Các bài toán về giá trị lớn nhất và giá trị nhỏ nhất

512
C101B
NGUYEN
THAI HOE
CAC
BAI
TOAN
TRI
LON
NHA^
CD
NHA
XUAT BAN GIAO DUG VIET NAM
NGUYEN THAI HOE
ve
CAC
BAI
TOAN
GlA TRILON NHAT
VA GlA
TRI NHO NHAT
NHA
XUAT BAN GIAO DUG VIET NAM
Nhitng ki hi$u va each viet gon dx^gfc dung trong sach
:
- MXD
:
Mien xac dinh
- GTLN
:
Gia tri

Idrn
nhat
- GTNN
:
Gia tri nho nhat
-
:
Gia tri eye dai
-
•
tri
etie
tieu
- Vx
:
Vdri
moi
x
- Dau
xay ra
:
DSng
thuTe xay ra
f
V
Cong
ty Co ph^n D^u ta va Phat
trig'n
Giao dgc Phirong Nam -
Nha xucit bin Giao due

Viet
Nam giO quyin eong bo tac pham.
Ma so':
8I598P9-DTN
05-2009/CXB/87-2170/GD
PHAN
MOT
GIAI
CAC BAI TOAN
GTLN,
GTNN
BANG
PHUONG PHAP CAO CAP (dung dao hdm)
§1. PHifdNG PHAP DUNG BJ^O HAM DE
GIAI
BAI TOAN
I.
CAC
DINH
NGHIA
(,'ho ham y = fix) c6 mien xac dinh X.
- So M
dtroc
goi la GTLN cua ham f(x) trong mien X va viet la max fix)
neu
* vdi moi X e X thi f(x) < M va
* ton tai so XQ e X ,sao cho
fCxg)
= M .
- So m

diroc
goi la GTNN cua ham so fix) trong mien X va viet la
min
fix) neu
* vdi moi x e X thi fix) > m va
* ton tai so XQ e X sao cho
fCxg)
= m.
Cdc luu y khi sit dung cdc dinh nghia
- Phai phan biet ro rang hai khai niem :
Cifc
dai,
circ
tieu va GTLN, GTNN cua ham so.
Su phan biet do du'Oc mo ta mot
each
hinh hoc cho bori hinh ve sau day
la
do thi ciia ham so fix) tren [a ; b
|
Ta CO
- Gia tri bien : fla), fib)
- Cac gia tri eye dai : f(x2), fix^)
- Cac gia tri cUe tieu : fCx^), f(x3), f(xg).
5
max f(x)
= f(b)
min f(x) =
fcTi^u
= ^x^)

Luu
y :
-
Tai
diem
c6
hoanh
do
X3,
ham
khong
c6 dao ham
nhtftig
do vin la
diem
CLTC
tieu.
-
Tai
diem
c6
hoanh
do X4 thi
f'(x4)
= 0
nhifng khong
la
diem
CLTC
tri.

- Khong
the bo
quen
dieu kien
thuf
hai trong dinh nghia
: ton tai
XQ
sao
cho
f(xo) = M hoSc f(xo) = m .
Vi
nhii
ta
biet
:
sinx
< 15 la bat
dSng
thufc diing vo'i
moi x e R
nhimg khong
the ket
luan
max sinx
= 15 vi
trong
R
khong
c6

XQ
de
sin
XQ
= 15.
TCr dinh nghia
va cac
liAi
y da neu ta c6
II.
PHtfOfNG
PHAP
CHUNG
DE TIM
GTLN
vA
GTNN
CUA
HAM
SO
Cho
ham so y =
fix).
Buac
1. Tim MXD
ciia
ham so (neu bai
toan khong
cho
trifdc)

va goi
mien
xac
dinh
la X.
Buac
2.
Tim trong
X cac
diem
ma
tai
do
thi
'f'(x)
= o , ^ :
f'(x) khong
ton tai
Goi
tat ca cac
diem
do la
x^,
X2, ,
x^^
Buac
3. • ^1
-
Tinh
gia

tri
cua fix)
tai
cac
bien
cua
fix)
(neu c6) . ;., .
6
r
-
Max f(x) = so 16'n
nhat trong
-
min fix) = so nho
nhat trong
•
- Tinh
gia tri cua fix) tai moi
diem
thuoc
mien
xac
dinh
ma f'(x) = 0
hoac
f'(x)
khong
xac
dinh

