Calculo II C Difrerencial

8/18/2019 Calculo II C Difrerencial

1/186

ESCUELA DE CIENCIAS BSICAS E INGENIERA

ASIGNATURA: C II (C D)

CORPORACIN UNIVERSITARIA REMINGTON

DIRECCIN PEDAGGICA

E aa dad d a Cac Uaa R (CUR), aa da d a CUR

d a.

2011


2/186

Cac Uaa R Dcc Pdaca

Cc II (Cc Dca) P. 5

Corporación Universitaria Remington - Calle 51 51-27 Conmutador 5111000 Ext. 2701 Fax: 5137892. Edificio Remington

Página Web: www.remington.edu.co - Medellín - Colombia

CRDITOS

E d d d d a aaa Cc II (Cc Dca) dad d a Cac UaaR. La ada d d aca dc d a a ca a bbaa. E cd d d d a d dc d a a a.

E aa dca d a c cc cca.

AUTOR

E C GI Ecca d a Udad Naca d Cba d Md. Dad D Cca.Cac Uaa R Ecaa Maca Acada Pa C. CacUaa R Dc d a aac R Dc d La Cac Uaa d Cca Da (UNICIENCIA)Dc d aca dcac bca da a Ic Edcaa Kd . ...

N: a cc (d aa ba ca) N ab cd ad cc, a c d aa; ca ca d da abdad a a Cac Uaa R, dca c cab.

RESPONSABLES

D E D C B ID. Mac Sda

D POca T [email protected] M MAca Rca [email protected]

GRUPO DE APOO

P U M MEDICIN Y MONTAJE

Pa . Fb d 2011. Dc Rad

Ea ba bcada ba a cca CaC. RccN CcaCa Ia 2.5 Cba.


3/186

Cac Uaa R Dcc Pdaca

Cc II (Cc Dca) P. 6



TABLA DE CONTENIDO

1. MAPA DE LA ASIGNATURA ............................................................................................. 8

2. FUNCIONES ................................................................................................................. 10

2.1. Cc dc ...................... ............................ ....................... ........................ . 12

2.1.1. Cc d c. ....................... .......................... ......................... ............................ . 12

2.1.2. D a. ........................... .......................... ......................... ............................ .... 13

2.1.3. Ic: (://.b.c/ac?=1WS4FKbA) .......................... .......... 29

2.1.4. Cdad: (://.b.c/ac?=Ad84VSc) ........................... ....... 32

2.2. Cacac d a c ................... ........................... ....................... ....................... 39

2.2.1. Fc Pca ......................... ........................... ........................ .......................... .... 39

2.2.2. Fc a c d ad: ...................... .............................. .................... . 40

2.2.3. Fc Cadca Fc D Sd Gad ............................ ..................... ........... 45

2.2.4. Fc Raca ......................... ....................... ............................ ................... ............... 52

2.2.5. Fc Iaca ......................... ........................... ........................ ........................ ......... 60

2.2.6. Fc Dda P Ta ................... ........................... ....................... ....................... 85

2.3. Acac ........................ ...................... ............................ .................... ....................... 91

3. LMITES ..................................................................................................................... 109

3.1. Dc Ia d L ................... ........................... ....................... ...................... 111

3.2. L aa Ea L ...................... ........................... ....................... ...................... 118

3.2.1. E d a ca a a a ca ........................ ....................... ............. 119

3.2.2. L aac abaca. .................. .............................. .................... .............. 122

3.3. L cdad ....................... ...................... ............................... ............................ 131

3.3.1. Cdad : ..................... ............................ ..................... .......................... 131

4. DERIVADA ................................................................................................................. 136

4.1. Cc dc acad c a dada ............................ ...................... ...... 137

4.1.1. Ra d cab d a d cab a. ................................ ........... 137

4.2. L aa da .......................... ...................... ............................... ............................ 142

4.2.1. Dada d a ca: ..................... ........................... ....................... ...................... 142


4/186

Cac Uaa R Dcc Pdaca

Cc II (Cc Dca) P. 7



4.3. Acac ac d a dada. ......................... ............................ ................. 157

4.3.1. Acac a: ...................... ........................... ....................... ...................... 1575. PISTAS DE APRENDIAJE ............................................................................................ 185

6. GLOSARIO ................................................................................................................. 187

7. BIBLIOGRAFA............................................................................................................ 188


5/186

Cac Uaa R Dcc Pdaca

Cc II (Cc Dca) P. 8



1. MAPA DE LA ASIGNATURA

NOMBRE DE LA ASIGNATURA

PROPSITO GENERAL DEL MDULO

E cc a aa d cda a abac d d a ac

da d c. E a d a ccaac acac d d

cc d a acc a d a d cc.

