Giáo trình giải tích 1 part 4 ppt

✲
x
✻
y
0
s
a
1
= a
s
a
2
s
a
3
s
b
4
s
a
4
s
b
1
= b
b
2
b
3
f(a)
f(b)

f(a) < 0 <f(b)
c f(c)=0
[a, b] t =
a + b
2
f(t)=0 c = t
f(t)f(a) < 0 a
1
= a, b
1
= t
f(t)f(b) < 0 a
1
= t, b
1
= b
f(a
1
) f(b
1
) [a
1
,b
1
]
c f(c)=0
[a
n
,b
n

],n ∈ N b
n
− a
n
=
b − a
2
n
f(a
n
) < 0 <f(b
n
) a
n
<c<b
n
, ∀n ∈ N
f(c)=0
b
n
− a
n
=
b − a
2
n
→ 0 n →∞ lim a
n
= lim b
n

= c f c
f(c) = lim
n→∞
f(a
n
) ≤ 0 f(c) = lim
n→∞
f(a
n
) ≥ 0
f(c)=0
F (x)=f(x) −γ F [a, b] F (a)F ( b) ≤ 0
c F (c)=f(c) −γ =0 f(c)=γ 
√
2 10
−1
x
2
−2=0 [1, 2]
f [a, b] a, b f(x)=0
f (a, b)
f [a, b]
f
−1
[f(a),f(b)] [f(b),f(a)]
f [a, b] [a, b]
f[a, b] f[a, b]
f(a),f(b) f
−1
:[c, d] → [a, b]

f
−1
y
0
∈ [ c, d] (y
n
) y
0
x
0
= f
−1
(y
0
) x
n
= f
−1
(y
n
) x
n
→ x
0
(x
n
k
) x

= x

0
f f(x

) = f(x
0
) f
f(x
n
k
) → f(x

) f(x
n
k
)=y
n
k
→ y
0
= f(x
0
) 
f(x)=a
0
+a
1
x+···+a
n
x
n

a
n
=0 n lim
x→−∞
f(x)=− (a
n
)∞ lim
x→+∞
f(x)= (a
n
)∞
a<0 <b f(a) f(b)
c ∈ (a, b) f(c)=0 c f(x)=0
f :[a, b] → [a, b] c : f (c)=c c
f
F (x)=f(x) − x F [a, b]
F (a)=f(a) −a ≥ 0 F (b)=f(b) − b ≤ 0
c ∈ [a, b] F (c)=f(c) −c =0 f(c)=c
f [a, b]
f max min α, β ∈ [a, b]
f(α)=max{f(x):a ≤ x ≤ b} f(β) = min{f(x):a ≤ x ≤ b}
f n ∈ N
x
n
∈ [a, b] |f(x
n
)| >n (x
n
)
(x

n
k
)
k∈N
c ∈ [a, b] f
|f(c)| = lim
k→∞
|f(x
n
k
)| = lim
k→∞
n
k
=+∞
M =sup{f(x):a ≤ x ≤ b} m =inf{f(x):a ≤ x ≤ b}
α, β f(α)=M f(β)=m
sup n ∈ N x
n
∈ [a, b] M −
1
n
<f(x
n
) ≤ M
(x
n
k
) α
f k →∞ f(α)=M

β f(β)=m 
arctan x
π
2
−
π
2
max, min R
f
X
∀>0, ∃δ>0:∀x, x

∈ X, |x −x

| <δ ⇒|f(x) −f(x

)| <
•
• X X
• f(x)=
1
x
(0, +∞) f X
∃>0, ∀δ>0:∃x, x

∈ X, |x −x

| <δ, |f(x) −f(x

)|≥

 =1 0 <δ<1 x
δ
= δ, x

δ
=
δ
2
∈ (0, +∞)
|x
δ
− x

δ
| <δ |f(x
δ
) − f(x

δ
)| =
1
δ
≥  =1
• X δ 
x ∈ X a δ
a a δ
f(x)=
1
x
a  δ

>0 δ>0 a a 0 δ
f [a, b] f
f [a, b]
>0 n ∈ N x
n
,x

n
∈ [a, b]
|x
n
− x

n
| <
1
n
|f(x
n
) − f(x

n
)|≥ (∗)
(x
n
) (x
n
k
) c ∈ [a, b] |x


n
k
−c|≤
|x

n
k
− x
n
k
| + |x
n
k
− c|→0 k →∞ (x

n
k
) c f
lim
k→∞
f(x
n
k
) = lim
k→∞
f(x

n
k
)=f(c) |f(x

n
k
) − f(x

n
k
)| k
∗ 
f :(a, b) → R f x
0
f
x
0
L ∈ R
f(x
0
+∆x)=f(x
0
)+L∆x + o(∆x) , ∆x → 0
f(x)=f(x
0
)+L(x −x
0
)+o(x − x
0
) , x → x
0
✲
x
✻

y
s
x
0
f(x
0
)
✡
✡
✡
✡
✡
✡
✡
✡
✡
✡
✡
y = f (x
0
)+L(x − x
0
)
f x
0
L = lim
∆x→0
f(x
0
+∆x) − f(x

