Calculo de Diversas Variables

7/23/2019 Calculo de Diversas Variables

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Rn





(Rn, +, ·) Rn

Rn

Rn

Rn

Rn

C1

C k.

Rn





(Rn, +, ·) Rn

Rn = {u = (x1, · · · , xn) : x j ∈ R, j = 1, · · · , n}

+ : Rn × Rn → Rn

(x1, · · · , xn) + (y1, · · · , yn) = (x1 + y1, · · · , xn + yn),

· : R× Rn → Rn

λ(x1, · · · , xn) = (λx1, · · · , λxn),

(Rn, +, ·)

Rn

0 ∈ Rn

0 = (0, · · · , 0)

uv

u− v

u+ v

Rn = { p = (x1, · · · , xn) : x j ∈ R, j = 1, · · · , n}

n

T : Rn×(Rn, +, ·) → Rn p = (x1, · · · , xn) u = (y1, · · · , yn)

T ( p, u) = (x1 + y1, · · · , xn + yn) T ( p, u) = p + u

Rn

p, q

→ pq

p q

q p → pq = ”q − p” p

→0 p

Rn

p q Rn 2 p−q

T (0, 2→0 p − →

0q ) = ”2 p−q ”

2 p−q T (0, 2→0 p − →

0q )



Rn

• x, y z f : Rn →Rm

f (x) x = (x1, · · · , xn)

f (y) f (z ) R2

f (x, y) (x, y)

R3 f (x,y,z )

•

A B Rn, λ ∈ R,

∗ A

∩B =

{x ∈Rn : x

∈ A x

∈ B

}.

∗ A ∪ B = {x ∈ Rn : x ∈ A x ∈ B}.∗ Ac = Rn \ A = {x ∈ Rn : x /∈ A}.∗ A + B = {z ∈ Rn : z = x + y x ∈ A, y ∈ B}.∗ A − B = {z ∈ Rn : z = x − y x ∈ A, y ∈ B}.∗ λA = {z ∈ Rn : z = λx, x ∈ A, λ ∈ R}.∗ A \ B = {x ∈ A : x /∈ B}.

{Aα}α∈A

α∈A

Aα

α∈A

Aα

∗ (Ac)c = A

∗ (A ∩ B)c = Ac ∪ Bc

α∈A

Aα

c

=α∈A

Acα

∗ (A ∪ B)c = Ac ∩ Bc

α∈A

Aα

c

=α∈A

Acα

∗ A \ B = A ∩ Bc

B ∩Ac = B \A

A ∩Bc = A \B

A ∩B = (Ac ∪Bc)c

Ac ∩Bc = (A ∪B)c

A

B



(Rn, +, ·) Rn

·

A

B

C

· (a, b)(x − a)2

A2 +

(y − b)2

B2 = 1

(a, b)A(x − a)2 − (y − b) = 0

(a, b)(x − a)2

A2 − (y − b)2

B2 = 1

· (a,b,c)(x − a)2

A2 +

(y − b)2

B2 − (z − c)2

C 2 = 0

· (a,b,c)(x − a)2

A2 +

(y − b)2

B2 +

(z − c)2

C 2 = 1

· (a,b,c)A(x − a)2 + B(y − b)2 − (z − c) = 0

·

(a,b,c)(x − a)2

A2 +

(y − b)2

B2 − (z − c)2

C 2 = 1

·

(a,b,c)(x − a)2

A2 − (y − b)2

B2 − (z − c)2

C 2 = 1

x,y,z x2 − y = 0 x − y2 = 0



Rn

x = (x1, · · · , xn) y = (y1, · · · , yn)

Rn

x, y = x · y =

nk=1

xkyk = x1y1 + · · · + xnyn.

x,y,z ∈ Rn λ ∈ R

x, y = y, x

x, x ≥ 0, x, x = 0 x = 0

x + z, y = x, y + z, y

λx,y = λx, y

E

·, · : E × E → R

(x, y) → x, y x,y,z ∈

E λ ∈ R

E ·, ·

x ∈ Rn

x =√

x · x =

nk=1

x2k

1/2

.

E

· : E

→ R

x → x

x ≥ 0 x = 0 x = 0

λx = |λ|x

λ ∈ R x ∈ E

x + y ≤ x + y

x, y ∈ E

E ·

x x = 1

x = 0 y = x

x

x · y x = √

x · x

|xk

| ≤ x

k = 1,

· · · , n

√ λ2 =

|λ

|

x = 0 y = 0

0 x y x − y2 = x2 + y2 −2xy cos θx,y θx,y x y

x − y2 = x2 − 2x · y + y2

x · y = xy cos θx,y

|x · y| ≤ xy



Rn

x + y2 = x2 + 2x · y + y2 ≤ x2 + 2xy + y2 = (x + y)2,

x + y ≥ 0 x + y ≥ 0

x·y = xy cos θx,y x, y

x · y = 0

x, y E ·, ·

x, y = 0 x y x ⊥ y

x, y

∈ E x y Πyx

Πyx = λy (x − Πyx) ⊥ y.

y = 0 Πyx = x,yy,y y

u

v

Πvu

u− Πvu

u = (2, 2, 0) v =(2, 0, 2)

Πvu = u·vv2 v = 4

8(2, 0, 2) = (1, 0, 1)

Rn

·, ·

E

x = x, x1/2 E

x, y ∈ Rn |x, y| ≤ xy

x, y = xy

x = λy y = λx

λ ≥ 0

·



· x

E x, y ∈ E

| x − y | ≤ x − y

x = y + (x − y) ≤y + x − y x − y ≤ x − y x y y − x ≤y − x = x − y x − y ≥ −x − y

x y

• x y x + y2 = x2 + y2.•

· x + y2 + x − y2 = 2x2 + 2y2.

Rn

x

1 = n

k=1

|xk

| x∞ = maxk=1,··· ,n |xk|

x∞ ≤ x ≤ x1 ≤ nx∞,

|xk| ≤ maxk=1,··· ,n

|xk| ≤

nk=1

x2k

1/2

≤n

k=1

|xk| ≤ n maxk=1,··· ,n

|xk|.

Rn

d(x, y) = x−y

Rn

dist(x, y) =

→xy

E

X d(·, ·) : X × X → R

d(x, y) = d(y, x) d(x, y) ≥ 0 d(x, y) = 0 x = y d(x, y) ≤ d(x, z ) + d(z, y)

x,y,z ∈ X. X (X, d(·, ·)).

|d(x, y) − d(y, z )| ≤ d(x, z ) .

E d(x, y) = x − y

E.

d(x + z, y + z ) = d(x, y)



Rn

Rn

{xk = (xk

1, . . . , xkn)}k

Rn

{xk}k ⊂ Rn x = (x1, . . . , xn), lim

k→∞xk − x = 0, ε > 0 kε

xk − x < ε k > kε.

limk→∞

xk = x, xk → x k → ∞, xk → x

Rn

R

xk → x xk

j → x j 1 ≤ j ≤ n.

0 ≤ |xk j − x j| ≤ xk − x ≤

n j=1

|xk j − x j|

limk→∞

(xk1, . . . , xk

n) =

limk→∞

xk1, . . . , lim

k→∞xkn

,

Rn

n

R

limk→∞(1/k,k sin(1/k)) = (0, 1).

·

R

x,

0 ≤

x−

y ≤

x−

xk

+

xk

−y

< ε k > K ε

ε > 0. x − y = 0 x = y.



Rn

Rn

x ∈ Rn r > 0

B(x, r) = {y ∈ Rn : d(x, y) < r} = {y ∈ Rn : y − x < r}

x r

A ⊂ Rn Rn

{B(x, r)}r>0 A Ac

A Rn

• x ∈ Rn A r > 0 B(x, r) ∩ A = ∅

x A

∗

B(x, r) Ac

x ∈ Rn A r > 0

B(x, r) ⊂ A r > 0 B(x, r) Ac

A A

Ao

x ∈ Rn A B(x, r)

A Ac

A F r(A)

F r(A) = A \ Ao

A = Ao ∪ F r(A) Ao ∩ F r(A) = ∅

∗ A \ {x}

x

∈Rn

A B(x, r)

A x A A

A \ A x

A r > 0 B(x, r) ∩ A = {x}

P I A = A∪P I A∩P I =∅

• x ∈ Rn

A x ∈ (A)c r > 0 B(x, r) ⊂ Ac

A Ext(A)

p0

p1

p2

p3

Ao

Fr(A) \ { p3}

Ext(A)

A

p3punt aıllat ∈ F r(A)



Rn

A A F r(A)

A Rn

x ∈

A

{yk

}k

⊂ A

yk → x

x ∈ A

{yk}k ⊂ A \ {x}

yk → x

x ∈ F r(A) {yk}k ⊂ A

{z k}k ⊂ Ac yk → x z k → x

A

{yk}k ⊂ A x ∈ Rn

x ∈ A

x ∈ A

k ∈ N

yk

∈ A ∩B(x, 1/k) yk − x < 1/k → 0 k → +∞

{yk}k ⊂ A yk → x ε > 0 kε

yk − x < ε k > kε ε > 0 B(x, ε) ∩ A = ∅

x ∈ A k ∈ N

yk ∈ (A \ {x}) ∩ B(x, 1/k) yk − x < 1/k → 0 k → +∞

{yk}k ⊂ A \ {x} yk → x ε > 0 kε

0 < yk − x < ε k > kε ε > 0 B(x, ε) ∩ A \ {x}) = ∅

F r(A) = A ∩ Ac

A =

{(x, y)