-
So
sanh
cac gia
tri bien
va moi
f(xi),
i =
l,2, ,k
Buac
4.
Ket luan
bai
toan.
cac gia
tri bien
va
c^c f(xi) (i =
l,2, ,k)'
cac gia
tri bien
va
cac
f(Xi),(i
=
l,2, ,k)"
Lilu
y :
Neu
bai

toan
chi doi hoi tim
GTLN
va
GTNN ciia
ham so thi
khong
nen
tim
gia
tri
ctfc
tri
de suy ra
GTLN
va
GTNN
cua ham so.
III.
CAC BAI
TOAN MINH
HOA
Bai
loan
1.
Tim
GTLN
va
GTNN
cua y -

2sinx sin"^
x
trong
[0; n].
Ldl
GlAl
isin'^
=
S'cos
X.
COS
2x
9 2
Cdch i.
Ta
CO
y' = 2 cos x - 4
sin
x cos x = 2 cos
x(l
- 2
sin
x)
y'
= 0o
cosx
= 0
cos 2x = 0
(k,
I e Z)

Trong
doan
[0;
TI]
, y = 0
khi
x = -, x
=^
-, x = —
4 2 4
X
= — +
krt
2
x = - + l-
4 2
Tinh
y(0), yin), y
Ta
CO
y(0)
- y{Tt) = 0
f
—
.
y
,
y
[
4 >

V
2.
=
y
371
I
4
(2
goc bu
nhau
c6 sin
bang
nhau)
^7:^
2
7
—
=
y
V4; [
4 J
Ta
ket luan dugfc : max y
=
y
min
y
=
y(0)
=

y(7r)
=
0
Cdch
2. Dat t = sinx
Wdi
moix e [0;
TI]
thi t e [0 ; 1]
Ta
CO
y(t)
= 2t t^.
3
Tim GTLN
va
GTNN
cua y(t)
trong
[0, 1] :
Ta
CO y'(t)
=
2 - 4t2 - 2(1 - 2t'^)
^/2
272
y'(t)
=
0 t
=

±
Trong
doan [0 ; 1] , y'(t) = 0 chi khi t =
72
Tinh
y(0), y
V2
v2,
va yd) :
273
Tac6y(0)
= 0, y(l)
=
|, y
.2;
Va
ket luan :
273
3
max y
=
, dat tai moi xma t
=
sinx
=
— x
=
-
72
2 "4

miny
=
0, dat tai moi x ma t
=
sinx
=
0=>x
=
0, x
=
7i
Bai
toan 2.
Tim GTLN
va
GTNN
cua ham so
1
9 Q
y =
—
In
X
trong
[1; e ]
x
Ta
CO y' =
x.21nx ln^x
.

X
^
Ldl
GIAI
In
x(2 -
In
x)
=>
y' = 0 o
X
"In
X = 0
lnx =
2
X
x-1
x-e^
8
Tinh
yd), y(e^), y(e^)
Taco
yd) = 0, y(e^) =
4 l^^^^
"4
y(e^) =
4ln2e^=4
Ta
ket
luan

: max y = yd) = 0
mmy
Bai
toan
3.
Tim
GTLN
va
GTNN
cua ham so
x + 1
y
=
trong
[-1 ; 2]
Ldl GIAI
Ta
CO y' =
Vx^+l-(x + l) J- •2x
2Vx2
+1
1
- X
x2+l
(x2+1)3^2
y' = 0 o X = 1.
Tinh
y(-l),
yd), y(2) va y(-l) = 0, y(l) = V2 , y(2) =
Ket luan