E bca c CLCULO DIFERENCIAL d d a dada,

ac acac a d a d cc, dd a Dada

c a a d cab.

E da cc a d cca aa da c acac a c

aa d ac bca.

E da d cc dca daa cca c a d

ac bca dd a aa daa cc.

C d d a a da a aa caa a

d d a aa da a a d cc.

OBJETIVO GENERAL

Eda ca c, abac aac acad a d d

ac d ac bca cc aac dca,

dcad a caacca d d d ac c a cadad c

a a da da d a abdad a da d dcac d ac

bca d cdadad d a c a d a.

OBJETIVOS ESPECFICOS

Aaa cc d c da ac, c aaac a a dac d ac bca da a

ac.

Ed cc d acac c a aac a d d a

dada.


6/186

Cac Uaa R Dcc Pdaca

Cc II (Cc Dca) P. 9



UNIDAD 1

F

Ta a aba caba.

S ca cc d: Fc, ac d c, d, a, a d a c, cdad,

cc dcc, d c. S ca a d ac d aa c

c; cc d c a a d da c d a c.

C

S ca a cacac d a c, a a d dca cada aa d c, c da

d, c aa ac ca.

2.1.1 Acac

S a a a d da, c c, d ac ca ba

ad a dad d dca c.

UNIDAD 2

L

Ta a aba caba.

Raa a dcc b d cc d , dd cc d da

dc a.

S ca a a d aa ad a d , a a d a a a

a d a dac d a a c b c ad a

acac a acaac.

UNIDAD 3

D

Ta a aba caba.

S ca cc d a d cab d a d cab a c a aac a

cc d dada. S ca a a d caca a dada d c ad a a d d

cc ca. S ca a d ac ada aa a a dada, a c a

ac ada aa a dada d d .S ca a d a ada aa da, aa a dac d a.S ca cc d dada c a a d cab aa d a c aca cc

aa ca ac bca d a d cab a, ca ( aa, c

aa) a d a c, d , aad d ca

acad.


7/186

Cac Uaa R Dcc Pdaca

Cc II (Cc Dca) P. 10



2. FUNCIONES

://.b.c/ac?=ZMAU&a=ad

O G

Aaa cc d c da ac, c a aac

a a dac d ac bca da a ac.

O E

Cd cad d c, cc d a; cc

dcc, cdad, aad a c aa d a d a aab

dd.

Idca d d c, dad d, a, abad

ca d aa c ad d a dca cc, dcc,

cdad, dcdad, d c c , .

Mda d d ac bca caa ad c.


8/186

Cac Uaa R Dcc Pdaca




P I

Paa cada a d a a c a dad, a a

aa.

1. La c d a cac:

( ) ( )V F

2. L aca a d cb c cc d

, dad a d d c.

( ) ( )V F

3. L aa ad a a aa d a.

( ) ( )V F

4. La a d a .

( ) ( )V F

5. E a d a .

( ) ( )V F

6. E ad d .

( ) ( )V F

7. A dd d d c, c, ad a a c.

( ) ( )V F

8. Paa cada a cc a c cca:

9. E ad d :

a. 6−


9/186

Cac Uaa R Dcc Pdaca




b. 9−

c. 9

d. 9

1

1. La cac c c:

a. = 5

b. = 5

c. = 5 = 5

d. = 25

2. La : a a:a. 8

b. 0

c. 1.

d. Idad.

3. Dada a :0

25 d aa :

a. E dada.

b. N .

c. E a a 25

d. E a a c.

2.1. C

2.1.1. C .

://.b.c/ac?=_3c8&a=ad

://.b.c/ac?=2NdC4NUU&a=ad

Ua c a a dcb a a a cadad dd d a; , a da , a daca cda a c d . (Sa,La, & Wa, 2001, .130).


10/186

Cac Uaa R Dcc Pdaca




S d dc, a da a c a abaca dca a

ac d aab. E cc dca da a c

aca d aab.

Ua d a aab a c aab dd adca aa a

a . Oa a a aa a aab dd : cad aa d

dcc, aa .

La a aab a aab dd adc aa a a .

2.1.1.1 N .

Paa dca a aab dd ca d a aab dd

( , dd d a aab dd ), a a ac:

)( x f y =

Q a a d

S d a a a.

)( xg y = )(t c y =

)( z f y = )(qc y =

La aab dd a d d a.

2.1.2. D .

DOMINIO: E d aa ca c cd d a

d a a aab dd (a ). Td a ad d

a.

RANGO: E a aa ca c cd d

d a a aab dd (a ). A a ab cc c a a

d a c

Q cadad c a a?