∈ R2; 1 < x2 + 2y2

≤ 4

} ∪{(0, 0), (1, 0)}. P 0 = (0, 0) P 0

A δ > 0

p − P 0 ≥ δ p ∈ A \ {P 0}

p = (x, y) ∈ A \ {(0, 0)} p − P 0 = (x, y) = (x2 + y2)1/2 ≥

(x2 + 2y2)1/2/√

2 ≥ 1√ 2

δ = 1/√

2

P 1 = (1, 1) A

{ pk} ⊂ A pk → P 1 1 < 12 + 2 12 = 3 ≤ 4 P 1 ∈ A

pk = P 1

P 1 = (1, 1) A

{ pk} ⊂ A \ {P 1}

pk → P 1

pk = (xk, yk) = (1 + 1/k, 1)

pk → P 1 x2k + 2y2

k = (1 + 1/k)2 + 2 1 < x2k + 2y2

k ≤ 4 k ≥ 3

P 2 = (1, 0) F r(A) P 2 ∈ A ∩ Ac

{ pk} ⊂ A

{q k} ⊂ Ac pk → P 2

q k → P 2

pk pk = (1, 0) (1, 0) ∈ A q k =(z k, wk) = (1 − 1/k, 0) k ≥ 2 q k =∈ Ac

0 <z 2k + 2w2

k = (1 − 1/k)2 < 1 k ≥ 2



A

R n

A o

A

A o

A o

• p ∈ A o

ε >

0

A

B

( p , ε ) ⊂ A

A

• p ∈ A

ε >

0

A

A

B ( p , ε ) ∩ A

• p ∈ A

ε >

0

A \ A o

• p ∈ A

B ( p , ε ) ∩ A

= ∅

F r ( A )

A

{ p k } k ∈ N

⊂ A \ { p }

p k

→

p

• p ∈ A

• p ∈ F r ( A )

{ p k

} k ∈ N

⊂ A

p k

→

p

ε >

0

B ( p , ε ) ∩ A

= ∅

B ( p , ε ) ∩

A c

= ∅

A i l l ( A )

A i l l ( A )

R n

• p ∈ F r ( A )

A

{ p k

} k ∈ N ⊂

A

{ q k

} k ∈ N

⊂ A c

• p ∈ A i l l ( A )

ε > 0

E x t ( A )

p

k

→

p

q k

→

p

B ( p , ε ) ∩ A

=

{ p }

E x t ( A )

A

E x t ( A )

E x t ( A )

E x t ( A )

• p ∈ E x t (

A )

ε >

0

B ( p , ε ) ∩ A

=

∅

B (

p , ε ) ⊂ A c

R n

=

A o

∪ F r ( A ) ∪ E x t ( A

) ,

A o ,

F r ( A )

E x t ( A )

A

=

A o

∪ F r ( A )

E x t ( A ) =

A c

E x t ( A )

=

( A o ) c

F r ( A ) =

A ∩ E x t ( A ) = ( A

\ A o ) ∪ A i l l ( A )



•

A Ao = A

• ∅

• A Ac A = A

• A

Rn A = Rn

• p A p ∈ Ao

A = {(x, y) ∈ R2 : 0 < x2 + 4y2 − 8y ≤ 5}

A

Ao

A A

F r(A)

A = {(x, y) ∈ R2 : (x2 + 4y2)(x2 + y2 − 4) ≥ 0}.

Ao A

Q R Q ⊂ R = Q Qn

Rn

Qn ⊂ Rn = Qn

(Qn)o = ∅ F r(Qn) = Qn = Rn

A Rn

Rn ∅

A ⊂ B Ao ⊂ Bo A ⊂ B

A Ac

Ao A C ⊂ A

C ⊂ Ao

Ext(A) Ac

A A

A Rn B

C

C ⊂ A ⊂ B

D = B \ C ⊂ F r(A)

A = B Ao = C F r(A) = D



A A Ao

A

A ⊂ B = C ∪ D ⊂ Ao ∪ F r(A) = A,

A = B

Ao

F r(A)

C ⊂ Ao

D ⊂ F r(A)

C = Ao F r(A) = A \ Ao = B \ C = D

z ∈ Rn r > 0 A = B(z, r)

A B(z, r)

z r

A = A = B (z, r) = {x ∈ Rn : x − z ≤ r} B(z, r)

z r

F r(A) = S (z, r) = {x ∈ Rn : x − z = r} S (z, r) z

r

B = B(z, r) C = B(z, r) D = S (z, r)

C = A ⊂ B = C ∪ D

• C x ∈ B(z, r) ε > 0 B(x, ε) ⊂B(z, r) y − x < ε y − z < r ε < r − x − z

y − z ≤ y − x + x − z < r

• B Bc = {x ∈ Rn : x − z > r}

x ∈ B c ε > 0 B(x, ε) ⊂ B c

y − x < ε y − z > r ε < x − z − r

y − z ≥ x − z − y − x > x − z − ε > r

• D

⊂ F r(A) x

∈ D B(x, ε) A

Ac A y

y − z < r

y − x < ε y y = x − δ (x − z ) δ > 0

y − z = |1 − δ |r y − x = δr 0 < δ < min{1,ε/r}

B(x, ε) Ac x ∈ Ac

B(x, r) B(x, r) = B (x, r) F r(B(x, r)) = S (x, r)

B(z, r) B(z, r)

x ∈ B(z, r) B(x, ε) A x

y y − z < r 0 < y − x < ε

x = z 0 < δ < min{r, ε} y = z + δw w

x = z

y = x − δ (x − z )

0 < δ < min{1,ε/r}



{Aα}α

A = ∪αAα x ∈ A

ε > 0 B(x, ε) ⊂ A

x ∈ A x ∈ Aα α Aα

ε > 0

B(x, ε) ⊂ Aα ⊂ A

A = ∩m j=1A j A j A j = ∅

A = ∅

A A j x ∈ A

x ∈ A j j ε j > 0 B(x, ε j) ⊂ A j

ε = min{ε j : 1 ≤ j ≤ m} > 0 B(x, ε) ⊂ A

{Bα}α B = ∩αBα

Bc

Bc = ∪αBcα Aα = Bc

α

B = ∩m j=1B j Bc = ∪m

j=1Bc j Bc

•

A j = (−1/j, 1/j) j ∈ N ∩ jA j = {0}

•

B j = [−1 + 1/j, 1 − 1/j] j ∈ N ∪ jB j = (−1, 1)

• τ

Rn

τ X X X

∅ X τ X

τ X τ X

· {Ai}i Rn

α

Aα

o

⊂α

Aoα

α

Aα

o

⊃α

Aoα

α

Aα

⊂α

Aα

α

Aα

⊃α

Aα.

(∩αAα)o ⊂ Aoα α

A j = (−1/j, 1/j) ⊂ R j ∈ N

∩αAo

α

(∪αAα)o ⊃ Aoα α

A1 = [−1, 0] A2 = (0, 1]

(∩αAα) ⊂ Aα α

A1 = [−1, 0] A2 = (0, 1]

(∪αAα) ⊃ Aα α

A j = [−1 + 1/j, 1 − 1/j] ⊂ R ∪αAα



A Rn

B Rm

A × B

Rn+m

A Rn B Rm

A × B

Rn+m

c = (a, b) a ∈ A b ∈ B

A×B ε > 0 B(a, ε) ⊂ A B(b, ε) ⊂ B B(a, ε)×B(b, ε) ⊂A × B B(c, ε) ⊂ B(a, ε) × B(b, ε) c

(x, y) − (a, b) = (x − a, y − b) < ε

x − a < ε y − b < ε x − a ≤ (x − a, y − b)

y − b ≤ (x − a, y − b)

A Ac Ac × Rm

Rn+m

Rm × Bc

Rn+m

(A × B)c = (Ac ×Rm)

∪(Rn

×Bc) (A

×B)c

• A = {(x,y,z ) ∈ R3 : x2 + y2 < 9, −1 < z < 2} A = B((0, 0), 3) × (−1, 2)

• A = {(x,y,z ) ∈ R3 : x2+y2 ≤ 9} A = B ((0, 0), 3)×R

A ⊂ Rn r > 0 x ≤ r

x ∈

A, A ⊂

B (0, r).

A M ≥ 0 |x j | ≤ M

x = (x1, · · · , xn) ∈ A R > 0 A ⊂ B(0, R)

•

A = {(x, y) ∈ R2; x4 + y6 < 4}

x4 < 4 y6 < 4 x2 + y2 ≤ 2 + 3√

4.• A = {(x, y) ∈ R2; xy < 4} {(k, 1/k)}k ⊂ A

(k, 1/k) =

k2 + 1/k2 → ∞

k → ∞.

Rn

Rn {xk}k Rn

K Rn

{xk}k ⊂ K K.



K Rn

K

K

· K ⊂ Rn A ⊂ Rn

A∩ K

A ⊂ K A

A ∩ K

· K Rn

K Rm, K × K

Rn+m.

K × K

K

{Aα}α K K ⊂ ∪αAα

K α j , 1 ≤ j ≤ J K ⊂Aα1 ∪ · · · ∪ AαJ

(X, d) K X

K

K





Rn

f : A ⊂ Rn → Rm Graf (f ) = {(x, y) ∈ A × Rm : y =

f (x)} f

f Rn+m

f : R → R

f : R2 → R

R2 z ∈ R

C z = {(x, y) ∈ R2 : f (x, y) = z }



RN

f : A ⊂ Rn −→ Rm

x = (x1, . . . , xn) −→ f (x) = (f 1(x), . . . , f m(x))

f : A ⊂ Rn −→ Rm a A.

f a b, ε > 0 δ = δ ε,a > 0

f (x) − b < ε, x ∈ A 0 < x − a < δ.

limx→a

f (x) = b, f (x) → b, x → a.

f : A

⊂Rn

−→Rm

a A.

limx→a f (x) = b

ε > 0 δ = δ ε,a > 0 f ((B(a, δ ) \ {a}) ∩ A) ⊂ B(b, ε)

limx→a

f (x) − b = 0.

{xk}k ⊂ A \ {a}

a, {f (xk)}k ⊂ Rm

b. f = (f 1, · · · , f m) b = (b1, · · · , bm) lim

x→af k(x) = bk k =

1, · · · , m

aδ bε

f A

•

limx→ax∈A

f (x) = b, f (x) → b, x → a, x ∈ A \ {a}.