:
min
y = y(-l) = 0
max
y = yd) = \/2
Bai
toan
4.
Tim
GTLN
va
GTNN
cua ham so y = if^ + ^1-x
Ldl GIAI
MXD
= [0 ; 1]
=
4x 4
+i-(l-x)
4 4
^/5
f
n
t
1"
Ta
CO y' =
x4
+
a-x)4

4
ila-x)Kx^
Giai bat phircfng
trinh
y' > 0 de xac
dinh
dau y'
Ta dtro-c y' > 0 <^ ijil - >
(1 - x)^ > x^
<=> 1 - X > X
C:> 0 < X < -
2
Ta CO dau y' va bien thien ciia y cho hdi
bang
:
0
Ta CO ket luan
max y = y
'1'
[2)
-4/
miny
= y(0) = yd) = 1
(Lau y : - Vi ham y c6 dang phufc tap nen ta dung
bang
bien thien de
hieu ro hcfn ve bien thien cua ham so do, tat nhien khong dung
bang
ta van
ket luan dirgrc bai toan.

- Chuyen cac ham
chufa
cSn thufc ve ham luy thCra de lay dao ham thi
dofn
gian hcfn) r ,
Bai
toan
5.
Tim
GTLN va GTNN cua ham so
y
= X +
10
Ldl
GIAI
Cdch 1. MXD = [-2 ; 2]
1
Ta CO y' = 1 +
-
(-2x)=
=> y' = 0 o yji-x^ = x <=>
V4-x^ -X
X
> 0
4-x2=x2
<=> X
= V2
Tinhy(-2),
y(V2),y(2).
Ta CO y(-2) =-2; y(2) = 2 ; y (N/2) = 2V2

Vay : maxy = y(V2) = 2N/2
min
y = y(-2) = -2 .
Cdc/i 2. Ap dung bat dang thufc Bunhia copxki cho 4 so : x, 1, ^4 - x^ , 1
ta
duoc
x^
+(4-x2)
[l2+l2] = 8
X
'J
4t — /—
dau = xay ra khi : j = x = \j2 .
Vay ta
duoc
y^ < 8, dau = xay ra khi x = V2
TCr do -2V2 < y < 2N/2 va khi x = V2 thi y = V2 + V2 = 2%/2 ta chi thu
diroc
-2V2 < y < 2N/2 , dau = xay ra khi x = N/2 va ta chi ket luan
duoc
maxy = y(72) = 2V2
{Li/u
y : nhieu hoc sinh ket luan min y = -2N/2 la sai vi khong c6
xo e[-2; 2] de y(xo) =
-2V2).
De tim miny, ta nhan xet rkng :
Khi
X = -2 thi ca 2 so hang x va V4-x^
dong
thcfi

dat GTNN, vay
min
y = y(-2) =-2 + 0 =-2
Cdch 3. (Liio'ng
giac
hoa ham y)
Do -2 < X < 2 ta
chon
x = 2 cos
cp
Khi
do
y = 2 cos (p> ^4 - 4 cos^ (p = 2 cos
cp
+ 2
|sin
cp
11
Chon
(pe[0;
7i]
khi
do
x-2cos9e[-2;
2]
sin
cp
> 0
sin
(p

=
sin
cp
va
y =
2(sin
cp
+ cos
9)
=
2^2 sin
doO<x<7i:o-<x
+ -<^
4
4 4
n
X
+
-
4
^/2
<
sin
x +
—
4
2V2.
V2I
<1
<y

<2V2.1
-2<y
<272
(xem hinh
ve)
(diem ngon
M
cua cung
x +
—
chay tren cung PBP'
4
=> diem
K
chay trong doan
[I ; B] )
Vay
maxy
- 2V2
miny
= -2
12
Bai
toan
6.
Tim
GTLN va GTNN cua ham so y = J + sin^ x trong
2' 2
Ldl
GlAl

