2)(x + 1) - (-x' - 2x + 3) -2x' - 2x - 2x - 2 + x' + 2x - 3
y =
(x + lf
(X +
._ -x' - 2x - 5
(x +
• Bang bien
thien
<
0, Vx e R\{-11
X
—00 —
+x
y'
- -
y
•
—
Vay ham so nghich bien

Vay h^m so
dong
bien
tren
R
. (Bdi 4 trang 8 SGK)
Gidi
ftx)
= ax - x^
Ham
so xAc
dinh tren
R
Ta c6: f '(x) = a - Sx^
* N6u a < 0 thi f '(x) < 0, Vx e R. H^m so nghich
bi§'n
trgn
* Ng'u a = 0 thi f '(x) < 0, Vx e R. DAng thiJc chi xay ra khi x = 0
Vay ham so' nghich bien
tren
R.
* Neu a > 0 thi f '(x) = 0 cs- x = ±
j-
V3
Bang bien
thien
x
—00
fa
~ V3

R
f
'(x) = x^ + 2ax + 4, c6 A' = a^ - 4
* Neu a^ - 4 < 0 hay -2 < a < 2 thi f '(x) > 0, Vx e R. H^m so
dong
bien
tren
R
• Neu a = 2 thi f '(x) = (x + 2f >
0,\fx*
-2. Ham so
dong
bien
tren
R.
• Neu a = -2 thi f '(x) = (x - 2f > 0, Vx 2. H^m so
dong
bien
tren
M.
• Neu a > 2
hoac
a < -2 thi f '(x) = 0 c6 hai nghiem xi, X2 (xi < X2)
Bang bien
thien
X -_co Xi X2 +00
f'(x)
f(x)
Ham
so nghich bien

3 3
H^m
so xdc
dinh
va
lien
tuc
tren
R
y' = -4x^ + 12x - 9 = -(2x - 3)^ < 0, Vx € R
H^m
so nghich bien
tren
K
c) y =
x'' - 8x + 9
x-5
Ham
so xac
dinh tren
R \}
(2x - 8)(x - 5) - (x' - 8x + 9) _ 2x' - ISx + 40 - x' + 8x - 9
y =
(x-5)^ (X - 5)^
x' -lOx + 31
(x-5)^
Bang bien
thien
>0,\fx^5
x

+
0
y
Ham
so
dong
bien
tren
khoang (0; 1), nghich bien
tren
khoang (1; 2)
')y = Vx' -2x + 3
Ham
so xac
dinh tren
R (do x^ - 2x + 3 > 0)
, _ 2(x -1) _ x-1
~ 2N/X' - 2X + 3 Vx' - 2x + 3
y' = 0 O X = 1
Bang bien
thien
X
—00
1
+00
y'
0
+
V-—
y

tren
m5i khoang (-00; -1) (-1; +00)
7.
iBdi
7
trang
8 SGK)
f(x)
= cos2x - 2x + 3
Ham
so' xac
dinh
tren
R
Gidi
f
'(x) =
-2sin2x
- 2 =
-2(sin2x
+ 1) < 0, Vx e K
f
'(x) = 0
<=> sin2x
= -1 o 2x = - ^ +
2k7t <:=>x
= -^
+k7r,
ke
Ham

tren
Do
do ham so
dong
bi§'n
tren
Ta
CO fix) >
f(0).
Vx e
:
f '(x) = 1 - cosx > 0, Vx e , 0; -
V
2.
2]
(
Nghia
la X -
sinx
> 0, Vx e 0; -
V
2
j
hay
X >
sinx,
Vx e
0;-,
2)
Mat

cdu a, g'(x) > 0, Vx > 0, do d6 h^m so g
dong big'n
tren
[0; +«),
Va
ta CO g(x) > g(0), Vx > 0
x'
Nghia
la cosx + 1>0,
Vx>0
2
TCr
d6 suy ra
vdi mpi
x < 0, ta c6:
(-x)^
cos{-x) + 1 > 0 hay cosx + 1>0,
Vx<0
2 2
x"
Vay
cosx > 1 - y , Vx 0
c) Xet h^m so h(x) = x - — -
sinx
6
CO
h'(x)
= 1 - — - cosx
Theo
cau b) h'(x) < 0, Vx * 0

tanx
- 2x
lien
tuc
tren
niira
khoang
1
0;
Co f '(x) = cosx +
Do
cos^x +
cos^x
2 > cos X +
cos^x
-
2, Vx 6
>
2, Vx e
cos'x
Do
do ham so fix)
dong bien
tr§n
Ta
c6: f(x) > f(0;, Vx e
Vay sinx
+
tanx
- 2x > 0

+10 660
f(25)
=
=
22
25
+ 5 30
So dan cua
thi
tran
nam 1995 la 22
nghin
ngKbi
b)
Ta c6: f '(x) =
26(t
+ 5) -
(26t
+ 10)
120
(t
+
5)^
(t + 5f
Vay
ham so'
dong bien
tren
[0; +oo)
>

5)^
= « t + 5 »
31
t = 26
(t
+ 5)'
0,125
Vao nam 1996
toc dp
tang dan
so
ciia thi
tra'n
la
0,125.
§2.
CaC TR!
CUA HAM
SO
1.
Dinh nghia:
Gia
suT ham
so' f
xac
dinh tren
tap
c
M
va