A

f : (0, +∞) −→ R, limt→0 f (t) limt→0+ f (t), t < 0

f : A ⊂ Rn → Rm a ∈ A, ε > 0

δ = δ ε,a > 0

f (x) − f (a) < ε, x ∈ A x − a < δ.

f a ε > 0 δ = δ ε,a > 0

f (B(a, δ ) ∩ A) ⊂ B(f (a), ε).



•

0 < x−a < δ x − a < δ

a ∈ A f

a.

•

a ∈ A ∩ A

f

a

limx→a f (x) = f (a).•

limx→a f (x) = b

f (x) : A ∪ {a} → Rm

f (x) =f (x), x = a

f (a) = b, f A \ {a} a.

• A δ ε a δ ε

A

f : A ⊂ Rn → Rm A f

A. A C(A).

·

f, g : A ⊂ Rn → Rm, a A.

limx→a

f (x) = b limx→a

g(x) = c.

limx→a

(f ± g)(x) = b ± c.

m = 1, limx→a

(f g)(x) = b c.

m = 1 c = 0, limx→a

(f /g)(x) = b/c.

f g A f ± g, f g f/g

g ·

f : A ⊂ Rn → Rm, a A

limx→a f (x) = b. g : B ⊂ Rm → Rk, b ∈ B. f (A) ⊂ B,

limx→a

(g ◦ f )(x) = g(b).

f A g B g ◦ f A.

•

f j(x) = x j

j = 1, · · · , n

• h(x, y) = sin(xyz )

xyz xyz = 0 h(x,y,z ) = 1 xyz = 0

R3 f (x,y,z ) = xyz g(t) =

sin t

t t = 0 g(0) = 1,

• f : A ⊂ Rn −→ Rm A, f : A ⊂ Rn −→ R



RN

f x → a

f (x) − b ≤ ϕ(x)

U ∩ A \ {a}

U a ∈ A limx→a

ϕ(x) = 0 limx→a

f (x) = b

• · : Rn → R | x − a | ≤x − a.

• M : Rn −→ Rm M (x) − M (y) =

M (x − y) ≤ C M x − y.

• f (x, y) =

4x2y3

x6 + y4 (x, y) = (0, 0),

0

(x, y) = (0, 0),

(0, 0).

|x| ≤ (x6 + y4)1/6

|y| ≤ (x6 + y4)1/4 |f (x, y)| ≤ 4(x6 +

y4)2/6+3/4−1 = 4(x6 + y4)1/(12) → 0 (x, y) → (0, 0)

• f (x, y) = y3 + x2 + y2 − 2x + 1

(x − 1)2 + y2 (x, y) (1, 0),

y3 + x2 + y2 − 2x + 1

(x − 1)2 + y2 − 1

=

y3

(x − 1)2 + y2

≤ |y| → 0 (x, y) → (1, 0)

• lim

(x,y)→(0,0)xy=0

sin3(xy)

x5y + xy3 = 0.

limt→0

sin t

t = 1 t = xy lim

(x,y)→(0,0)

sin(xy)

xy = 1

sin3(xy)

x5y + xy3 =

sin3(xy)

(xy)3x3y3

x5y + xy3 =

sin3(xy)

(xy)3x2y2

x4 + y2

0 ≤ x2y2

x4 + y2 ≤ x2

lim(x,y)→(0,0)xy=0

x2y2

x4 + y2 = 0

lim(x,y)→(0,0)

xy=0

sin3(xy)

x5y + xy3

= lim(x,y)→(0,0)

xy=0

sin3(xy)

x3y3

lim(x,y)→(0,0)

xy=0

x2y2

x4 + y2

= 0.



f : A ⊂ Rn → Rm. B

A b B. b f B l

limx→bx∈B f (x) = l.

B ⊂ A a B limx→a

f (x) = b

limx→ax∈B

f (x) = b.

A =

N j=1

A j limx→ax∈Aj

f (x) = b, j = 1, . . . , N ,

limx→ax∈A

f (x) = b

N = ∞

f : A ⊂ Rn −→ R,

limx→a

f (x)

limt→0a+tv∈A

f (a + tv), v

Rn.

l

|f (x) − l| l

B ⊂ A limx→ax∈B

f (x) = l

f (x, y) = xy3

x2

+ y4

A = R2

\ {(0, 0)

}.

lim(x,y)→(0,0)

xy3

x2 + y4.

t = 0, f (at, bt) = abt2

a2 + b4t2

limt→0

abt2

a2 + b4t2 = 0. a = 0 b = 0 f (at,bt) = 0



RN

|f (x, y) − 0| =

|f (x, y)| → 0 (x, y) → (0, 0)

|x| ≤

x2 + y4 |y| ≤ 4

x2 + y4, |f (x, y)| ≤ (x2 +

y4)1/4

→ 0

f (x, y) = xy

x2 + y2 A = R2 \ {(0, 0)}.

lim(x,y)→(0,0)

xy

x2 + y2.

f

R2 \ {(0, 0)}

·

f (at,bt) = ab/(a2 + b2).

limt→

0 f (at,bt) = ab/(a2 + b2). a = 0 b = 1

a = 1 b = 1

lim(x,y)→(0,0)

xy

x2 + y2

f : R2 → R

f (x, y) =

x6y3

x2 − y, y = x2,

0, y = x2.

lim(x,y)→(0,0) f (x, y).

f (at,bt) =

a6b3t9

a2t2 − bt, bt = a2t2,

0, bt = a2t2.=

a6b3t8

a2t − b, bt = a2t2,

0, bt = a2t2.

Am = {(x, y) ∈ R2 : y =x2 + xm} m ∈ N Am

y = x2

·

f Am

f (x, x2 + xm) = x6(x2 + xm)3

−xm = −x12−m(1 + xm−2)3

m > 12 f (x, x2 + xm) → −∞ = 0 x →= 0+

lim(x,y)→(0,0)

f (x, y)

(0, 0)



f (x, y) = (x − 1)

|y| log(x2 + y2)

(x − 1)2 + y2 A = R2 \

{(0, 0), (1, 0)}. lim(x,y)→(1,0)

f (x, y).

limt→0 log(1 + t)t = 1 lim(x,y)→(1,0)

log(x

2

+ y

2

)x2 + y2 − 1 = 1. lim(x,y)→(1,0)

f (x, y)

lim(x,y)→(1,0)

(x − 1) |y|(x2 + y2 − 1)

(x − 1)2 + y2

g(x, y) = (x − 1)

|y|(x2 + y2 − 1)

(x − 1)2 + y2 (x, y) = (1, 0)

t → 0

g(1 + at, bt) = a |b| |t|((1 + at)2 + (bt)2 − 1)

t =

a |b| |t|(2at + t2)

t

= a |b| |t|(2a + t) → 0.

(1, 0)

g

|x−1| ≤ ((x−1)2+y2)1/2

|y| ≤ ((x−1)2+y2)1/2 |x2+y2−1| =

|(x−1)(x+1)+y2| ≤ 2|x−1|+y2 ≤ 2|x−1|+|y| |x−1| < 1 |y| ≤ 1

|(x − 1) |y|(x2 + y2 − 1)| ≤ 3((x − 1)2 + y2)1/2+1/4+1/2−1 = ((x − 1)2 + y2)1/4 → 0

(x, y) → (1, 0)

f (x, y) =

x2

x + y, x = −y

0, x = −y

f (x, y) =

x2

|x| + |y| , (x, y) = (0, 0)

0, (x, y) = (0, 0)

f (x,y,z ) =

1 − cos(x2 + y3 + z 2)

x2 + y4 + z 4 , (x,y,z ) = (0, 0, 0)

0, (x,y,z ) = (0, 0, 0)

Rn

X Rn. A X

X U Rn

A = U ∩ X.



RN

X = (0, 1] ⊂ R, (1/2, 1] (0, 1],

(0, 1/3] (0, 1].

f : A ⊂ Rn −→ Rm.

f A. f −1(V ) V Rm

A. f −1(T ) T Rm

A.

(1) ⇒ (2) y = f (x) δ x > 0 f (B(x, δ x) ∩ A) ⊂B(y, ε) ⊂ V, B(x, δ x)∩A ⊂ f −1(V ). U = ∪x∈f −1(V )B(x, δ x)

f −1(V ) = U ∩ A.(2) ⇒ (1) f a ∈ A. f (a) = b

f −1(B(b, ε)) = U ∩ A U Rm. a ∈ U δ > 0

B(a, δ ) ⊂

U f (B(a, δ )∩

A) ⊂

B(b, ε)

(2) ⇔ (3)

(2) (3) T Rm \ T

Rm (2) U Rn

f −1(Rm \ T ) = U ∩A.

f −1(T ) = A \ f −1(Rm \ T ) = A \ (U ∩ A) = (Rn \ U ) ∩ A. Rn \ U

Rn

•

f −1(V ) = f −1(V ∩ f (A)),

f (A)

A.

f (A) A.•

•

f (x) = e−x [0, ∞) (0, 1] R.

f : Rn → Rn

A Rn

A f (A) f −1(A)

A f (A) f −1(A)

(f (A))o = f (Ao) f (A) = f (A) F r(f (A)) = f (F r(A))

A f (A) f −1(A)

· Rn

f (x1, · · · , xn) = f (x1 + a1, · · · , xn + an)

A

f (A)

f (x1, · · · , xn) = (λ1x1, · · · , λnxn) λ j = 0 j = 1, · · · , n



A · x2+4y2 < 4 A = f (B((0, 0), 1))

(x, y) = f (s, t) = (2s, t) Ao = f (B((0, 0), 1)) = A A = f (B((0, 0), 1) ={(x, y) ∈ R2 : x2 + 4y2 ≤ 4}

F r(A) = f (S (0, 0), 1)) = {(x, y) ∈ R2 : x2 + 4y2 = 4}

A = {(x, y) ∈ R2; xy < 1, x > 0} R2

(−∞, 1) × (0, ∞) f (x, y) = (xy,x).