Ta CO y' = i + 2 sin X cos X = ^ + sin 2x .
2 2
=> y' = 0
sin2x
= <=>
2x = + k2ji
6
6
n
12
771 ,
xo = — + ln
^ 12
Trong doan
•K
7t 71
2' 2
thi
y' = 0 chi khi
571
12
X2 =-
Tinh
y
Do
•2J
, y
77t
^71^
12

12
57t
12
12' 12
hcfn, ta bien doi
khong la cac goc dac biet, vi vay de
tinh
toan de dang
X . 2 1
y
=
—
+ sin X = —
•^2 2
X
+ (1 -
cos2x)
Khi
do thi
2
J
2
v2.
- — + 1'-
C0S(-7l)
2
= 1-^
- + 1 - cos 7t
2
4^

12,
57t
12
+ 1 - cos
12
571 ,
+ 1 - cos
12
V
6,
12 2
571
V
571 , VS
+ 1
12 2
Ta ket luan : max y = y
min
y = y
2j
:.1
+ ^
4,
57t^ _ 1
1 _
571
V3I
12
>
~ 2

12
2
Bai
toan
7.
Tim
GTLN
va
GTNN
cua hkm so
y
= 5 cos
X
- cos 5x
trong
4'
4
Ldl
GIAI
Ta
c6 y' = -5
sin
x + 5
sin
5x =
5(sin
5x -
sin
x)
y'

= 0 o
sin5x
=
sinx
<=>
5x
= X + k2n
5x
=
Jt
- X + I2n
Xi
= k
—
^
2
Xo
= - + / -
.6
3
Trong
doan
71
n
4'
4
,
y' = 0 chi
khi
x = 0, x = ±

6
De
tinh
gia
tri
cua y, ta de y
rkng
cosx,
cos5x.la
ham so
chin
cho nen:
672
\
71
—
=
y
—
V 4J
/
N
n
/
7t
.6,
=
y
6.
_

71 57t _ V2
=
5
cos cos — = 5
4
4 2
71
-cos
—
4
53t
73
- 5
cos — cos — = 5
6
6 2
7t
-cos
—
6y
=
372
'I
=
6^ = 373
2
va
y(0) = 4
TCf
do ta ket

luan diJorc
:
max
y = y ±
—
= 373 ;
min
y = y(0) = 4
I
6j
Bai
toan
8.
Tim
GTLN
va
GTNN
cua ham so y = 2
sin^
x + cos'^ 2x
LOI
GIAI
Ta bien
doi
+
cos'* 2x
y
=
2(sin''
x)^

+ cos"* 2x = 2.
^l-cos2x^^^
(l-cos2x)^+cos^2x
8
Chon
phep
bien
doi
t =
cos2x
t e [-1 ; 1]
14
Voi bien
t, ham y =
^(1
-
t)^
+
8
Ta tim GTLN
va
GTNN
cua y(t)
trong
[-1 ; 1]
(2t)^
-
(1
-1)^
Taco

y'(t) =
^(l-t)3.(-l)
+
4t3=^
2 ^ -
1
2t
-
(1
-1)
4t2
+
2t(l
-1) +
(1
- if
(3t-l)
(St^ +1)
2
Do 3t^
+ 1 > 0
vdri
moi t nen
y'(t)
=
Q
o = o
•
o
Tinh

y(-l),
yd) va y
3j
Ta
CO
y(-l)
= 3, yd) = 1, y
3;
27
Va
ket
luan
:
maxy
= 3, dat
tai
moi x ma cos2x = 1 x = krc
min
y = — , dat
tai
moi x ma cos 2x =
—
= cos 2a => x = ± a +
/:t
•^27
3
Bai
toan
9.
AC