*
Xo
difpc gpi la mot diem cxic tieu
cua
ham
so' f
neu:
Ton tai mot khoang
(a; b)
chufa
XQ
sao cho (a; b) e D
va f(x) > f(xo) vdi mpi x e (a; b) \.
Khi
do
f(xo)
dupe
gpi
la gia tri
cij^c tieu ciia ham
so' f
* Diem
cifc
dai
va
diem cUc
tieu
gpi chung
la
diem

ham so f
dat
CLTC
tri
tai diem XQ. Khi d6, neu
f
c6
dao ham
tai
XQ
thi
f
'(xo)
= 0
3. Dieu ki^n
dii de ham so dat
ci^c
tri
Gik sijf
ham
so f
lien
tuc
tren
khoang
(a; b)
chiJa diem xo va
c6 dao ham
tren
cae

> 0
vdi mpi
x e (a; XQ) va f
'(x)
< 0
vdi mpi
x e
(XQ;
b)
thi
ham
so dat
cue
dai tai diem
XQ.
4. Qui
tSc
tim
c\ic tri
* Qui
t^c 1:
• Tim
f
'(x)
• Tim
cac
diem
Xi
(i = 1, 2, )
tai

tai
Xj.
•f'(X)
Xi
f'(x)
f(x)
X,
f(X)
I f(Xi)
cUc
tieu -
* Qui
tac 2
• Tim
f
'(x)
• Tim
cac
nghi^m
Xi
(i = 1, 2, ) cua
phupng
trinh
f
'(x)
= 0
• Tim
f
"(x) va
tinh

dinh tren
K
f
(X)
=
x^
+ 4x + 3;
f
(X)
= 0 o
x^
+ 4x + 3 = 0 o
X
= -1
hoac
x =
Bdng bien
thien
X
1-00 -3 -1
-3
+0C
Vay
ham
so da cho
dat cUc dai tai diem
x = -3,
gia
tri
eUe dai la f(-3)

=
0c:.x^
+ 4x + 3 = 0ox =
-l
hofic
x = -3;
f
"(x)
= 2x + 4
Vi
f
"(-1)
=
2(-l)
+ 4 = 2 > 0 nen
ham
so dat cUc
tieu
tai diem
x
7
f(-l)
=
3
f
"(-3)
= 2(-3) + 4 = -2 < 0 nen
ham
so dat cUe dai
tai diem

3

=
-1;
=
-3,
c) f(x)
= X + —
X
Ham
so xac
dinh tren
f
'(x)
= 1 - =
R
\1
x^-1
2
•
f
'(X)
=
0«x'-l
= Oc:>X = ±l
Bang bien
thign
X
—oc
-1

Ham
so' xac
dinh
va
lien
tuc
tren
R
f-x(x + 2) v(Ji X < 0
|x(x + 2) vdi X > 0 "
• V6i X < 0, f '(x) = -2x - 2; f '(x) = 0 x = -1
• Vdi X > 0, f'(x) = 2x + 2 > 0
Bang bien
thien
Ta c6 f(x) =
X
—00
-1
0
f'(x)
+
0
_
fix)
Vay ham so' dat
cifc
dai tai x = -1, gia tri cUc dai fi-1) = 1 va dat ciic
tieu
tai
x = 0, gia tri

0
+
fix) 32
15
15
Vay ham so dat
ciTc
dai tai x = -1, gia tri cvtc dai fi-1)
28
32
15
va ham so
dat
cUc
tieu
tai x = 1, gia tri
ciTc tieu:
fil) =
15
f)
fix) =
x^ - 3x + 3
x-1
Ham
so' xac
dinh tren
M \|
(2x - 3)(x - 1) - (x' - 3x + 3) x' - 2x
f
'(x) =

12 trang 17 SGK)
Gidi
a) y = x V4-x^
Ham
so xac
dinh
va
lien
tuc
tren
[-2; 2]
y' = ^ ~ v6i moi X e (-2; 2); y' = 0 « x = ± V2
V4-x'
Bang bien
thien
x
-2 -72 72 2
y'
0
+
0
y
0
—*. -2-—-
^2—
~ ^0
Ham
so dat
cifc
tieu

• •
Ham
so dat
cifc
dai tai x = 0, gia tri
cxic
dai: 2 72
c) Ap dung qui tAc 2
y = x - sin2x + 2
Ham
so xac
dinh tren
R
Ta c6: y' = 1 -
2cos2x
y' = 0 <=>
2cos2x
= 1 o
cos2x
= i<=>2x = ±|^ + k27x, k e Z
2 o
ox = ±—
+k7t,
keZ
6
Ngoai ra: y" = 4sin2x
* Vi y"
— + kn
=
4sin

=273
>0
3 2
Do do h^m so dat
cUc
tieu tai cAc diem x = - + ku, k e Z
6
* Vi y"
— + k7t
=
4sin
- + k27I
k6
U
J
Gid
tri ciTc tieu y
—
+ kn
6
- + k27I
=
— + k7i - sin
6
7t ,* 73 -
=
-+k7t
+2
6 2
+

• Vi y"
±^ + k27:
3
=
2cos
±^ + k27r
3
+
4cos
3
0 271
=
2cos
—
. 4rr
+
4cos
— =
3 3
=
-2cos
—
-
4cos
—
= -
3 3
n
^
n