0
)
∆x

f x
0
f x
0
f(x) −f(x
0
)=L(x − x
0
)+o(x − x
0
) → 0, x → x
0
f(x)=|x|
f
x
0
f x
0
f

(x
0
)
f

(x

0
) = lim
∆x→0
f(x
0
+∆x) −f(x
0
)
∆x
= lim
x→x
0
f(x) −f(x
0
)
x − x
0
f

(x
0
) L : R → R, ∆x → L(∆x)=f

(x
0
)∆x
f x
0
df (x
0

)
f(x)=x f

(x
0
) = lim
∆x→0
(x
0
+∆x) −x
0
∆x
=1, ∀x
0
∈ R.
dx(∆x)=∆x
df (x
0
)=f

(x
0
)dx f

(x
0
)=
df
dx
(x

0
)
f x
0
f

+
(x
0
) = lim
∆x→0
+
f(x
0
+∆x) −f(x
0
)
∆x
f

−
(x
0
) = lim
∆x→0
−
f(x
0
+∆x) − f(x
0

)
∆x
f

(x
0
) f

+
(x
0
),f

−
(x
0
) f

+
(x
0
)=f

−
(x
0
)
f(x)=e
x
f


(x)=e
x
lim
∆x→0
e
x+∆x
− e
x
∆x
= lim
∆x→0
e
x
(e
∆x
− 1)
∆x
= e
x
f(x)=sinx f

(x)=cosx
lim
∆x→0
sin(x +∆x) − sin x
∆x
= lim
∆x→0
2cos(x +

∆x
2
)sin
∆x
2
∆x
= lim
∆x→0
cos(x +
∆x
2
) lim
∆x→0
sin
∆x
2
∆x
2
=cosx
f(x)=|x| x
0
=0
f(0 + ∆x) −f(0)
∆x
=
|∆x|
∆x
∆x → 0
f


+
(0) = lim
∆x→0
+
|∆x|
∆x
=1 f

−
(0) = lim
∆x→0
−
|∆x|
∆x
= −1
f(x)=
3
√
x 0 f

(0) = lim
∆x→0
3
√
∆x
∆x
= ∞
y = f(x) x
0
x x

0
∆x = x − x
0
x
0
∆y = f(x
0
+∆x) − f(x
0
)
•
f x
0
∆y = f

(x
0
)∆x + o(∆x)
f

(x
0
)∆x ∆y x
0
y = f(x
0
)+f

(x
0

)(x − x
0
)
y = f(x) x
0
•
M
0
(x
0
,f(x
0
)) M(x
0
+∆x, f(x
0
+∆x)) f
∆y
∆x
= M
0
M M
0
M Ox
f x
0
M
0
f


(x
0
)= f (x
0
,f(x
0
))
y = f(x
0
)+f

(x
0
)(x −x
0
)
•
f(x) x
∆y
∆x
= ∆x
f x
0
f

(x
0
) x
0
• f


(x
0
) = lim
∆x→0
∆y
∆x
y = f(x) x x
0
• f,g x
0
f ± g fg
f
g
g(x
0
) =0 x
0
(f ± g)

(x
0
)=f

(x
0
) ± g

(x
0

)
(fg)

(x
0
)=f

(x
0
)g(x
0
)+f(x
0
)g

(x
0
)
(
f
g
)

(x
0
)=
f

(x
0

)g(x
0
) − g

(x
0
)f(x
0
)
g(x
0
)
2
• f x
0
g f(x
0
) g ◦f x
0
(g ◦f)

(x
0
)=g

(f(x
0
))f

(x

0
)
• f f

(x
0
) =0
f
−1
y
0
= f(x
0
)
(f
−1
)

(y
0
)=
1
f

(x
0
)
f(x+∆x)g(x+∆x)−f (x)g(x)=(f(x+∆x)−f(x))g(x+∆x)+f(x)(g(x+∆x)−g(x))
∆x ∆x → 0
f(x +∆x)

g(x +∆x)
−
f(x)
g(x)
=
(f(x +∆x) −f(x))g(x) −f(x)(g(x +∆x) − g(x))
g(x +∆x)g(x)
g(f (x +∆x)) −g(f(x)) =
g(f (x +∆x)) − g(f (x))
f(x +∆x) − f(x)
(f(x +∆x) −f(x))
y = f(x), ∆y = f(x +∆x) − f(x)
g(f (x +∆x)) −g(f(x))
∆x
=
g(y +∆y) − g(y)
∆y
f(x +∆x) −f(x)
∆x
∆x → 0 ∆y → 0
f
−1
◦f(x)=x (f
−1
)