Tim
GTLN
cua ham so y = sin xcos x
Ldl
GIAI
Cdch
1 :
(dung
bat
dSng
thufc
Cosi)
Ta bien
doi
•
2 2 1-2 1-2 I2 I2 I2
1
= sin X + cos X = —
sm
x +
—
sin
x +
—
cos x +
—
cos x +
—
cos x
2 2 3 3 3

3
fl
. 2 '
2
1
2
—
Sin
X
- COS
X
u U J
va
y =108
Ap dung
bat
dSng
thufc
Cosi
doi vdi
tich
5
thiTa
so
khong
am,
trong
do c6
—
sin*^

X va 3
thiifa
so'
—
1
2 3
2
thCfa
so —
sin^
x va 3
thiTa
so
— cos^
x, ta
diTo'c
•
2
2
2
— sin
X
— cos
X
U ;
13
J
1-2 1-2 1 2 1 2 1 2
-Sin
x+-sin x+cos

x+ cos x+ -cos x
2 2 3 3 3
15
/' 1 >
1.2
2
fl
2 '
' 1
1
—
sm X -
COS
x
U ;
u , 5^ 3125
va y <
108
3125
Dau
= xay ra khi :
—
sin^
x
—
cos^ x
2 3
1
1 - cos 2x 11 + cos 2x
<=> =

2 2 3 2
o cos 2x -
CO
nghiem
5
Vay : maxy = dat tai moi x la nghiem cua phifcfng
trinh
cos2x
= -
olzo 5
Cdch
2. (dung
each
bien doi y thanh hkm dai so de dung dao ham)
Ta bien doi y = (sin^ x)^ (cos^ x)^ = (sin^ x)^ (1 -
sin^
x)^
roi
chon
phep
bien doi u = sin^ x => u e [0 ; 1,]
Vdri
bien u, ham y c6 dang
y(u) = u2(l-u)^
vdriu
G [0 ; 1]
Ta tim GTLN va GTNN cua y(u) trong [0 ; 1]
Taco
y'(u) = u2.3(l-u)2.(-l) + (l-u)^.2u
=

u(l - uf [2(1 - u) - 3u;
=
u(l-u)2(2-5u)
y' = 0 khi u = 0, u = 1, u = -
5
Ta
tinh
y(0) = 0, yd) = 0, y
108
3125
108
Ta
CO
ket luan : maxy =
-r—;r,
dat tai moi x la nghiem cua phucfng
trinh
3125
2 2 1
u
= sm X = — o cos 2x =
—
.
16
Bai
toan 10.
Trong doan
,
hay
tim GTLN

va
GTNN
cua
ham
so y =
sin^ x.
cos*^
x
v6i
p e
N"^,
p > 2, q > 2.
Ldl GlAl
Ta
CO
y' =
sin^ x.q
cos^"^
x(- sin
x) + cos*'
x.p sin^"^
x cos x
= p
cos^"^^
X
sinP"-^
X
- q
sin^"^^
x cos^~^ x

=
sinP~^
x cosP"^
x.[q
cos^ x -
p sin^
x]
y'
= 0<=>
sin
x = 0
cos X = 0
2
• 2
q
cos X =
p sm
x
x
= 0
71
X
= —
2
2
• 2
q
cos X =
p sm
x

Ta tim nghiem
cua *)
trong doan
Ta
CO
*)<=>
p sin^
x =
q(l
-
sin^
x)
<=> (p +
q) sin
X = q
o
sin^
X = ^
p
+ q
Goi
Xo
la
nghiem
cua
phircfng
trinh
*) (chSc ch^n c6
nghiem)
Khi

do
sin^
XQ =
cos XQ =
p
+ q
P
p
+ q
Ta
CO y(0) = 0, y
f-1
.2J
=
0,
y(xo)
=
(sin2xo)2.(cos2xo)2
=
Vay maxy
= y(xo)
q
(p)2
P
q
(p
+ q)2 (p + q)2
>0
min
y