+ cos- = 3 +
3cos^
= ^
3 3 3 3 3 2
13. (BM 13 trang 17 SGK)
f(x) - ax^ + bx^ + cx + d
Giai
flam
so xac dinh tren K
f
'(x) = 3ax^ + 2bx + c
• Ta CO f(0) = 0 d = 0.
Ham so dat cifc tieu tai diem x = 0, nen:
f
(0) = 0
* Ta CO f(l) = 1 => a + b + c + d = 1 r.r> a + b = 1
Ham so dat cUc dai tai diem x = 1 nSn:
f
(1) = 0 =;
=> c = 0
3a + 2b + c = 0
3a + 2b = 0 .
Gidi
he phiTOng
trinh
3a + 2b = 0
a + b = l
a? + 2b = 0
-2a - 2b = -2
3a + 2b = 0

-4a + b + 12 = 0
a+b+c+l=0
Giai
he phUcfng
trinh:
a+b+c+l=0
o ]-4a + b + 12 = 0
3a - 3b - 9 - 0
a+b+c+l=0
o -12 + b + 12 = 0
a = 3
Kiem
tra lai ket qua
f(x) = X-' + 3x^ - 4
f
'(x) = 3x? + 6x, f '(x) = 0 o X =
a+b+c+l=0
-4a + b + 12 = 0
-9a+ 27 = 0
3 + c + 1 = 0
b
= 0
a = 3
c = -4
b
= 0
a = 3
r\\
/ n tt ^ ^ /
Ham so dat ciTc tri tai

(x - m + l)(x - m - 1)
y = ; 75 , X m
(x - m)'
y' = Oox = m- l
hoSc
x = m + 1
Bang
bien
thien
X
—oc
m - 1
m m + 1 +00
y'
+
0
-
0 +
y
^ -m^ + m - 2 \
\
^-m' + m +.2^
Vay vdi moi gia tri ciia m, ham so da cho dat ciTc dai tai
diem
x m - 1 v;
dat ciTc tieu tai
diem
x = m + 1.
§3. GIA TR! L6N
NHAT,

hieu m = minf(x)
•r'^^^itac
tim gia tri Idn nhat, gia tri nho nhat
* Tim cac
diem
x,, x-,, x^
thuoc
(a; b) tai do h^m so f c6 dao ham bfing
0
hoac
khong
c6 dao ham.
, Tinh f(xi), f(x2), fix
J,
f(a), f(b)
* So s^nh cdc gia tri tim
di^cfc
_ So Idn nhat trong cac gia tri do la gia tri Idn nhat ciia f tren [a; b]
_ So nho nhat trong cac gia tri do la gia tri nho nhat ciia f tren [a, b]
J
BAITAP
16. (Bdi 16 trang 22 SGK)
Giai
f(x) =
sin''x
+
cos''x
Ham so xac dinh tren K
* f(x) =
(sin^x

(x) =
0o2x
+ 2 =
0<=>x
= -l
J^ng
bien
thien
[
X
—
X
-2
-1
3
+
x
• f'(x)
0
+
-5
-6 ^
^10
' f(-l) = (-1)' + 2(-l) - 5 = -6
' f(-2) = (-2)- + 2(-2) - 5 = -5
' f(3) = 3^ + 2.3 - 5 = 10
/'ay min f(x) = -6; max f(x) = 10
X6|
2: 3]
xt[~2;

^
-4 '
^^^^
f(-4)
=
3
+
2(-4)' + 3.(-4)
3
«-3)
=
(-3)^
3
+
2.{-Zf
+ 3.(-3)
-
4 = -4
•
«0)
=
-4
fi:-i)
=
3
+
2{-\f + 3(-l) -
4 1^
3
1

-
0
+
f(x)
f(l)
= 1 + i = 2
Vay
min f(x) =2
Ham so khong dat gia tri Idn
nhat
tren (0; +«>)
d)
f(x) = -x' + 2x + 4
Ham so xac dinh tren [2; 4]
f
(X)
= -2x + 2;
f
(x)
= 0<=>-2x + 2 =
0ox=l
Bang bien thien
X
-X
1 2
4
f'(x)
+
0 -
f(x)

2x''
+ 8x
+
6
(x
+
2)' ~ (x
+
2)'
f
(x)
= 0 » 2x^
-i-
8x + 6 = 0 •»
X
= -1
hoac
x = -3
Bang bien thien:
X
—
y:
-3 2 -1
0 '
1
f'(x)
+
0 -
-
0

fix)
^^^^^^^^^^^^^^^^
f(2)
. 2 - i = -
2 2
Vay
max f(x) = —
xul0;2l
2
Ham so khong dat gia tri nho
nhat
tren nOfa khoang (0; 2J
18. (Bai 18 trang 22 SGK) Gidi
a) y = 2sin^x + 2sinx - 1
H^m
so xdc dinh tren
M
Dat t = sinx, -1 < t < 1
y
= fit) = 2t^ + 2t - 1
Tim
gid tri Idn
nhS't
wk gid tri nh6
nhS't
cua ham so y = fit) tren
doan [-1; 11,.do cung la gid tri nhd
nhat
cua ham so da cho trSu
f(t)

Vay min
f
(t)
= - - ; max
f
(t)
= 3
. t€(-l; II 2 II
3
Do d6 miny = —, maxy = 3
"•^^ 2 X€i
b) y =
cos^2x
-
sinxcosx
+ 4
Ham so xac dinh tren M
Ta
CO
y = 1 - sin^2x - - sin2x + 4 = -sin^2x - — sin2x + 5
2 2
Dat t = sin2x, -1 < t < 1
Ta CO y = f(t) =
-t^
- -1 + 5
2
1
1 li
f'(t)
= -2t - - ; f'(t) =