A = {(x,y,z ) ∈ R3; x2 + y4 + z 6 ≤ 5}

(−∞, 5] f (x,y,z ) = x2+y4+z 6.

|x| ≤

x2 + y4 + z 6 ≤ √

5, |y| ≤ 4

x2 + y4 + z 6 ≤ 4

√ 5, |z | ≤ 6

x2 + y4 + z 6 ≤

6√

5,

A1 = {(x, y) ∈ R2; x2 + 4y4 − 6x ≤ −5},

A2 = {(x, y) ∈ R2; x2 + 2xy + 2y2 < 4},

A3 = {(x,y,z ) ∈ R3; x2 + xz + yz + z 2 < 4},

A4 = {(x, y) ∈ R2; y2e−x < 1}.

K Rn f : K −→ Rm f (K ) Rm.

{yk}k ⊂ f (K )

f (K ) {xk}k ⊂ K yk = f (xk). K

xk → x ∈ K. f f (xk) = yk → f (x) =y ∈ f (K ).

f (x) = c R f −1({c}) = R

K Rn

f : K −→ R f K, a, b ∈ K

f (a) ≤ f (x) ≤ f (b) x ∈ K.

f (K )

R f (K ) sup(f (K )) inf(f (K ))

f (K )



RN

sup(f (K )) inf(f (K )) f (K ) f (K ) f (K ) sup(f (K )) = max(f (K )) inf(f (K )) = min(f (K ))

∅ = K ⊂ Rn

∅ = A ⊂ Rn

c ∈ K a ∈ A d(K, A) = inf {x − y : x ∈ K, y ∈ A} = d(c, a) = c − a

K

K A K ×A R2n

f (x, y) = x − y

Rn × Rn

(c, a) K × A d(K, A) = c − a

A K r K ⊂ B (0, r) B(0, r) ∩ A = ∅ d(K, A) ≤ 2r

B = B(0, 4r) ∩ A B

d(K, A) = d(K, B)

d(K, A∩Bc

) ≥ 3r

d(K, A) =c − a c ∈ K a ∈ B ⊂ A

A

d(K,A)

A ∩B(0, 4r)

K



o

f, g : A ⊂ Rn → R A a

f o g a f = ox→a(g) U a

ϕ f = ϕg ϕ(x) → 0 x → a

f = ox→a(g)

|g| ≤ h

f = ox→a(h)

f (x) = x − aα ox→a(x − aβ) β < α

f (x, y) = x2+2y3 |f (x, y)| ≤ (x, y)2+2(x, y)3

f = o(x,y)→(0,0)((x, y)α) α < 2

n = 1 f : (a, b) → R t0 ∈ (a, b)

limt→t0

f (t) − f (t0)

t − t0 f (t0).

f t0 limt

→t0

f (t) − f (t0) − f (t0)(t − t0)

t

−t0

= 0,

limt→t0

|f (t) − f (t0) − f (t0)(t − t0)||t − t0| = 0.

o f t0

f (t) − f (t0) − f (t0)(t − t0) = ot→t0(t − t0).

y = f (t0) + f (t0)(t − t0)

f (t0, f (t0))

(f

±g) = f

±g

(f g) = f g + f gf

g

=

f g − f g

g2

(g ◦ f ) = f · (g ◦ f ) (sin(f )) = f cos(f )



A Rn, f : A ⊂ Rn −→ R, a A

f x j a

limxj→aj

f (a1, . . . , a j−1, x j, a j+1, . . . , an) − f (a1, . . . , a j−1, a j, a j+1, . . . , an)

x j − a j=

∂f

∂x j(a).

f x j a

g(t) = f (a + te j) 0,

∂f

∂x j(a) = lim

t→0

f (a + te j) − f (a)

t .

•

∂f

∂x j(a), f xj (a), D jf (a).

•

x ∈ A

∂f

∂x j(x) x ∈ A j−

• f (x, y) = x2y3, f x(1, 2)

f (x, 2) = 8x2 x = 1. f y(1, 2)

f (1, y) = y3

• f (x, y) =

xy

x2 + y2 (x, y) = (0, 0) f (0, 0) = 0, f x(0, 0)

f (x, 0) = 0 f y(0, 0) = 0. (x0, y0) = (0, 0)

f x(x0, y0) = y0(y20 − x20)(x2

0 + y20)2

, f y(x0, y0) = x0(x20 − y20)(x2

0 + y20)2

.

f x(x, y) =

y(y2−x2)(x2+y2)2 , (x, y) = (0, 0)

0, (x, y) = (0, 0), f y(x, y) =

x(x2−y2)(x2+y2)2 , (x, y) = (0, 0)

0, (x, y) = (0, 0).

f

f a,

f

∇f (a) = Grad(f )(a) = (f x1(a), . . . , f xn(a)).

f : A ⊂ Rn −→ R, a A u

f u a

limt→0

f (a + tu) − f (a)

t = Duf (a) = d f a(u).

j = 1, . . . , n , f xj (a) = (Dejf )(a).



f (x, y) =

xy2

x2−y y = x2,

0 y = x2.

u = (a, b) t 0 t = 0, f (at, bt) = abt2

a2t − b.

limt→0

f (ta,tb) − f (0, 0)

t = 0, Duf (0, 0) = 0 u.

f (0, 0) f (t + t8, t2) → ∞ t → 0.

f (x, y) = xexy

(1, 0).

A Rn, f : A ⊂ Rn −→ Rm, a A.

f a df a : Rn −→ Rm,

limx→a

f (x) − f (a) − df a(x − a)

x − a = 0.

df a

f a. Df (a) df a.

f

A

A.

•

limx→a

f (x) − f (a) − df a(x − a)x − a = 0.

•

f (x) − f (a) − df a(x − a) = ox→a(x − a).

• df a Rn Rm,

f : R3

−→ R2

f 1, f 2 f,

f 1(x,y,z )f 2(x,y,z )

−

f 1(x0, y0, z 0)f 2(x0, y0, z 0)

−

m11 m12 m13

m21 m22 m23

x − x0

y − y0

z − z 0

(mij)

{e j}n j=1 Rn.



A Rn, f : A ⊂ Rn −→ Rm, a A.

f a.

f j : A −→ R, 1 ≤ j ≤ m,

f

f : A ⊂Rn → Rm

A Rn

f : A −→ Rm a

f j a, Duf j(a) =d(f j)a(u). df a f a

J (f )a =

∂f 1∂x1

(a) . . . ∂f 1

∂xn(a)

∂f m∂x1

(a) . . . ∂f m

∂xn(a)

.

u = 0 Au = {a+tu ∈ Rn; t ∈ R}. f a

limx→a

x∈D∩Au

|f j(x) − f j(a) − d(f j)a(x − a)|x − a = 0.

limt→0

|f j(a + tu) − f j(a) − td(f j)a(u)||t|

= 0, Du(f j)(a) =

d(f j)a(u).

· f : A ⊂ Rn −→ R a ∈ A Du(f )(a) df a(u) = (∇f )(a), u.

•

Rn df ≡ 0.

• T : Rn −→ Rm Rn

dT ≡ T .•

π j(x) = x j Rn

dπ j ≡ e j.

•

f (x, y) =

x2y2

x2 + y2

(x, y) = (0, 0)

f (0, 0) = 0 (0, 0). f

(0, 0) df (0,0) = 0(= (0 0)).

|f (x, y)|(x, y) ≤ (x, y) (x, y) → 0 f

(0, 0) df (0,0) = 0.



A Rn

f : A −→ Rm

a ∈ A. M = M a < ∞ r = ra > 0 f (x)−f (a) ≤ M x−a

x

−a

< r. f a.

f (x) − f (a) = f (x) − f (a) − df a(x − a)

x − a x − a + df a(x − a).

f ε > 0 δ = δ ε,a > 0

f (x) − f (a) ≤ C x − a, x − a < δ,

A

Rn

f, g : A −→ Rm

a ∈ A.

f + g a, d(f + g)a = df a + dga. m = 1, f g a, d(f g)a = g(a)df a + f (a)dga.

m = 1 g(a) = 0, f

g a d

f

g

a

= g(a)df a − f (a)dga

g2(a) .

•

Rn,

• f (x, y) = xy/(x2 + y2) (0, 0)

A Rn

U Rm.

f : A −→ U a g : U −→ Rk f (a), g◦f

a, d(g ◦ f )a = dgf (a) df a.

b = f (a),

∂ (g1 ◦ f )

∂x1(a) . . .

∂ (g1 ◦ f )

∂xn(a)

∂ (gk ◦ f )∂x1

(a) . . . ∂ (gk ◦ f )

∂xn(a)

=

∂g1

∂y1(b) . . .

∂g1

∂ym(b)

∂gk

∂y1(b) . . .

∂gk

∂ym(b)

∂f 1

∂x1(a) . . .

∂f 1

∂xn(a)

∂f m

∂x1(a) . . .

∂f m

∂xn(a)

.