=^(0) = y
=
0
17
Bai
toan 11.
Tim
GTLN va GTNN cua ham so y = sin^^ x +
cos^°
x
Ldl
GlAl
Chon
u - sin x => u e [0 ; 1]
Khi
do : y ^u^^ +
(l-u)^°
Tim
GTLN va GTNN cua ham y(u) trong [0 ; 1].
Ta CO y' = lOu^ - 10(1 - ur = 10
ap dung
cong
thufc
^2n.l
_ i^2n.l ^ _
b)[a2"
+
a^"-^b
+
••

+
ab^""^
+ b^"1
CO
2n+l
so hang
Ta thu
dtfoc
f
y' = [u - (1 - u)] [u^ + u"(l - u) +
•••
+ u(l -
u)"^
+ (1 - u)^] = (2u
' ; 1—V '
SO
9 so hang = A
trong do A gom 9 so hang > 0 do cac so hang la luy thiTa u va
khong
dong
thoi
bSng
0 nen A > 0
=> y' = 0 khi u =
2
Ta CO y(0) = 1, yd) = 1 va y
Va ket luan
diroc
maxy
= 1 dat

di/pc
khi
1
2] 512
miny =
sin x = 0
sin^
x = 1
1
u
=0
u
= 1
X
= kit
X = — + In
2
(k, /eZ)
—— dat
diToc
khi sin^ x = —
512 • 2
1
- cos 2x 1 _
o = — <=>
cos2x
= 0
2 2
<=> x = — + m— (meZ)
4 2

18
Bai
toan 12.
Tim
GTLN
va
GTNN
cua ham so
y
=
+
2x - 3
—
In X tren doan
2
LOI
GIAI
Ta
CO y =
x^
+ 2x - 3
+
-
In X tren
(1; 4
2
^
-X
- 2x
+

3
+
—
In X tren
Khido
y'-
2x + 2
+
-
tren
(1; 4]
2
X
3 1
-2x - 2 +
tren
2
X
Chu
y
rkng
:
3
v(Ji
moi X e (1 ; 4] thi 2x + 2 + — > 0 (de
thay)
2x
vo'i
moi
x e

2^^
thi
-2x - 2
+
— < 0 *)
2x
Ta
CO *) o
-4x^
-4x + 3
2^^ ^
<0
o
4x^ +4x-3 > 0
C5>
X < — hoac X >
—
2
2
=>
-2x - 2
+
— < 0
vo'i
moi x e
2x
;
1
Bien
thien ciia

ham y cho boi
bang
sau :
19
Taco
y(4) = 21
+
-ln2
2
yd)
= 0
/1 \
Vay
4 2 2 2
maxy
= y(4) = 21 + -ln2
2
miny
= y(l) = 0
Bai
toan 13.
Tim
GTLN
va
GTNN
cua ham so : y =
xlnx
-
xln5
tren

[1 ; 5]
LOI
GIAI
1
5
Ta
CO y' = x
—
+ In
X -
In 5
=
In
x -
In
—
X
e
5 5
=:>
y' = 0 o
In
X =
In
- <=> X = - e
[1;
5]
e e
Ta
CO yd) = - ln5,

y(5)
= 0 va y
=
-
In In
5.
e e e
=
-(ln5-lne) ln5
=
e e e
va
CO
ket
luan
max
y = y(5) = 0
min
y = y
'5^
5
e
20
Bai
toan 14.
Tim GTLN cua ham so
y =
2x^ - 3x^ - 12x + 1
trong
doan

[-3 ; 3]
Ldl
GIAI
Goi z = 2x^ - 3x2 - I2x + 1, khi do y =
Ta xet
bien
thien cua ham so z tren [-3 ; 3] roi suy ra
bien
thien cua
ham so y de tii do c6 ket luan cua bai toan.
Ta CO z' = 6x2 _ 6x - 12 = GCx^ - x - 2)
=> z' = Oox = -l
hoac
x = + 2
De hieu bai toan mot
each
tiTdng minh, ta nen
dung
bang
bien
thien sau :
Luu
y tren
bang
bien
thien :
Bien thien cua z Ja diidng net lien.
Bien thien cua y la difcfng net duft.
21
TCr M, I, N bien thien ciia y va z trCing nhau.