2
+
5
81
16
81
16
Gidt
9.
(Bdi 19 trang 22 SGK)
pat BM = x (0 < X < -)
2
Ta c6: AQMB = APNC
Nen BM = CN = X
Do do MN = a - 2x
QM
= BMtan60" = x x/3
Dien tich hinh chQ" nhat
MNPQ
la:
S(x) = MN.QM = (a - 2x).x N/3 = -2 x/3 x% a 73 x
Bai
toan qui ve tim gid tri Idn nhat cua S(x) tren
Ta c6: S'(x) =-4 ^/3 x + a
V3
S'(x) =
0«-4V3x
+ a^/3
=Oc>x=
^

8
20. (Bai 20 trang 22 SGK) Gidi
Tren moi don vi dien tich ciia mat ho c6 n con ca thi sau mot vu, so' ca tren
moi don vi dien tich mat ho trung binh can nang
f(n) = nP(n) =
480n
- 20n''^ (gam)
Xet ham so f(x) =
480x
- 20x^ x e (0; +x)
'Bien so n e N* dUOc thay bkng bien x e (0; +«) )
f
(x) = -40x + 480
f'(x) = 0 » -40x + 480 = 0 « X = 12
Bang bien thien
X
0 12
+»
_ f'(X)
+
0
b ^^^^
?ssn
^
f(12) =
480.12
-
20.12^
=
2880

33 SGK)
Gidi
a) f(x)
x^ +1
H^m
so xac
dinh tren
K
, x^ + 1 - x.2x -x^ + 1
fx)
= =
fix)
= 0 o -x^ + 1 = 0 o X = ±1
Bang bien
thien
,
X
-x^
-1
1
f X
f'(x)
0
+
0
f(x)
* ~2
1
^ 2
Vay h^m so dat

-00
3
2
I
0
,
f
(x)
0
+
+
0
+
fix)
27 ^
4
H^m
so dat cUc
tieu
tai diem x =
tilu
la
f
f I = ^ ,
-—, gia tri cUc
tieu
la f
I
2
, gid tri

d) f(x) = X + x/x' -1
Ham
so xac
dinh tren
(-oo; -1] u [1; +oo)
X
f
(x) = 1 +
Bang bien
thien
7x^-1
vdi
X < -1
hoac
x > 1
X
-oo -11 +x
f'(x)
-
1"
II
f(x)
- —-—
f(x)
•—•
- —-—
Ham
so' nghich bien
tren
(-oo; -1),

CO
cure
dai va cUc
tilu
khi va chi khi phifang
trinh
(1) c6 hai
nghi§m
phan biet kh^c 1.
f
(x) =
f
(x) = 0 o
\1
Nghia
1^
A'
= 1 - (1 - m) = m > 0
<=> m > 0
m
;^ 0
23. {Bdi 23
trang
23 SGK)
Gidi
X6t
hkm so G(x) =
0,75x^
-
0,025x\ > 0

lieu
lucfng thuoc can
tren
cho benh nhan de huyet ap giam nhieu nhat
la
20 mg. Khi do do giam huyet ap la 100.
24.
(Bai 24 trang 23 SGK)
Gidi
Goi M(x; x^) la diem bat ky cua parabol (.f)
Khoang
each
AM: AM = V(x + 3)^ + (x^ - 0)^
hay AM^ = (x + 3)^ + x^ = x^ + x^ + 6x + 9
Khoang
each
AM dat gia tri nho nhat khi va chi khi f(x) = AM^ dat gia tri
nho nhat
Ta c6: f(x) = x" + x" + 6x + 9
f'(x) = 4x'^ + 2x + 6 = 4x^' + Ay? - 4x=^ + 6x - 4x r '
=
4x2(x
+ 1) - 4x(x + 1) + 6(x + 1) = (x + l)(4r
f
(x)
= 0 o X + 1 = 0 (do 4x^ - 4x + 6 > 0) « X = -1
4x + 6)
X
—00
-1

tieu
hao cua ca khi
wxidi
300 km la
E(v)= cvl
300
v-6
=
300 e.
v-6
(jun),
V > 6
Taeo
E'(v) =
300c.
3v^(v-6)-v^
(v-6)^
,3 1Q.,2
=
300 e.
3v2 - 18v^ - v^
{v-6)^
=
300 c. 2X1^= 600 cv^ ^^-^^
(v-6)^
(v-6)^
E'(v) = 0 <o V = 0
hoac
v = 9 (v = 0 loai do v > 6)
Bang bien

v6i t e [0,25]
CO
f
(t)
= 90t - 3t2 - 3t (30 - t)
a) Toe do truyen benh vao ngay
thuf
5 la:
f'(5) =
3.5(30
- 5) = 375 ngifdi/ngay
b) Xet ham so' xae
dinh
to'e do truyen benh
y ii: fit) rz 90t - 3t^
y' =
f
"(t)
= -6t + 90
y' = 0 « t = 15
Bang bien
thien
t
0
15
25
y'
+
0
y

+
fit)
f'(t)
> 0, Vt e (0; 25)
Vay ham so f dong bien
tren
[0; 25]
27.
(Bdi 27 trang 29 SGK)
Gidi
a) a) fix) = V3-2x
Ham
so' xac
dinh
tren
[-3; IJ
-1
f'(x) = =
<
0. Vdi moi x e
V3-2x
Ham
so' f nghich bien
tren
doan [-3; 1;
3
-3
3
2
f(-3)