∂ (g j ◦ f )

∂xl(a) =

mi=1

∂g j

∂yi(b)

∂f i

∂xl(a)



• f (x,y,z ) = (sin(x2 +

yz ), xy) R3 g(x,y,z ) = (x2 + yz,xy)

h(u, v) = (sin u, v). p = (x0, y0, z 0)

J (f ) p = cos(x2

0 + y0z 0) 0

0 1 2x0 z 0 y0

y0 x0 0

2x0 cos(x20 + y0z 0) z 0 cos(x2

0 + y0z 0) y0 cos(x20 + y0z 0)

y0 x0 0

• f (x, y) =

(1 − cos(xy))/(xy), xy = 0

0 xy = 0

R2 g(x, y) =

xy R2, h(t) =

(1 − cos(t))/t t = 0

0 t = 0 R, f = h ◦ g.

df (0,0) = h(0)d(xy)(0,0) = (0 0).

f : R4 → R, u : R2 → R

v : R2 → R

g : R2 → R g(x, y) = f (x,y,u(x, y), v(x, y)).

g f, u v.

gx(x, y) = (D1f )(x,y,u(x, y), v(x, y))

+ (D3f )(x,y,u(x, y), v(x, y)) ux(x, y) + (D4f )(x,y,u(x, y), v(x, y)) vx(x, y).

gy(x, y) = (D2f )(x,y,u(x, y), v(x, y))

+ (D3f )(x,y,u(x, y), v(x, y)) uy(x, y) + (D4f )(x,y,u(x, y), v(x, y)) vy(x, y).

g : R2

→ R

R2

f (x, y) =

sin(g(x2 + y3, xy))

f x(x, y) = cos(g(x2 + y3, xy)

2x(D1g)(x2 + y3, xy) + y(D2g)(x2 + y3, xy)

f y(x, y) = cos(g(x2 + y3, xy)

3y2(D1g)(x2 + y3, xy) + x(D2g)(x2 + y3, xy)

.

C1

f

f f

f : A ⊂ Rn → Rm

C1 A

f ∈ C1

(A)

f j

A

C 1(Rn)

A Rn f : A −→ Rm

C 1(A). f

A.



C1.

R f (t) = t2 sin(1/t2) t = 0 f (0) = 0 R

n = 2, f (x, y) = xy sin(1/(x2 + y2)), (x, y) = (0, 0)

0, (x, y) = (0, 0)

R2 (0, 0)

(0, 0) |f (x, y)| ≤ |xy| ≤ (x, y)2

(x, y) = (0, 0),

f x(x, y) = y sin(1/(x2 + y2)) − 2x2y

(x2 + y2)2 cos(1/(x2 + y2)).

f x

1/(√

4kπ, 1/(√

4kπ)

→ ∞, f x (0, 0).

A Rn,

f ∈ C1(A) ⇒ f A ⇒ f A⇓ f A

⇓ f A

A Rn

f : A −→ R a ∈ A.

{e j} j=1,··· ,n e j = (0,...., 0, 1 j, 0, ....0)

u j = (e j, D jf (a))

∈ Rn+1

j = 1,

· · · , n

f p = (a, f (a)) ∈ Rn+1

(∇f (a), −1)

f p

(x − a, y − f (a)), (∇f (a), −1) =n

j=1

(x j − a j)D jf (a) − (xn+1 − f (a)) = 0,

xn+1 = f (a) +n

j=1

D jf (a)(x j − a j).

Duf (a) f u

Duf (a) = u · ∇f (a) = ∇f (a) cos θu,∇f (a) u

(a, f (a)) u = ∇f (a)

∇f (a)

∇f (a) = 0

∇f (a)

f (x, y) = 3 + sin(2y − x). f x(0, 0) = −1

f y(0, 0) = 2 f p = (0, 0, 3) −x+2y−(z −3) = 0, z = 3 − x + 2y



p v1 = (1, 0, −1), v2 = (0, 1, 2). p ∇f (0, 0) =

√ 5

f (x,y,z ) =2x2 + y4 + z 3 p = (1,

−1, 1, 4)

p

A Rn, f : A −→ R

x, y A, θ ∈ (0, 1) f (y) − f (x) = df (1−θ)x+θy(y − x).

h : [0, 1] −→ R h(t) = f ((1 − t)x + ty).

h(t) = df (1−t)x+ty(y − x).

f (y) − f (x) = h(1) − h(0) = h(θ) = df (1−θ)x+θy(y − x).

· A

Rn

f A

df x = 0 x ∈ A, f

A

x, y ∈ D γ : [0, 1] −→ A. df x ≡ 0 f

f f (γ (t))

f (x) = f (y) x, y ∈ A.

f :

D ⊂Rn

−→Rm

.

f : [0, 1] −→R2

f (t) = (cos(2πt), sin(2πt)),

f (1) − f (0) = df θ 0 ≤ θ ≤ 1, (0, 0) =2π(− sin(2πθ), cos(2πθ))

f : D ⊂ Rn → Rm



C k.

f : A ⊂Rn −→ R

D jf x A. D jf

xk f

f

n = 2

f

∂f

∂x = f x = D1f,

∂f

∂y = f y = D2f

∂ 2f

∂x2 = f xx =

∂f x∂x

= (f x)x = D21f,

∂ 2f

∂y∂x = f xy =

∂f x∂y

= (f x)y = D2D1f,

∂ 2f

∂x∂y = f yx =

∂f y∂x

= (f y)x = D1D2f, ∂ 2f

∂y2 = f yy =

∂f y∂y

= (f y)y = D22f.

∂ 3f

∂x3 = f xxx =

∂f xx∂x

= (f xx)x = D31f,

∂ 3f

∂y2∂x = f xyy =

∂f xy∂y

= (f xy)y = D22D1f,

∂ 3f

∂y∂x∂y = f yxy =

∂f yx∂y

= (f yx)y = D2D1D2f.

•

g(x, y) = x2y3 gx(x, y) = 2xy3

gy(x, y) = 3x2y2, gxx =2y3, gxy(x, y) = 6xy2, gyx(x, y) = 6xy2, gyy = 6x2y.

f (x, y) = xy(x2 − y2)

x2 + y2 , (x, y) = (0, 0) f (0, 0) = 0, f x(0, y) = −y

f y(x, 0) = x. f xy(0, 0) = −1 f yx(0, 0) = 1.

A Rn

f : A −→ R f

Ck(A) k

f A. f : A −→ Rm

Ck

A



A Rn

f, g : A → R

Ck

A

f ± g, f g Ck(A)

f /g A g ·

A Rn

B Rm

f : A → B Ck(A)

g : B → Rk

Ck(B) g ◦ f Ck(A)

A Rn f : A −→ R

C2

(A),

1 ≤ j, k ≤ n,

(D jDk)f

(DkD j)f

f ∈ Ck(A)

k

f :(a, b) −→ R Ck, t0 ∈ (a, b)

f (t) =

k j=0

f ( j)(t0)

j! (t − t0) j + ot→t0(|t − t0|k).

k

k f t0. (T k,t0f )(t).

f k t0. f Ck+1

Rk,t0(f )(t) = f k+1((1 − θ)t + θt0)

(k + 1)! (t − t0)k+1,

θ ∈ (0, 1). k f t0

k

f (t) − (T k,t0f )(t) = ot→t0(|t − t0|k),

q k(t) = (T k,t0f )(t) q k(t) − (T k,t0f )(t) = 0 k ot→t0(|t − t0|k)



t0 = 0

T k,0

1

1

−t

=

k

j=0

t j = 1 + t + t2 + · · · + tk.

T k,0

1

1 + t

=

k j=0

(−1) jt j = 1 − t + t2 − t3 + · · · + (−1)ktk.

T k,0

1

(1 − t)m

=

k j=0

(m + j − 1)!

(m − 1)! j! t j

= 1 + mt + m(m + 1)

2! t2 + · · · +

m(m + 1) · · · (m + k − 1)

k! tk.

T k,0(log(1 + t)) =k

j=1

(−1) j−1

j t j = t − t2

2 +

t3

3 − t4

4 · · · +

(−1)k−1

k tk.

T k,0(et) =k

j=0

t j

j! = 1 + t +

t2

2! + · · · +

tk

k!.

T 2k+2,0(sin t) =k

j=0

(−1) j t2 j+1

(2 j + 1)! = t − t3

3! +

t5

5! − t7

7! + · · · + (−1)k

t2k+1

(2k + 1)!.

T 2k+1,0(cos t) =k

j=0

(−1) j t2 j

(2 j)! = 1 − t2

2! +

t4

4! − t6

6! + · · · + (−1)k

t2k

(2k)!.

α = (α1, . . . , αn) α1, . . . , αn ≥ 0,

|α| = α1 + . . . + αn α! = α1! · · · αn!. α = (α1, · · · , αn) α j ≥ 0 |α| = α1 + · · · + αn = k

Ck

∂ kf (x1,· · ·

, xk)

∂xα = Dα

f (x1, · · · , xk) = ∂ kf (x1,

· · · , xk)

∂xα11 · · · ∂xαn

n.

yα = yα11 · · · yαn

n

α = (2, 3, 0), |α| = 2 + 3 + 0 = 5, α! = 2!3!0! = 12,

Dαf (x,y,z ) = ∂ 5f (x,y,z )

∂x2∂y3 (x,y,z )α = x2y3.



A Rn, p A,

f : A −→ R Ck(A),

T k,p(f )(x) = |α|≤k

(Dαf )( p)

α! (x

− p)α

k

Rk,p(f )(x) = f (x) − T k,p(f )(x) = ox→ p(x − pk).

T k,p(f )(x) k f

p Rk,p(f )(x) k f

p

f Ck+1(A), p

|f (x) − T k,p(f )(x)| ≤ C x − pk+1 C > 0

g(t) = f ( p+t(x− p)/x− p) t = x− p

v Rn

g(t) = f ( p+tv)

g(t) = v · ∇(f )( p + tv) =n

j=1

v jD j(f )( p + tv) g(t) =n

j=1

v j

ni=1

viDi(D j(f ))( p + tv)

n = 2 p = (a, b)

T (a,b),k(f )(x, y) =k

l=0

i,j≥0i+ j=l

1

i! j!

∂ lf

∂xi∂y j(a, b)(x − a)i(y − b) j.

k = 3

T (a,b),3(f )(x, y) = f (a, b) + ∂ f

∂x(a, b) (x − a) +

∂ f

∂y(a, b) (y − b)

+ 1

2!0!

∂ 2f

∂x2(a, b) (x − a)2 +

1

1!1!

∂ 2f

∂x∂y(a, b) (x − a)(y − b) +

1

0!2!