La a a a c a d c,

aa (Ra a d a), a d a d c.


11/186

Cac Uaa R Dcc Pdaca




Paa aa b a d ca a a :

L ac cd a d dd d cc d c dd

a ca.

://.b.c/ac?=BF6La9Y

://.b.c/ac?=P4VQLI03U&a=ad

://..c//b/b/a.?=&a=143421

://.b.c/ac?=cNV6Q&a=ad

2.1.2.1 R

S ca aa b aa a a c:

Vba (c a dcc d aaba)

Nca (c a aba d a)

Va (c a ca)

Abaca (c a a ca)

S a a c d a d a ca aa, a d a aa d

a ac a a, aa ad cc adca d a c. (Sa,

1999, .15).

Paa d cc d c a d a d a c,

aac a ac cca cdaa.

S 1:

Paa a ac a da aca a 40 da d Cada

Pbca aa a CORPORACIN UNIVESITARIA RMINGTON,

d:

Q a d a a aab a da aca a?

S:

Ca c cc.

Q a d a a aab da?


12/186

Cac Uaa R Dcc Pdaca




Ca 0 40.

La a dd d da da dd d a a?

S:

La a dd d da, a cc b d da d ab

c a ad a aa.

P a a , ccd a a, d ab a da

c, , a a 3.5 a da c? La a a

d a da.

S aa D a d a d a a aab dd, c

d a ac?

S:

S a a a 40 da c d cd a d

0 40, dc,

[ ]40,0. ∈ x Dom

S aa Ra a d a d a a aab dd, C a ca?

S:

E a cd a a d da a a d b da,

dc, ca 0 5,

[ ]5,0∈ y Rango

E b da a d a d aca

a?


13/186

Cac Uaa R Dcc Pdaca




S:

N b, a a a Maca ca aa ca.

E b a a a cda a d da d?

S:

S b, a , d cd d da a a a a d 5.0 3.0,

2.5, ca a a a.

S 2:

E a bca a a 835 ad, d aa a

ca 1 12 aa a.

1. C aab aca a ac bca?

S:

2. Vaab dd: Ead.

3. Vaab dd: Saa d ad.

4. E b ad a d d d? P ?

S:

5. N b, a a a aa d c aa

aba.

6. E b d cda a d ad d? P

?

S:

7. S b, d a ca a ad .

8. C a aab dd?


14/186

Cac Uaa R Dcc Pdaca




S:

Saa d d cada ad.

C a aab dd?

S:

Ead abaad.

1. C d?

S:

Ca c 835, 835 abaad.

2. C ?

S:

Ca 1 12 aa , d dca .

S 3:La a adaac d a ac aada a (Ub & O, N

ccad, .169)

E c a c d c a aba

P C

Haa 20 U.S. $ 0.20

E 20 50 U.S. $ 0.26

E 50 110 U.S. $ 0.39

E 100 250 U.S. $ 0.85E 250 500 U.S. $ 1.70

E 500 1000 U.S. $ 2.35

E 1000 2000 U.S. $ 3.20

Ca Maa cb a a J, Naaa, La Sba. La caa d J a 15

, La d Naaa a 85 , a d La 90 a d Sba a 525 . Ca:

1. C ca cada caa?


15/186

Cac Uaa R Dcc Pdaca




S:

La caa d J ca U.S. $ 0.20

La caa d Naaa ca U.S. $ 0.39.

La caa d La ca U.S. $ 0.39.

La caa d Sba ca U.S. $ 2.35.

2. E b a d caa cda a?

S:

S b, d c a caa d Naaa La, a aba ca U.S $ 0.39

a d .

3. A a a caa d cd c d?

S:

N b, a a c aa d c a a caa.

4. C d a d aac cca? Jca:a. E d a caa dd d c d a a.

b. E c d a caa dd d d a a.

S:

E c d a caa dd d , a c d a aba, a aa aa

c d caa caa dada d .

5. Q a d a a aab c d ?

S:

Ca a U.S. $ 0.20 U.S. $ 3.20

6. Q a d a a aab d a caa?

S:

Ca a 0 2000 , ba c a a a ca, a

, cd a a a caa, d aaa a a caa.

Ec, a a cca : D c a ada ( c c) aa 2000

a.


16/186

Cac Uaa R Dcc Pdaca




S 4:

La ac a adaac d d a ca d a (Ub & O,

N ccad, .170)

Ca a a caa d ca d 30 c d a 20 c d ac. Rca

cadad d ad cm x a ca a aa c a caa aa, c a

a cca a a1 .