(y
0
)f


(x
0
)=1 y
0
= f(x
0
) 
dg
dx
=
dg
dy
dy
dx
g

x
= g

y
y

x
g = g(y) y = f(x)
x
(x
α
)

= αx

α−1
(a
x
)

= a
x
ln a (e
x
)

= e
x
(log
a
x)

=
1
x ln a
(ln x)

=
1
x
(sin x)

=cosx
(cos x)


= −sin x
(tan x)

=
1
cos
2
x
( x)

= −
1
sin
2
x
(arcsin x)

=
1
√
1 − x
2
(arccos x)

= −
1
√
1 − x
2
(arctan x)


=
1
1+x
2
( x)

= −
1
1+x
2
e
x
sin x 
f(x)=e
ax
sin bx
f

(x)=(e
ax
)

sin bx + e
ax
(sin bx)

= ae
ax
sin bx +e

ax
b cos bx = e
ax
(a sin bx + b cos bx)
g(x)=x
x
g

(x) ln g(x)=x ln x
g

(x)
g(x)
=lnx + x
1
x
=lnx +1
g

(x)=g(x)(ln x +1)=x
x
(ln x +1)
f :(a, b) → R x
0
f
x
0
f

(x

0
)=0
f x
0
−f
∆y = f(x
0
+∆x) −f(x
0
) ≤ 0 ∆x
f

+
(x
0
) = lim
∆x→0
+
∆y
∆x
≤ 0 f

−
(x
0
) = lim
∆x→0
−
∆y
∆x

≥ 0
f

(x
0
)=f

+
(x
0
)=f

−
(x
0
) f

(x
0
)=0 
✲
x
✻
y
❝
❝
❝
x
0
f(x

0
)
f

(x
0
)=0 f x
0
f(x)=x
3
f [a, b] (a, b) f(a)=f(b)
c ∈ (a, b): f

(c)=0
f [a, b] x
1
,x
2
∈ [a, b]
f(x
1
)= max
x∈[a,b]
f(x)=M f(x
2
) = min
x∈[a,b]
f(x)=m
m = M f f


(x)=0 x ∈ (a, b)
m<M f(a)=f(b) x
1
x
2
a, b
f

(x
1
)=0 f

(x
2
)=0 
✲
x
✻
y
s
c
s
a
s
b
✏
✏
✏
✏
✏

✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
✏
f,g [a, b]
(a, b) c ∈ (a, b)
f(b) − f(a)
g(b) − g(a)
=
f

(c)
g


(c)
c ∈ (a, b) f(b) − f(a)=f

(c)(b −a)
F (x)=(f(b)−f(a))(g(x) −g(a)) −(g(b)−g(a))(f(x)−f(a))
F [a, b] (a, b) F (a)=F(b)=0
c ∈ (a, b) F

(c)=(f(b) − f(a))g

(c) − (g(b) − g(a))f

(c)=0 
f(x
0
+∆x) −f(x
0
)=f

(x
0
+ θ∆x)∆x,
x
0
,x
0
+∆x ∈ (a, b) 0 <θ<1 x
0
, ∆x

|f(x) − f(y)|≤ sup
c∈(a,b)
|f

(c)||x −y| x, y ∈ [a, b]
f

(x)=0, ∀x ∈ (a, b) f (a, b)
f

(x)=g

(x), ∀x ∈ (a, b) f − g f = g+
(sin x)

=cosx
|sin x −sin y|≤|x − y|, ∀x, y ∈ R
(arctan x)

=
1
1+x
2
≤ 1
|arctan x −arctan y|≤|x − y|, ∀x, y ∈ R
f x
0
f(x
0
+∆x)= n ∆x + o(∆x

n
) , ∆x → 0
f (a, b) f

(a, b) f

x
0
f

(x
0
)=(f

)

(x
0
)
n f x
0
f
(0)
(x
0
)=f(x
0
),f
(n+1)
(x

0
)=(f
(n)
)

(x
0
)
n f x
0
d
n
f(x
0
)=f
(n)
(x
0
)dx
n
n n f
(n)
(x
0
)=
d
n
f(x
0
)

dx
n
C
n
(a, b) f n (a, b) f
(n)
(a, b) f
C
n
f,g n x
0
α
(f + g)
(n)
(x
0
)=f
(n)
(x
0
)+g
(n)
(x
0
)
(αf)
(n)
(x
0
)=αf