Doc tren
bang
bien thien ta thu duac
max y = y(-3) = 44
Chu y rkng, bai toan thiTc ra chi can giai
ngSn
gon nhiT sau :
- Tim z' va giai z' = 0 tim cac nghiem
- Tinh z (-3), z(3) va gia tri ciia z(x) tai cac nghiem ma a day la
z(-3) = - 44, z{-l) = 8, z(2) = - 19, z(3) = -8
- Do y =
|z|
cho nen GTLN ciia ham y chi la so Ion nhat trong cac so sau :
z(-3)|
= 44 , |z(-l)| = 8 ,
|z(2)|
= 19,
|z(3)|
= 8
suy ra maxy =
|z(-3)|
= 44
Nhifng
de hieu ro
tinh
chat bien thien ciia ham y thi
each
diing
bang
bien thien tot ho'n. ,

Bai
toan 15.
Goi Xj , X2 la cac nghiem ciia phiTdng
trinh
:
x^ + px + = 0
1) (p ^ 0)
Hay tim p de : u = xf + x^
dat gia tri nhd nhat va tim GTNN cua u.
Ldl GIAI
Quy
trinh
giai bai toan bao gom cac
bifdc
sau :
Bi/ac
1. Tim dieu kien ciia p de phiTcrng
trinh
1) c6 nghiem
(Buoc
nay de
bo sot va ddn den sai lam)
Budc
2. Tim u
theo
p.
Budc
3. Tim GTNN ciia u vdi moi p da tim dJcfc trong
bi/dc
1.

+
Dieu kien de phifong
trinh
1) c6 nghiem la
A > 0 *)
p > V2
Ta CO *) o p'
\>0
<=>p'^>4op^>2
o
<-^/2
+
Bien doi u ve dang thuan
Icfi
de dung
diToc
dinh li Viet cho phifofng
trinh
1).
22
Ta CO u = (x? + X2)^-2x?x| =
i2
(xi
+X2)^ - 2x^X2 - 2x'fx2
Vol
moi p thoa man dieu kien *) thi
xi
+ X2 = -p
X1X2 = -Y
P

Khi
do u =
-2
=> u = p +
+
Tim
GTNN
ciia
u(p) vo'i moi p thoa man dieu kien *)
(co the giai khong dung,
nhiT
sau :
Do p'^ + > 2 => min
u
= 2 - 4 = -2 la sai vi voi dieu kien *) thi
P
pn-L>2)
P
Ta
chon : t =
p'^
^ t > 4.
Khi
do ta diipc u(t) = tt 4
Ta
tim min u(t) voi t > 4
Ta
CO u'(t) =
1
- -77 > 0 vdi moi t > 4

=:> u(t) la ham so dong thien
vdti
t > 2 va khi do thi miny(t) =
—
4
Ta
ket luan diToc :
minu
=^
-, dat khi t =
p''
= 4 => P =
±N/2
4
Bai
toan 16.
Ydi
0, cho ham so
f
(x) = 3a^x^ - Sa^x + a^ - a^ + -
4
Hay tim a de ham u = xf + x| dat GTNN, GTLN va tim min u, max u
trong do x^ va X2 la cac nghiem cua ham f(x).
Ldl
GlAl
+
Dieu kien de f(x) c6 nghiem : A > 0 *)
Ta CO *) <=> 9a^ -12a^
4 2 3
a -a +