+
Vr"4 = -2
ft2)
=
2 + vT^' = 2
Vay
max f(x) =2^,; min f(x) = -2
>'sl-2, 2]
xtl-2;2)
c) fix) =
sin''x
+ cos^x + 2
Ham
so xac
dinh tren
E
f(x)
=
sin^x
+ 1 -
sin^x
+ 2 =
sin''x
-
sin"x
+ 3
Dat
t =
sin''x,
0 < t < 1

k e Z
o 6
71
Tt ^
y^i_
< X <
7i;
f
(x)
- 0 khi v4 chi khi x = - - hoSc x = g
571
hoac X =
Ta
c6 f
71
71
~
-
sin
—
6y
6 V
3;
6
71
3
71
6
71^
^571^

sin27r
= n
57t 73
Vay
max f(x) = -~- + — va
min
f(x) = —
'
' 2
28.
(Bai 28
trang
24 SGK)
Gidi
Gpi
X (cm) la
chieu
d^i
hinh
chOt
nhat
0 < x < 20
Chieu
rong
hinh
chCi
nhat
1^ (20 - x) cm
Dien tich hinh chil nhat
la x(20 - x)cm^

Idn
nhat
la 100 cm^
Vay hinh chiJ nhat
dat gia
tri lorn nhat
khi va chi khi
chieu
dai
hinh chijf nhat
=
chieu
rpng hinh
chOf
nhat
= 10 cm
Hay trong
tat ca c^c
hinh
chuf
nhat, hinh
vuong
c6
canh
tlai
10 cm c6
dien
tich lorn nhat.
§4.
DO TH! CUA HAM SO

theo
thil
tu c6
cung vectot
dan
vi
i , j
vdi hai
true
Ox, Oy.
Goi
M la
diem
bat
ki eiia mat ph^ng
(x;
y) la toa do
diem
M doi
vdi
he
tpa
do Oxy
(X;
Y) la toa do
diem
M doi
v6i
he
tpa

Gpi
y - f(x) la
phuong
trinh
cua do
thi
{'^) do'i v6i he tpa dp Oxy,
khi
do
phJOng
trinh
ciia
do
thi
{'f) doi v6i he tpa dp
IXY
la Y = f(X + x,,) - yo.
Ta
CO
y
y
Y
Y
M
y
y
y.o
A
0
Xo

Phi/otng
trinh
duorng Parabol
doi
vdi
he
tpa
dp
IXY la:
Y
- - =2
8
x
=
y
=
X +
—
4
Y-i
8
(
3^
2
(
3^
X
+ -
-
3

Dinh
I
X,
=1
1
, 7 . Ta CO I
y,
= 1^-1-3
=
^'2
2
2)
Cong thufc chuyen
he
tpa
dp
trong phep
tinh
tien
vectof
01 la
Phucng
trinh
Parabol do'i vdi
he
tpa
dp
IXY \k:
Y-
- = -(X+

y.4-4
1^
16
.
Ta CO I
1_
1_
8'
16
Cong thufc chuyen
hf
tpa
dp
trong phep
tinh
tien
vectcf
01 la
Phuorng
trinh
Parabol
doi
vdi
he
tpa
dp
IXY 1^:
x2
X
=

Dinh
I
X,
= 0
y,
=-5
.
Ta CO
1(0;
-5)
Cong thiJc chuyen
he
tpa
dp
trong phep
tinh
tien
vectof
01 la
Phuong
trinh
Parabol
doi
vdi
he
tpa
dp
IXY la:
Y
- 5 - 2X' - 5

diTcfng
cong
doi vdi h$ toa dp IXY la
Y - 1 = (X + If - 3(X
+1)2+1
hay Y = X^ + 3X2 + 3X + 1 - STX^ + 2X + 1) + 1 + 1
^ Y = X^ + 3X2 + 3X + 1 - 3X2 - 6X - 3 + 2 => Y = X^ - 3X
'
Taco
Y = f(X) =
X^ 3Xxdcdinhtr§nR
VX € R, -X e R
f(-X) = (-X)^ - 3(-X) = -X^ + 3X = -(X^ - 3X) = -f(X)
Vay h^m so Y = X^ - 3X la h^m
so'le,
nen goc toa do I la tam doi xufng
ciia do thi
c)
Phuong
trinh
tiep tuyen (d) vdi CiO tai 1(1; -1) 1^:
y + 1 ^ f'(l) (x-1)
hay y + 1 = -3(x - 1)
Vay (d): y = -3x + 2
Dat g(x) = -3x + 2, ta c6
f(x) - g{x) = x' - 3x2 _^ J _ (_3J^ + 2) = X^ - 3x2 + 3x - 1 = (X -
• Neu x <-1 thi \x) - g(x) < 0 <=> fix) < g(x)
Vay tren
khoang
(-oo;