∂ 2f

∂y2(a, b) (y − b)2

+ 1

3!0!

∂ 3f

∂x3(a, b) (x − a)3 +

1

2!1!

∂ 3f

∂x2∂y(a, b) (x − a)2(y − b)

+ 1

1!2!

∂ 3f

∂x∂y2(a, b) (x − a)2(y − b) +

1

0!3!

∂ 3f

∂y3(a, b) (y − b)3.

n = 3 p = (a,b,c)

T (a,b,c),k(f )(x,y,z ) =k

l=0

i,j,k≥0i+ j+k=l

1

i! j! k!

∂ lf

∂xi∂y j∂z k(a,b,c)(x − a)i(y − b) j(z − c)k.



k = 2

T (a,b,c),2(f )(x,y,z ) = f (a,b,c) + ∂ f

∂x(a,b,c) (x − a) +

∂ f

∂y(a,b,c) (y − b) +

∂ f

∂z (a,b,c) (z − c)

+ 1

2!0!0!

∂ 2f

∂x2 (a,b,c) (x − a)2 + 1

1!1!0!

∂ 2f

∂x∂y (a,b,c) (x − a)(y − b)

+ 1

1!0!1!

∂ 2f

∂x∂z (a,b,c) (x − a)(z − c) +

1

0!2!0!

∂ 2f

∂y2(a,b,c) (y − b)2

+ 1

0!1!1!

∂ 2f

∂y∂z (a,b,c) (y − b)(z − c) +

1

0!0!2!

∂ 2f

∂z 2(a,b,c) (z − c)2.

n = 2 n = 3

f C2

J (f ) =∂f

∂x∂f ∂y

J (f ) =

∂f ∂x

∂f ∂y

∂f ∂z

, H (f ) =

∂ 2

f ∂x2 ∂ 2

f ∂x∂y∂ 2f ∂y∂x

∂ 2f ∂y2

, H (f ) =

∂ 2f

∂x2

∂ 2f

∂x∂y

∂ 2f

∂x∂z∂ 2f ∂y∂x

∂ 2f ∂y2

∂ 2f ∂y∂z

∂ 2f ∂z∂x

∂ 2f ∂z∂y

∂ 2f ∂z2

.

n ≥ 4

C2

f

f p

T p,2(f )(x) = f ( p) + J (f ) p (x − p) + 1

2! (x − p)t H (f ) p (x − p) (vt

v).

n = 2 n > 2

T (a,b),2(f )(x, y) = f (a, b) +

∂f ∂x

∂f ∂y

(a,b)

x − ay − b

+ 1

2!

x − a y − b

∂ 2f ∂x2

∂ 2f ∂x∂y

∂ 2f ∂y∂x

∂ 2f ∂y2

(a,b)

x − ay − b

.

· g : (−δ, δ ) → R Cl P l(t)

l t0 = 0

Q : Rn → R x ∈ B ( p, 1)

|Q(x)| ≤ C x − pm

C > 0 m ∈ N

f (x) = g(Q(x)) U = {x ∈ Rn : |Q(x)| < δ }

p f (x) = P l(Q(x)) + ox→ p(x − plm)



L = limx→ p

f (x) − P l(Q(x))

x − pml = 0

limt→0

g(t) − P l(t)

tl = 0 G : U → R

G(x) =

g(Φ(x)) − P l(Φ(x)Φ(x)l

, Φ(x) = 0

0, Φ(x) = 0

U G( p) = 0 x ∈ B( p, 1) ∩ U f (x) − P k(Q(x))

x − pml

= |G(x)| |Φ(x)|x − pml

≤ C |G(x)| → 0, x → p,

L = 0

k = 5 f (x, y) = ex4−y2

p = (0, 0)

f (x, y) = g(Q(x, y)) g(t) = et Q(x, y) = x4 − y2 f

C∞(R2)

|Q(x, y)| = |x4 − y2 ≤ x4 + y2 ≤ (x, y)4 + (x, y)2 ≤ 2(x, y)2

(x, y) < 1 m = 2 f

et l = 3 ≥ k/m = 2.5

P 3(t) = T 3,0(et) = 1 + t + t2

2! + t3

3!

f (x, y) = P 3(Q(x, y)) + o(x,y)→(0,0)((x, y)6)

P 3(Q(x, y)) = 1 + (x4 − y2) + (x4 − y2)2

2 +

(x4 − y2)3

6

= 1

−y2 + x4 +

y4

2

+ o(x,y)

→(0,0)(

(x, y)

5).

f (x, y) = 1 − y2 + x4 + y4

2 + o(x,y)→(0,0)((x, y)5)

T 3,(0,0)(f )(x, y) = 1 − y2 + x4 + y4

2

h(x, y) = ex4−y2+5 h(x, y) = e5f (x, y) f (x, y)

(T 3,(0,0)h)(x, y) = e5(T 3,(0,0)h)(x, y) = e5

1 − y2 + x4 +

y4

2

.

(0, 1, 0)

f (x,y,z ) = (x + 2z )cos(x + y)

y .

T 2,(0,1,0) = x + 2z − x(y − 1) − 2(y − 1)z



t0 ·

t0

f

f (t0) = 0 f (t0) > 0, t0

f (t0) = 0 f (t0) < 0, t0 f (t0) = 0

f t0 f (t) = t4 t0 = 0

f (t) = t3 t0 = 0

f (t0) = 0 f (t) =f (t0) + 1

2f (t0)(t − t0)2 + ot→t0((t − t0)2) f (t0) > 0

t0

12f (t0)(t − t0)2 + ot→t0((t − t0)2) > 0 f (t) > f (t0) t = t0

f t0

f (t0) > 0 t0

12

f (t0)(t−t0)2+ot→t0((t−t0)2) < 0 f (t) < f (t0) t = t0 f

t0

A Rn f : A −→ R

p, r > 0 f (x) ≤ f ( p) x ∈ B( p, r) ⊂ A

f p, r > 0 f (x) ≤ f ( p)

x ∈ B( p, r) ⊂ A

f A

A Rn

f : A ⊂ Rn −→ R

p ∈ A Jf p = 0

A

Rn

, f : A −→ R

A

p

f p f.

g(t) = f (a + tu) (−ε, ε) ε > 0

t = 0. g(0) = df a(u) = 0.

u df a = 0.

f : A → R C2

A ⊂ Rn J (f ) p = 0 p ∈ A

H (f ) p H (f ) p(x − p, x − p) > 0

x = p f p

H (f ) p

H (f ) p(x − p, x − p) < 0

x = p f p

f p H (f ) p

H (f ) p(x − p, x − p) ≥ 0

f p H (f ) p

H (f ) p(x − p, x − p) ≤ 0

H (f ) p f

p



f p H (f ) p f

A

•

•

•

•

•

n ∆1, · · · , ∆n A n × n

∆1 = a11, ∆2 = det

a11 a12

a21 a22

, · · · , ∆n = det(A) = det

a11

· · · a1n

an1 · · · ann

.

M

• M ∆ j M

• M (−1) j∆ j > 0, j = 1,...,n, ∆1 <0, ∆2 > 0, ∆3 < 0,...

• ∆ j

M

• 2×2 M

ai,i ai,j

a j,i a j,j

M ∆2 < 0 M



n = 2

M =

a bb c

, u = (s, t) = 0 M (u, u) = as2 + 2bst + ct2.

a > 0 ac

−b2 > 0, c > 0

M (u, u) = as2 + 2bst + ct2 = (√

as + b/√

at)2 + ac − b2√

a t2 > 0.

a < 0 ac − b2 > 0, c < 0

M (u, u) = as2 + 2bst + ct2 = −(√ −as − b/

√ −at)2 − ac − b2√ −a t2 < 0.

ac−b2 < 0, a > 0 M (u, u) = as2+2bst+ct2 = (√

as+b/√

at)2− b2−ac√ a t2,

a < 0 a = 0.

•

M = 1 2 0

2 5 00 0 4

.

∆1 = 1, ∆2 = 1 2

2 5 =

1, ∆3 = det(M ) = 4 M

• M =

−1 2 0

2 −5 00 0 −4

. ∆1 = −1, ∆2 =

−1 22 −5

= 1, ∆3 =

det(M ) = −4 M

• M =

1 2 0

2 6 00 0 −1

. ∆1 = 1, ∆2 =

1 22 6

= 2, ∆3 =

det(M ) =

−2 M

M

M

det

6 00 −1

= −6 < 0

M

f (x, y) = −x3 + 4xy − 2y2 + 1.

J (f )(x,y) = (−3x2 + 4y 4x − 4y), H (f )(x,y) =

−6x 44 −4

.

f x = −3x2 + 4y = 0

f y = 4x − 4y = 0

p1 = (0, 0)

p2 = (4/3, 4/3)



H (f ) p1 =

0 44 −4

, ∆2 = −16 < 0,

H (f ) p2 =−8 44 −4

, ∆1 = −8 < 0, ∆2 = 16 > 0.

p1 p2

f (x,y,z ) = x2z 2 + 6x3 + y2 −2yz.

J (f ) = (2xz 2 + 18x2 2y − 2z 2x2z − 2y), H (f ) =

2z 2 + 36x 0 4xz

0 2 −24xz −2 2x2

J (f ) = 0 f p1 = (0, 0, 0),

p2 = (−1, 3, 3)

p3 = (−1, −3, −3).

H (f ) p1 =

0 0 0

0 2 −20 −2 0

,

f (x, 0, 0) = 6x3 (0, 0, 0),

u = (0, s , t)

p1

H (f ) p2 =

−18 0 −120 2 −2

−12 −2 2

. ∆2( p2) = −36 < 0,

p2 H (f ) p2(e1, e1) = −18 < 0 H (f ) p2(e2, e2) =2 > 0.

p3

f (x, y) = 5x4 + 4x2y + y2 + 7.

J (f )(x,y) = (20x3 + 8xy 4x2 + 2y), H (f )(x,y) =

60x2 + 8y 8x

8x 2

.