—
4
>
0 « -3a^ + 12a'* -9a^ >0
Sa^Ca^
- 4a2 + 3) < 0 (do
33^
> 0)
(a^ -1) (a^ - 3) < 0 o 1 < a^ < 3 o 1 <
|a|
< 73
1
< a < VS '
-V3 < a < -1
+
Ta CO u = ixi + X2) -
3xiX2(xi
+ X2).
Vdri
moi a thoa man dieu kien *) ta c6
diTcfc
x^ + X2 = a
1 '
^^^^^^
4 2 3
a - a +
—
4
Khi
do u = a^ - 3

•
4 2 3
a -a + "
• 4
3a2
u
= a
4a
Xet bie'n thien ciia ham u(a) tren 2 mien 1; y/sl va T-Vs; -1
Ta CO u' = 1 +
4a'
>
0 vdi moi a
=> Tren 2 mien thoa man dieu kien *), ham u(a) la ham
dong
bien.
Bien thien cua u(a) cho bdi
bang
sau :
24
Ta thu
diTcfc
maxu = u (Vs j =
Sj3
minu
=
u^-\/3J
= -
3V3
Bai

toan 17.
Cho x>0,
y>Ovax
+ y = l
Tim
GTLN va GTNN cua ham so u = + ^
y + 1 X + 1
Ldl GIAI
Bien ddi u -
+
y^ +1
xy + 2 .
9
(x + y)^
—
2xy + 1 2
—
2t
Chon
phep
bien doi t = xy. Khi do u = =
xy + 2
t
+ 2
Tim
dieu kien cho t:do l = x + y> 2^^
=> t < —, dau = xay ra khi x = y = —
va t > 0 , dau = xay ra khi x = 0
hoSc
y = 0

Vay 0<t<
4
Ta tim GTLN va GTNN cua ham so u(t) = trong
doan
t
+ 2
Ta
CO u'(t) =
(t +
2)(-2)
- (2 - 2t)
(t + 2Y
(t + 2)2
<0
u(t) nghich bien
trong
khi
do ta c6
ngay
ket luan :
max u = u(0) = 1 dat khi t = 0 => x = 0
hoac
y = 0
min u
= u
2 11
=
—
dat khi t =
—

=> x = y = —
3 4 2
• LUu y : Trong cac bai toan minh hoa d
tren,
ta da stf dung djnh If sau
day de l<hang dinh ve sU c6 gia trj Idn nhat va gia trj nho nhat cua cac ham
so lien tuc. Dinh If do la :
Moi ham so lien tuc trong doan [a ; b] thi c6 trong doan do cac gia tri Idn
nhat va gia tri nho nhat.
Dieu dang luu y la dinh If nay chi cho ta dieu kien du de ham so c6 GTLN
va GTNN.
Dieu do c6 nghTa la, neu khong thoa man cac dieu kien cua djnh If, chang
han :
- ham so khong lien tuc trong doan [a ; b] ' I
- ham so lien tyc nhung khong trong ca doan [a ; b] ma chi la (a ; b); (a ; b];
[a,
b) thi khong the khang dinh la ham so khong c6 GTLN hoac GTNN
trong mien dang xet.
Khi do, de khang djnh ta phai xet cu the cho tCmg ham so trong cac mien
tudng ufng da cho.
De
giiip cac em hoc
sinh
tranh nhkm Ian, tac gia gidfi thieu cac bai
toan
minh
hoa sau :
Bai
toan
18.

Cho
ham so' y = 7x -1 + 79 - x .
Hay
tim
GTLN
va
GTNN
ciia
ham so
trong
cac mien sau day :
a)
[3 ; 6] ; b) (3 ; 6) ; c) [3 ; 6) ; d) (3 ; 6]
Ldl GIAI
Taco y' = _l=
+
—L=(-l)
= i
2Vx^ 2V9-X 2
V9 - X - Vx -1
V9-x.Vx-l
=> y'>0 o 79 - X > Vx -1 <^ x<5
Bien
thien cua ham y cho bdi bang sau
26