Tap x^c dinh ham so nay la 7 = R \1
VX G 7, -X e 7,- ft-X) = -1 = = -f(X)
X X
Do d6 h^ni so Y = - — 1^ h^m so 1§ nen do thi Cf)
nhSn
goc toa dp I 1^ tam
X
doi xijfng. f
32. (Bai 32 trang 28 SGK) Gidi
a)
Thifc
hifn
ph6p
tinh
tien
vectcf
OI vdi 1(1; 1)
Cong
thiJc
chuyen
h# toa do trong
ph6p
tinh
tien n^y 1^
X
= X + 1
y = Y + 1
Phi/cfng
trinh
cua do thi doi vdi hp toa dp IXY 1^

tinh
tien n^y \k
Phiforng
trinh
ciia do thi doi vdi he toa dp IXY 1^:
Y + 3 = 3
fx = X-l
y = Y + 3
-^=>Y = -A
X-1+1 X
DatY=f(X)
=
-:J
A
Tap xac dinh cua ham s6' nay la 7 = R \1
Ta CO VX G 7, -X G 7
f(-X) = -| = I
=-f(X)
X X
Vay ham so' Y = - — la ham so' le nen do thi ham so nhan gdc toa dp I la
X
tam doi
xuTng
Do d6 tam doi xilng cua ham s6' da cho la I(-l; 3)
(Bai 33 trang 28 SGK) Giai
Cong
thufc
chuyen
hp tpa dp trong
phep

\1
Vx
e 7, -X 6 (/
«-X) =
a(-X)
+
-X
ax
=
-f(x)
Vay
hkm
so
Y
= aX +
— la
ham so
le
nen nhan goc
toa
do
I la
tarn doi xiiCng.
§5.
DJdNG
TIEM
CAN CUA DO THj HAM SO
1.
Dadng th^ng
y

Du6ng thing
x =
XQ
dUdc
goi la
ducmg tiem
can
diJng (gpi
tit la
ti$m
can
drfng) ciia
do thi ham so y =
f(x)
neu:
limf(x) =+00 hoac limf(x) =+oo hoac
limf(x)
=-co
hoac limf(x) =
-3o
o
y^
= flx)
yi
y
= fix)
II
y
II
0

= fix)
X
Chii
y: De
xdc dinh cac
he so a, b
trong phifong trinh diTdng tiem
can
xien
y
= ax + b, ta
CO
the
sut dung
,.
f(x)
Cong thiTc
a = lim
f(x)
,
b =
lim [f (x)
-
ax]
hoac
a=
lim ,b=
lim[f(x)-ax]
(Khi
a = 0 thi ta c6

so' (khi
x ^
+co
3
va
khi X -co)
*
Vi lim y = +co va lim y = -QO
Nen ducfng thing
x = — la
difcfng tiem
can
dufng ciia
do thi ham so
3
x
^
2^
va
khi X ->
K
3J
b)y
=
-2x-2
x +
3
Ham
so xac
dinh tren K

(-3)^)
c) y = X + 2
X - 3
Hkm so xdc dinh tren R \|
*
Vi lim y = +x va
lim
y = -oo
Nen
dudng th^ng x = 3 1^ ti^m ogn diifng ciia do thi ham so' (khi x ^ 3
va
khi X 3*)
•
Vi Hm [y - (x + 2)] = 0
x-+±x
Nen
dirdng th^ng y = x + 2 Ik tiem can xien cua do thi ham so (khi x ->
+QO
va
khi X -ao)
x' - 3x + 4
d)y =
2x + l
Ham
so xac dinh tren K \j -—
i Vi lim y = -00 vk lim y = +oo
Nen
dxibng
thing x = -— la tifm can diJng ciia do thi hkm so (khi
2

4
1 7
Vay
difdng thang y = - x - - la ti^m can xien ciia do thi ham so (khi
x -> -00 va khi x +oo)
x + 2
e)y=^
Hkm s6 xkc dinh tren R \; 1|
*
Vi lim y = +oo va lim y = -x
x-»(-l)-
x (-l)*
Nen
dudng thing x = -1 la tiem can dufng ctia do thi ham so (khi
X (-1)- va khi x -> (-If)
*
Vi lim y = -oo va lim y = +oc
x->
I
x-> I"
Nen
difdng thing x = 1 la tiem can dufng ciia do thi ham so' (khi x ^ 1
va
khi X 1*)
*
Vi lim y = 0
Nen
dircmg thing y = 0 Ik tiem
CSLQ
ngang ciia do thi hkm so (khi x +« va

vk lim y = -co
x »"'
x->0*
Nen
Avcdng thing y = 0 Ik ti^m c^n dufng ciia do thi hkm s6' (khi x -> 0"
vk
khi x 0^)
*Vi
lim[y-(x-3)l= lim^^ = 0
Nen
dufmg thing y = x - 3 Ik ti^m c$n xien cua d6 thi hkm so (khi x -> +oc
vk
khi X -oo)
x' + 2
b)y =
x'-2x
Hkm
so xac dinh tren R \; 21
•
Vi lim y = -oo vk
lim
y = +oo
x->(r x-»o*
N§n
diibng
thing x = 0 Ik ti^m ckn dufng ciia do thi hkm so (khi x -> 0"
vk
khi X -> 0*)
•
Vi lim y = -oo vk lim y = +oo

=
lim-
*±=t X - 2x
=
2
c)y =
Nen
diiejng thing y = x + 2 la ti|m can xien cua do thi hkm so (khi x +»
va
khi X -» -oo)
x^ + X +1 •
X^-l
Hkm
so xac dinh tren R \; 11
*
Vi lim y =
+00
vk
lira
y =
-00
Nen
diXbng
thing
x = -llk
ti$m
cgn
diifng
cua d6
thi

*
Tiem
can
xien
c6
dang
y =: ax + b, ta c6
o
,.
y ,. x^ + x +1
a
=
lim
—
=
lim —5
= 1
x^.±« X
"-t* X -
X
b= lim[y-x]
= lim ^
1^
^
- X
1- 2x + l ^
=
lim = 0
x'
-1