(0, 0). H (f )(0,0) =

0 00 2

f (x, y) = x4 + (2x2 + y)2 + 7 ≥ 7 = f (0, 0).

f (0, 0)

f (x,y,z ) = 2x3 − 3x2 + 2y3 + 3y2 + z 2

(1, 0, 0) (0, 0, 0), (0, −1, 0) (1, −1, 0).

f (x, y) = 16x3 + 9x2y −30x + 9/y.



f

Ck

Rn

f : (a, b) ⊂ R → R Ck f (c) > 0

c ∈ (a, b) f (c) < 0) ε > 0

f (x) > 0 x ∈ (c − ε, c + ε) ⊂ (a, b) f (c − ε, c + ε)

f : (c − ε, c + ε) → (f (c − ε), f (c + ε))

f −1 : (f (c − ε), f (c + ε)) → (c − ε, c + ε) Ck

(f −1)(f (x)) = 1

f (x)

f (x) > 0 x ∈ (a, b)

f : (a, b) → (f (a), f (b)) Ck

A Rn f : A −→ Rn

Ck A k ≥ 1 a ∈ A det(df a) = 0

U ⊂ A a

• V = f (U ) Rn

• f : U → V

• f −1 : V → U

Ck V d(f −1)f (a) = (df a)−1

f −1 f (a)

f a

•

det(df a)

f

a

• U V

Rn. f : U −→ V Ck−

U V f f Ck(U ), f −1

Ck(V ).

f Ck

f a f Ck

a



• f : U → V

Ck k ≥ 1, det(df x) = 0, x ∈ U

1 = det(Id) = det(d(f −1 ◦ f )x) = det(d(f −1)f (x)) det(df x).

•

det(df x) = 0

f

Ck f : R → R

f (x) = x3 f : R2 → R2

f (x, y) = (x3, y)

· A Rn, f : A −→ Rn C1(A)

det(df x) = 0 x A, f

A Rn.

B A p ∈ f (B)

f (B) p = f (b) b

∈ B

U ⊂ B b f (U ) ⊂ f (B) f (b) f (b)

f (B)

A Rn, f : A −→ Rn

Ck(A) det(df x) = 0

x A f f Ck A f (A)

f (A) f −1

Ck

f (A)

•

f : (a, b) ⊂ R −→ R C

1

(a, b)

f : (a, b) −→ f (a, b)

f : A ⊂ Rn −→ Rn det(df x) = 0

A f : A −→ f (A)

f : R2 −→ R2 f (x, y) = (ex cos y, ex sin y),

det(df (x,y)) = det

ex cos y −ex sin yex sin y ex cos y

= e2x = 0,

f (x, y + 2π) = f (x, y).

f (x, y) = (x cos y, x2 + 2y + 1) C∞

(0, 0) (0, 1), f C∞(R2) det(df (0,0)) =

det

1 00 2

= 2.

(u, v) = f (x, y) (0, 1)

d(f −1)(u,v) =

cos y −x sin y

2x 2

−1

= 1

2cos y + 2x2 sin y

2 x sin y−2x cos y

.



d(f −1)(0,1) =

1 00 1/2

.

(0, 1)

u v.

∂ (f −

1)1∂u = 2

2cos y+2x2 sin y , ∂ (f −

1)1∂v = x sin y

2cos y+2x2 sin y ,∂ (f −1)2

∂u = −2x

2cos y+2x2 sin y, ∂ (f −1)2

∂v = cos y

2cos y+2x2 sin y.

r R2 (x, y) = (r cos(θ), r sin(θ)), −π ≤ θ < π.

ϕ : (0, ∞) × (−π, π) −→ Q2 Q2 = R2 \ {(−∞, 0] × {0}}, ϕ(r, θ) = (r cos(θ), r sin(θ)) = (x, y).

C∞

det(dϕ) =

cos(θ) −r sin(θ)sin(θ) r cos(θ)

= r = 0.

f C∞ (0, ∞) × (−π, π) Q2

f (x, y) = ey −2y −x2 = 0

p = (−1, 0) f C∞

p

R2

∂f

∂y( p) = ey − 2| p = −1 = 0 ε > 0

g : (−1−ε, −1+ε) → R

C∞ f (x, g(x)) = 0 x ∈ (−1−ε, −1+ε)

g(−1) = 0

eg(x) − 2g(x) − x2 = 0 g

−1 g(x)eg(x)−2g(x)−2x = 0 g(x) =

2x

eg(x) − 2

x −1 g(−1) = 2

g

A Rn×Rm f : A → Rm

Ck A p = (a, b) ∈ A f (a, b) = 0.

∂ (f 1, . . . , f m)

∂ (y1, . . . , ym)(a, b) = det

∂f j∂yi

(a, b)1≤i,j≤m

= 0,

U ⊂ Rn a V ⊂ Rm

b U × V ⊂ A

g : U −→ V, Ck(U ) x ∈ U,y = g(x) V f (x, y) = 0

f (x, y) = 0 y = g(x) Ck

U a g(a) = b

f (x,y,z ) = y + xez + z = 0.



• f (x,y,z ) = y + xez + z = 0 z = g(x, y) C∞, (0, 0, 0). x

x1 y x2 z y1. f C∞(R3), (0, 0, 0)

f z(0, 0, 0) = 1 = 0.

A (0, 0), B 0 g : A −→ B

C∞(A) z = g(x, y) f (x,y,z ) = 0 B.• g.

y + xeg(x,y) + g(x, y) = 0. z = g(x, y),

(y + xeg(x,y) + g(x, y))x = eg(x,y) + xeg(x,y)gx(x, y) + gx(x, y) = 0,

gx(x, y) = − ez

xez + 1.

(y + xeg(x,y) + g(x, y))y = 1 + xeg(x,y)gy(x, y) + gy(x, y) = 0,

gy(x, y) = − 1

xez + 1.

• g(x, y) (0, 0).

T 2,(0,0)g(x, y) = g(0, 0) + gx(0, 0)x + gy(0, 0)y + 1

2gxx(0, 0)x2 + gxy(0, 0)xy +

1

2gyy(0, 0)y2.

g(0, 0) = 0, gx(0, 0) = −1 gy(0, 0) =−1.

(y + xeg(x,y)

+ g(x, y))xx = 2eg(x,y)

gx(x, y) + xeg(x,y)

(gx(x, y))2

+ xeg(x,y)

gxx(x, y)+ gxx(x, y) = 0,

(y + xeg(x,y) + g(x, y))xy = gy(x, y)eg(x, y) + xeg(x,y)gx(x, y)gy(x, y) + xeg(x,y)gxy(x, y)

+ gxy(x, y) = 0,

(y + xeg(x,y) + g(x, y))yy = xeg(x,y)(gy(x, y))2 + xeg(x,y)gyy(x, y) + gyy(x, y) = 0.

(x, y) = (0, 0),

−2 + gxx(0, 0) = 0, gxy(0, 0) = 1, gyy(0, 0) = 0.

T 2,(0,0)g(x, y) = −x − y + x2 + xy.

f 1(x,y,u,v) = xu6 + y2v3 + 1 = 0f 2(x,y,u,v) = xy3 + uv2 = 0

.

• u = u(x, y), v = v(x, y)

p = (x0, y0, u0, v0) = (0, 1, 0, −1).



p

C∞(R4).

∂ (f 1, f 2)

∂ (u, v) ( p) =

6xu5 3y2v2

v2 2uv p = 0 31 0 = −3.

A (0, 1) B (0, −1) u v.

• ux(x, y) vx(x, y).

(xu6 + y2v3 + 1)x = u6 + 6xu5ux + 3y2v2vx = 0(xy3 + uv2)x = y3 + v2ux + 2uvvx = 0

.

ux = 3y5v2 − 2u7v

12xu6v − 3y2v4, vx =

u6v3 − 6xy3u5

12xu6v − 3y2v4.

uy(x, y) vy(x, y).

(xu6 + y2v3 + 1)y = 2yv3 + 6xu5uy + 3y2v2vy = 0(xy3 + uv2)y = 3xy2 + v2uy + 2uvvy = 0

.

uy = 9xy4v2 − 2u7v

12xu6v − 3y2v4, vy =

2yv5 − 18x2y2u5

12xu6v − 3y2v4 .

ux(0, 1), vx(0, 1), uy(0, 1), vy(0, 1), x = 0, y =1, u = 0, v = −1

ux(0, 1) = −1, vx(0, 1) = 0, uy(0, 1) = 0, vy(0, 1) = 2/3.

u

v (0, 1),

x

3vx(0, 1) = 01 + ux(0, 1) = 0

,

vx(0, 1) = 0ux(0, 1) = −1.

• h(x, y) = (u(x, y), v(x, y)) C∞−

(0, 1) (0, −1). h

C∞ (0, 1) h(0, 1) = (0, −1)

det(dh(0,1)) = 0.

det(dh(0,1)) = det−

1 00 2/3

= −2/3 = 0.

• u (0, 1)

du(0,1) = 0,

2xey+t + y + x = 0x + t + ey = 1

.



x = x(t) y = y(t). x(t)

x(t) t = 0

Rn

M ⊂ Rn Rn

0 < m < n Ck M Ck− Rn

k

p M U p n − m f j Ck(U ),

M ∩ U = {x ∈ U ; f m+1(x) = . . . = f n(x) = 0}.

f j

p, ∇f j( p), j = m + 1, . . . , n ,

•

f = (f m+1, . . . , f n) rang(df p) = n − m.• f (x)

p

∂ (f m+1,...,f n)∂ (xm+1,...,xn)

( p) = 0.

A × B p, f (x) = 0 xm+1, . . . , xn : A −→ B

x1, . . . , xm M ∩(A×B).

Ck

− Rn

Ck.•

∇f j( p), j = m + 1, . . . , n ,

p

•

ax + by + cz + d = 0 (a,b,c) = (0, 0, 0)

C∞

R3.