(n)
(x
0
)
(fg)
(n)
(x
0
)=
n

k=0
C
k
n
f
(k)
(x
0
)g
(n−k)
(x
0
)

n
(x
α
)
(n)

= α(α − 1) ···(α − n +1)x
α−n
(a
x
)
(n)
= a
x
(ln a)
n
(log
a
x)
(n)
=
(−1)
n−1
(n − 1)!
x
n
ln a
(sin ax)
(n)
= a
n
sin(ax + n
π
2
)
f n +1 (a, b) x

0
x ∈ (a, b) 0 <θ<1
f(x)=f(x
0
)+
f

(x
0
)
1!
(x−x
0
)+···+
f
(n)
(x
0
)
n!
(x−x
0
)
n
+
f
(n+1)
(x
0
+ θ(x − x

0
))
(n +1)!
(x−x
0
)
n+1
x M
f(x)=f(x
0
)+
n

k=1
1
k!
f
(k)
(x
0
)(x − x
0
)
k
+ M(x − x
0
)
n+1
.
h(t)=f(x) − f(t) −

n

k=1
1
k!
f
(k)
(t)(x −t)
k
− M(x −t)
n+1
.
h(x)=h(x
0
)=0 c = x
0
+ θ(x − x
0
), 0 <θ<1
h

(c)=0
−
1
n!
f
(n+1)
(c)(x −c)
n
+(n +1)M(x − c)

n
=0, M =
f
(n+1)
(c)
(n +1)!

n f x
0
T
n
f(x)=f(x
0
)+
1
1!
f

(x
0
)(x − x
0
)+···+
1
n!
f
(n)
(x
0
)(x − x

0
)
n
• f n f n
f(x)= T
n
f(x)+R
n
(x)
n R
n
(x)=f(x) − T
n
f(x)
f x
0
• R(x
0
)=R

(x
0
)=···= R
(n)
(x
0
)=0
R
n
(x)=o((x −x

0
)
n
) x → x
0
• f n +1
R
n
(x)=
f
(n+1)
(x
0
+ θ(x − x
0
))
(n +1)!
(x −x
0
)
n+1
, θ ∈ (0, 1)
n +1 M
|R
n
(x)|≤
M
(n +1)!
|x −x
0

|
n+1
f(x)=a
0
+ a
1
(x − x
0
)+a
2
(x − x
0
)
2
+ o(x − x
0
)
2
f 2 x
0
f(x)=1+x + x
2
+ x
3
D(x) D
x
0
=0
e
x

=1+
x
1!
+ ···+
x
n
n!
+
e
θx
(n +1)!
x
n+1
sin x = x −
x
3
3!
+ ···+(−1)
n−1
x
2n−1
(2n − 1)!
+
(−1)
n
cos θx
(2n +1)!
x
2n+1
cos x =1−

x
2
2!
+ ···+(−1)
n
x
2n
(2n)!
+
(−1)
n+1
cos θx
(2n +2)!
x
2n+2
ln(1 + x)=x −
x
2
2
+ ···+(−1)
n−1
x
n
n
+
(−1)
n
x
n+1
(n + 1)(1 + θx)

n+1
(1 + x)
α
=1+αx + ···+
α(α − 1) ···(α −n +1)
n!
x
n
+
α(α − 1) ···(α −n)(1 + θx)
α−n−1
(n +1)!
x
n+1
6 0
e
−x
2
=1+(−x
2
)+
1
2!
(−x
2
)
2
+
1
3!

(−x
2
)
3
+ O((−x
2
)
4
)=1− x
2
+
x
4
2
−
x
6
6
+ O(x
8
)
1
√
1+x
3
=(1+x
3
)
−
1

2
=1−
1
2
x
3
+
3
8
(x
3
)
2
+ O((x
3
)
3
)=1−
x
3
2
+
3
8
x
6
+ O(x
9
)
f n +1 f(x)

n x
0
f(x
0
+∆x) ≈ f(x
0
)+
1
1!
f

(x
0
)∆x + ···+
1
n!
f
(n)
(x
0
)∆x
n
|R
n
(∆x)| =
|f
(n+1)
(x
0
+ θ∆x)|

(n +1)!
|∆x|
n+1
= o(∆x
n
)
n
√
1+x x
n
√
1+x x
0
=1
n
√
1+x ≈
n
√
1+(
n
√
1+x)

|
x=1
x =1+
1
n
x

e <
e ≈ 1+
1
1!
+
1
2!
+ ···+
1
n!