5
Nen difdng thing
y = - ^ Ik
tiem
can
ngang
cua do
thi hkm
so
(khi
x ^
+co
o
vk
khi
X ->
-00)
*
Vi lim y =
+00
vk lim y =
-00
x->(-l)" x->(-l)'
N§n diremg
thSng
x = -1 Ik
tiem
can
dutng
cua do

5
vk
khi
X
-)•
—)
5
36.
(Bai 36,
trang
36 SKG)
a)y
=
Vx^-1
Gidi
Hkm
so xac
dinh tren
[-00;
-1] u
[1;
+x>).
- Ti$m
can
xien
c6
dang
y = ax + b
=
limjl-4

y
*
a =
lim—
=
lim
= lim
X
"^-^ X
-X,
1-
=
lim-Jl-^
= -1
*
b =
lini[y
+ x] =
lim(\/x^
-1 + x] =
lim-
-1
=
0
«7x^
-1 -X
V$y difdng thing
y = -x Ik
ti$m
can

lim
X-»+^X
X-»
+
X
=
lim
x-*+x
2
+
xJl-
= lim 2
+
=
3
b
=
lim[y
-
3x]
=
lim[2x
+
N/X'
- 1 - 3x]
=
lim[Vx^
- 1 -
x]
= li

2 +
lim—^-
— = 2-1 = 1
*
b =
lim[y
- x] =
lim[x
+
Vx'
-1] =
lim
^= = 0
''^-"x-Vx'-l
Vay dudng thing
y = x la
ti|m
can
xien
cua do
thi hkm
so
(khi
x ->
-00).
c)
y = X +
vx^n
Hkm
so xkc

0
"-"'Vx^+l
+ x
Vsly dudng thing
y = 2x Ik
ti^m
can
xien
ciia
do
thi hkm
so
(khi
x
+<»)•
Vi
limy
=
lim(x
+
Vx^
+1) =
lim-
-1
=
0
"^^x-Vx^+l
Vky dudng thing
y = 0 Ik
ti^m

1)
*
b =
lim[y
- x] =
lim(Vx^
+ x + l - x) = lim . ^ + ^
''"^Vx^x
+ l + x
=
lim-
X
ill,
1
+ - + -^ +1
X
X
1
2
V^y
diremg
thing
y = x + ^
lli ti^m c$n
xi§n
cua
d6
thi hkm
so
(khi

x^ x^
= lim-
1 1 1 1
X
X
J.
2
Vay dudng thing
y = -x - i Ik
tigm
can xien cua
do thi
hkm
so
(khi
x ->
-oc).
37.
(Bai 37,
trang
36 SGK)
Gidi
a)
y = X +
Vx^
-1 -
Hkm
so xkc
dinh tren (-oo;
-1] u

x ->
+oo).
* limy
=
lim(x
+
Vx^
-1) = lim } =0
_x + Vx'-l
Dudng thing
y = 0 Ik
tif
m
can
ngang
ciia
do thi
hkm
so
(khi
x ->
-co).
b)
y = Vx'
-4X-I-3
Hkm
so xkc
dinh tren (-oo;
1] u [3; +-0).
4 3_

x"
X
xUl-^
+ 4+1
X
x^
=
lim
x-»+»|
1 4 3 ,
X
X
= -2
Dudng thing
y = x - 2 la
tiem
can
xien
cua d6 thi hkm so
(khi
x
- -> +00).
,.
y ,.
VX^-4X
+ 3 "^^1^
*
a =
lim-
= lim = lim

X
'
A 3)
= 2
l-^
+ A-l
X
X
Bvtbng thing
y = -x + 2 Ik
tiem
can
xien
cua do thi
hkm
so
(khi
x ->
-00).
c)
y =
Vx^n
Hkm
so xkc
dinh tren
K.
=
lim
-^-lim,
1 + 4- = 1

(khi
x +«).
*
a =
limi
-
lim^^^^^—-
=
lim-
^
= lim
x-*-xx
x-»-x
-1
b
=
lim(y
+ x) =
lim(Vx^
+4 + x) = lim ^ -
x->-«.
x->-x
""'""vx
+
4 —;
Diforng
thSng
y = -x la tiem can xien ciia do thi h^m so' (khi x -oo).
d)y =
X

a) Ham so xac dinh tren E \.
* Vi limy =
-oo
va limy =
+oo
Nen
dudng
thang x = 3 la tiem can
dufng
cua do thi ham so (khi x -> 3
vk khi x ^ 3^).
5
Ta CO y = x + 1 +
x-3
lim[y - (x + 1)] = lim = 0
x^«x-3
Nen
dudng
thing y = x + 1 la tiem can xien cua do thi ham so (khi x +<x
va khi X -> -oo).
b) Toa do
giao
diem
I ciia hai tiem can la
nghiem
cua he
phu'ong
trinh:
y = X + 1 fy = 4
X

VX G V, -X e 'J'
(-X)' +5 X' + 5
f(-X) =
-X
X
=
-f
(X)
X^ + 5
Vay ham so' Y - la ham so' le nen do thi nhan go'c 1(3; 4) lam tam
X
doi xiifng.
39. (Bai 39, trang 36 SGK). Gidi
+X-4
Hkm so xac dinh tren M \.
* Vi lim y = +C0 v^ lim y = -oo
x->(-2)- x->(-2)*
Nfen diffmg thing x = 2 1^
ti§m
can
dufng
ciia do thi ham so (khi x (-2)
v^ khi x ^
(-2)").
2
Ta CO y = X - 1 -
x + 2
Vi
lim[y - (x - 1)] = lim
x + 2