• T : Rn −→ Rm, 0 < m < n,

m, M = {x ∈ Rn; T (x) = 0} C∞

Rn n − m.

rang(dT ) = rang(T ) = m.

C∞

• x − a = r > 0 n − 1 C∞

Rn. S = {x ∈ Rn; f (x) = x − a2 − r2 = 0},

∇f (x) = (2(x1 − a1), . . . , 2(xn − an)). x ∈ S, ∇f (x) = 2r

∇f (x) = 0.• f : A ⊂ Rn −→ Rm

Ck(A), Ck−

Rn+m n.

Graf (f ) = {(x, y) ∈ A × Rm; f (x) − y = 0}.



•

·

1 R2 Rn

• ·

Rn

f : [a, b] → R

C1 [a, b] f

[a, b] f

[a, b] (a, b)

f (a, b)

x1, · · · , x j f (a, b)

f [a, b] M = max{f (a), f (x1), · · · , f (x j), f (b)}

m = min{f (a), f (x1), · · · , f (x j), f (b)}

f (x) = x3 − 3x2 + 1 x1 = 0, x2 = 2

f [1, 3] M = max{f (1), f (2), f (3)} = max{−1, −3, 1} = 1 3 m = min{f (1), f (2), f (3)} =min{−1, −3, 1} = −3 2

h

C1 K

K

K

p1, · · · , p j h K o

K

q 1, · · · , q m

h pi

q i h K



M

Rn m Ck. U p ∈ M U ∩ M =

{x ∈ U ; f m+1(x) = . . . = f n(x) = 0} f j ∈ Ck(U )

p

h : U −→ R

C1(U ). h

M ∩ U p, λm+1, . . . , λn

∇h( p) = λm+1∇f m+1( p) + . . . + λn∇f n( p).

h(x, y) = c1

h(x, y) = c3=mınim absolut de h sobre A

h(x, y) = c2

h(x, y) = c4

h(x, y) = c5

h(x, y) = c6=maxim absolut de h sobre A

c1 < c2 < c3 < c4 < c5 < c6 < c7

h(x, y) = c7

A = {(x, y) ∈ R2 : f (x, y) = 0}

p

q

∇(f )( p)

∇(h)( p)

∇(f )(q )

∇(h)(q )

cor es e n vell e corresponents als valors

•

x = (x1, . . . , xn) λ = (λm+1, . . . , λn)

F (x, λ) = h(x) − λm+1f m+1(x) − . . . − λnf n(x).

h M

p λ p (∇F )( p, λ p) = 0.

h(x, y) = xy

(0, 0)

√ 2

h(x, y) = xy f (x, y) = x2 + y2 − 2 ≤ 0

h R2

B((0, 0), √ 2)

h

B((0, 0), √ 2)

B((0, 0),√

2) Jh(x, y) = (y x)

h p = (0, 0) B((0, 0),√

2)

x2+y2 = 2 R2



F (x,y,λ) = xy − λ(x2 + y2 − 2)

x y F

y − 2λx = 0x

−2λy = 0

−(x2 + y2 − 2) = 0 x2 − y2 = 0

x2

+ y2

= 2 q 1 = (1, 1), q 2 = (1, −1)

q 3 = (−1, 1), q 4 = (−1, 1) .

h f ( p) = 0 f (q 1) = f (q 4) = 1 f (q 2) =f (q 3) = −1

h B((0, 0),√

2) 1 (1, 1) (−1, −1)

h B((0, 0),√

2) −1

(1, −1) (−1, 1)

h(x, y) = xy

K = {(x, y) ∈ R2 : x2 + y2 ≤ 2, x ≥ 0, y ≥ 0}

h(x, y) = xy f 1(x, y) = x2 + y2 − 2 ≤ 0 f 2(x, y) = x ≥ 0 f 3(x, y) = y ≥ 0

h R2

K x2+y2−2 ≤ 0

x ≥ 0 y ≥ 0

h

K

h

K o

= {(x, y) ∈ R2

: x2

+ y2

< 2, x > 0, y > 0}

h p = (0, 0) K o

• h K 1 ={(x, y) ∈ R2 : x2 + y2 − 2 = 0, x > 0, y > 0}

F (x,y,λ) = xy − λ(x2 + y2 − 2) x y F

q 1 = (1, 1), q 2 = (1, −1), q 3 = (−1, 1), q 4 = (−1, 1).

x > 0 y > 0

q 1 = (1, 1)

• K 2 = {(x, y) ∈ R2 : x2 + y2 − 2 ≤ 0, x =0, y ≥ 0} h(x, y) = 0

• K 2 = {(x, y) ∈ R2 : x2 + y2 − 2 ≤ 0, x ≥0, y = 0} h(x, y) = 0

h f (1, 1) = 1 f (0, y) = f (x, 0) = 0



h K 1 (1, 1)

h K 0 (x, 0) (0, y) x2+y2 ≤ 2 x ≥ 0 y ≥ 0 (x, 0) 0 ≤ x ≤ √

2

(0, y)

0 ≤ y ≤ √ 2

h(x,y,z ) = x2 +y2 + z 2 + x + y + z K = {(x,y,x) ∈ R3 : x2 + y2 + z 2 ≤ 4, z ≤ 0}.

• K

(0, 0, 0) K = Φ−1((−∞, 4]×(−∞, 0]) Φ(x,y,z ) = (x2 + y2 + z 2, z ). Φ : R2 −→ R2

(−∞, 4] ×(−∞, 0]

R2, K K

h K,

K.

• K K Rn

Rn,

K = K 0 ∪ K 1 ∪ K 2 ∪ K 3

K 0 = {(x,y,z ) ∈ R3 : x2 + y2 + z 2 < 4, z < 0},

K 1 = {(x,y,z ) ∈ R3 : x2 + y2 + z 2 < 4, z = 0} ⊂ F r(K ),

K 2 = {(x,y,z ) ∈ R3 : x2 + y2 + z 2 = 4, z < 0} ⊂ F r(K ),

K 3 = {(x,y,z ) ∈ R3 : x2 + y2 + z 2 = 4, z = 0} ⊂ F r(K ).

K 0 Rn, K 1 R3

C∞−

K 2

C∞−

K 3 R3

C∞−

h

K 0 = {(x,y,z ) ∈ R3; x2 + y2 + z 2 <4, z < 0}. ∇h(x,y,z ) = (2x + 1, 2y +1, 2z + 1) = (0, 0, 0), p = (−1/2, −1/2, −1/2).

K 1 = {(x,y,z ) ∈ R3 : x2+y2+z 2 <4, z = 0}. K 1 h(x,y,z ) h1(x, y) = h(x,y, 0) =

x2 + y2 + x + y

x2 + y2 < 4.

∇h1(x, y) = (2x + 1, 2y + 1) = (0, 0), (x, y) =(−1/2, −1/2).

(−1/2, −1/2, 0). K 2 = {(x,y,z ) ∈ R3 : x2+y2+z 2 =

4, z < 0}. h



h(x,y,z ) =4 + x + y + z.

F (x, y, z, λ) = 4 + x + y + z − λ(x2 + y2 + z 2 − 4).

h

F

1 − 2λx = 01 − 2λy = 01 − 2λz = 0

x2 + y2 + z 2 − 4 = 0

(x, y, z, λ) = (2/√

3, 2/√

3, 2/√

3,√

3/4) (x, y, z, λ) =(−2/

√ 3, −2/

√ 3, −2/

√ 3, −√

3/4).

q 1 = (2/

√ 3, 2/

√ 3, 2/

√ 3), q 2 = (−2/

√ 3, −2/

√ 3, −2/

√ 3).

z < 0

K 3 = {(x,y,z ) ∈ R3 : x2+y2+z 2 =4, z = 0}. h h3(x, y) = 4 + x + y,

(x, y) x2 + y2 = 4.

h3 x2 + y2 = 4.

F (x,y,λ) = 4 + x + y − λ(x2 + y2 − 4). ∇F = (1 − 2λx, 1 −2λy, −x2 − y2 + 4) (x,y,λ) = (2/

√ 2, 2/

√ 2,

√ 2/4) (x,y,λ) =

(−2/√

2, −2/√

2, −√ 2/4),

(2/√

2, 2/√

2, 0) (−

2/√

2,−

2/√

2, 0).•

(x,y,z ) f (x,y,z )(−1/2, −1/2, −1/2)

(−1/2, −1/2, 0)

(−2/√

3, −2/√

3, −2/√

3) 4 − 2√

3(√

2,√

2, 0) 4 + 2√

2(−√

2, −√ 2, 0) 4 − 2

√ 2

• h K 4 + 2√

2

(√

2,√

2, 0)

h K −3/4

(−1/2, −1/2, −1/2).

p = (−2, 0) y = x2−2x+1. M = {(x, y) ∈ R2; f (x, y) = x2−2x+1−y = 0}

R2 C∞ R

M



d( p, M ) = inf q∈M p − q

d( p, M )2 (x + 2)2 + y2 x2 − 2x + 1 − y = 0

F (x,y,λ) = (x + 2)2 + y2 − λ(x2 −

2x + 1 − y).

2(x + 2) − λ(2x − 2) = 02y + λ = 0

−(x2 − 2x + 1 − y) = 0

,

(x + 2) + 2(x2 − 2x + 1)(x − 1) = 2x3 − 6x2 + 7x = 0

x = 0. (0, 1) d( p, M ) =

√ 5.

y x2

− 2x + 1 = (x

− 1)2 h(x) = d( p, M )2 =

(x + 2)2 + (x − 1)4 h(x) +∞ x → ±∞

h(x) = 2x+4−4(x−1)3 =4x3 − 12x2 + 14x = 2x(2x2 − 6x + 7) = 0 x = 0

h(0) = 14 > 0 h 0

d( p, M ) =

h(0) =√

5 (0, 1)



Ck

·



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Calculo de Diversas Variables

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