(X-2)'
+(X-2)-4
X-2 + 2
^ Y =
X'-4X + 4 + X- 2- 4
X
X' -3X-2 + 3X _
+ 3
X'-2
X ' * X
Dat Y = f(X) = ^—^, tap xac dinh = R \: VX € V, -X e V.
f(X)
=
k^ljzl
= =
_f
(X)
Vay ham
so'Y=
~^
-X
X
le nen nh$n goc tpa dp I(-2; -3) la tam doi xufng.
c) V = - 8x + 19
x-5
Ham so xac dinh tren R \.
r-s^
—
Vi
limy = -oo vk limy =

thu-
chuyen
he toa do
theo
ph6p
tinh
tien 01 F = + ^
ly
= Y + 2
c)
Phirong
trinh
m doi vdi h? toa dO IXY Ik:
Y
+ 2 =
(X +
5)'-8(X
+ 5) + 19
x + 5-5
^ Y = + IPX + 25 - 8X - 40 4 19 _ X' + 2X + 4 - 2X _ X^ + 4
X X ~ X
X^ + 4
Dat Y = fiX) = ^ , tap xac dinh 7 = K \: Vx € '/, -X e 7.
—A A
X^ + 4
Vay hkm so Y = ^ Ik hkm s6 1^ nen nhan g6c toa do 1(5; 2) lam tkm
doi xufng.
§6.
KHAO
SAT SI/

dien
ket qua vko
bang.
• BvCdc 3: Ve do thi cua ham so.
* Ve ckc
dufcfng
ti|m can cua do thi (neu c6).
* Xac dinh mpt so
diem
dac bi#t ciia do thi.
(giao
diem
vdi cdc true toa
dp, neu
phep
toan
khong
philc
tap).
* Nhan xet ve do thi (chi ra true vk tam doi xijfng cua do thi (neu c6)).
2. Ham so y = ax' + bx^ + cx + d (a 0).
*
Dang
do thi
M
1
/ ^
\
•y
V

Dang
1
Dang
2
b)
1 BAITAP
40. (Bai 40, trang 43 SGK) Gidi
a) Khdo sat sU bii'n
thien
va ve do thi ham s6 y = + 3x^ - 4.
1-
Hkm so xac dinh tren R.
2- Su
bien
thien cua hkm so:
a) Gidi han tai v6
ciTc
limy = -00 va limy =
+00
b) Suf
bi§'n
thien,
ciTc
tri
Ta c6: y' = Sx^ + 6x = 3x(x + 2)
y' = 0 o 3x(x + 2) =
0c:>x
= 0
hokc
x = -2.

^ 0
-4-—
»+oo
3. Do thi
• Diem uon:
Ta c6 y" = 6x + 6, y" = 0 o x = -1.
y" = 0 tai X = -1 va y" doi dau tCf am sang
diTcfng
khi x qua x = -1. Vay diem U(-l; -2)
la
diem uon
ciia
do thi.
•
Giao
diem ciia do thi vdi true tung la (0; -4).
• y =
0c=>x^
+ 3x^-4 = 0o(x- l)(x^ + 4x + 4
o (x - l)(x + 2r = 0 o
-2 -1
ji
=
0
-2/
-4
••x
x = 1
x = -2
D6

f(-X) = (-X)' - 3(-X) = -X^* + 3X = -(X^ - 3X) = -f(X).
Vay ham so Y = X^ - 3X la ham le nen nhan goc toa do U(-l; -2) \km
tam
doi xufng.
Do do U(-l; -2) la tam doi xufng ciia do thi.
41. (Bai 41, trang 44 SGK)
Gidi
a)
Khdo sat sU bien thien va ve y = -x^ + 3x^ - 1.
1.
Ham so' xac
dinh
tren
M.
2. Sii bien
thien
ciia ham so',
a)
Gidi
han tai v6 cUc.
• lim = +CO vk lim = -00
b) SU bien
thien,
cUc tri
• Ta CO y' = -Sx^ + 6x = -3x(x - 2)
y' = 0 <=> -3x(x -2) =
0<=>x
= 0
hoac
x = 2.

y
+00
__-*3-—
r
^ —00
3
y = m
1-
-••••ju
i
/I 1
J.1 2.
3
3. Do thi
• Diem uon:
Ta CO y" = -6x + 6, y" = 0 o x = 1.
y" = 0 tai x = 1 va y" doi dau tiT di/Ong
sang
am khi x qua x = 1. Vay diem U(l; 1)
la
diem uo'n cua do thi.
•
Giao
diem cua do thi va true tung la (0; -1).
Nhdn
xet: Do thi ham so nhan diem U(l; 1)
la
tam doi xufng.
b)
Bien

Gidi
a)
Khdo sat sil bien thien va ve do thi y = ^x^ - x^ - 3x - - .
3 3
1-
Ham so xac
dinh
tren
K.
2- Su bien
thien
ciia ham so'.
^)
Gidri
han tai v6
cifc:
•
limy
= -00 •
limy
= +00
b) Su bien
thien,
cUc
tri:
• Ta c6 y' = x^ - 2x - 3
y' = 0 c:> X = -1
hoac
x = 3
Ham