Fa 1: D d a caa

(A. E Cba G. Ma 11 d 2011)

S d :

E a d a caa :

E ac d a caa :

E a d a caa :

Ca a a:


17/186

Cac Uaa R Dcc Pdaca




1. La aa d a caa ?

S:

V = a * ac * a

Sa: .

T :

Ecad a cac cd dcd a:

2. Q aab a ac bca?

S:

I d aab :Aa d a caa, ac d a caa, a d a caa d a caa.

Pd a, ac d a caa dd d a aa d a caa.

Tab d dd d ac, a a aa d a caa.

Obad a c a, d aa dd d a aa d a caa.

La aab d a ac bca :

Vaab dd: Aa d a caa ( ad d cadad a a).

Vaab dd: V d a caa.

1. C a aab dd?

Aa d a caa.

2. C a aab dd?

V d a caa.

3. C d?

Paa da d aa a aba Ec.


18/186

Cac Uaa R Dcc Pdaca




P :

E a ca a d a , ca, d c a .

E a ca cd a () cb a a aa .

=4*A2^3100*A2^2+600*A2.

Dad b a ac d a aba.

A B

Va d

Va d

()

0 0

1 504

2 832

3 1008

4 1056

5 10006 864

7 672

8 448

9 216

10 0

11 176

12 288

13 312

14 224

15 0

16 384

17 952

18 1728

19 2736

20 4000


19/186

Cac Uaa R Dcc Pdaca




S:

Rcd d c d a d a a aab

dd, dc a .

Paa ca aca, db da a a a da abca a caa c

a a d ca d 30 c 20 c.

N ba dca :

C a d a d a a c, d ab caa, b c a caa c

d aa, dc, a aa, a ac d a caa d a a

.

Paa ca a a b ad d a aa Ec, dad a a

bad a a ().

S d aa = 0, b () = 0, a, = 0 c a d d

().

Tab ba () :

Dd = 1 aa = 9, ( = 0 = 10 () = 0)

Dd = 18 () .

Rcd d cd a d a d d a caa a aa.

S d = 18:

E a d a caa a:

E ac d a caa a:

E a d a caa a:

E cd a a d 18 ada.

La d aa da.

P a d d () :

E a dc c , ca a = 0 a

= 10, a d d c ada, aa 10 c 10.

1. C a?


20/186

Cac Uaa R Dcc Pdaca




S:

Rcd a c d a d a a aab dd,

ca a aab dd d a caa.

E d a caa d c, a.

Paa da d a ad Ec:

Paa a b a ca cd a () aa a d 0 10,

d () a a dd c aa 1056 a a a c.

Ra:

E cc a dc c , dc, dacaa 1056 c3.

2. E dd d aa d cadad a a aa d cadad a a

dd d ? J.

S:

S d dd d aa d cadad a a, a ddd d

a aa, a aa d, cd c .

S 5:

U aa a c d da:

Vd a a 2000 a dad c dc d 10 a cada.

Ca a a.

3. S d a a c ?

S:

Pc = 2000 10(1) = 1990.

I = 1990(1) =1990.

4. S d d a C ?

S:

Pc = 2000 10(2) = 1980. P.


21/186

Cac Uaa R Dcc Pdaca




I = 1980(2) = 3960 .

1. S d 10 a C ? C c d a d cada a?

S:

Pc = 2000 10(10) = 1900.

I = 1900(10) = 19000 .

2. Q aab a ac bca?

S:

Sa : N d a dda.

Sa = (): I bd a a d a a.

3. C a aab dd?

S:

I bd a a d a.

4. C a aab dd?

S:

Cadad d a dda.

5. E b a a ac bca ad d?

S:

S.

6. Ec d aca d aa.

S:

Paa c a c d a a d a a, a ca :

I = c d a cad a cadad dda.


22/186

Cac Uaa R Dcc Pdaca




E a aba ba a ccc d a c d aa a a d

a.

Cadad d a

dda

Pc d a I

1 2000 10(1) [2000 10(1)]*1

2 2000 10(2) [2000 10(2)]*2

3 2000 10(3) [2000 10(3)]*3

4 2000 10(4) [2000 10(4)]*4

Q 2000 10 (2000 10)*

C = () bd a d a:

Paa ca a a 9, 10, 11 12 ac a aa Ec.

Cadad

dda

I

bd

0 0

10 19000

20 36000

30 51000

40 64000

50 75000

60 8400070 91000

80 96000

90 99000

100 100000

110 99000

120 96000

130 91000


23/186

Cac Uaa R Dcc Pdaca




140 84000

150 75000

160 64000

170 51000

180 36000

190 19000

200 0

210 21000

220 44000

230 69000

240 96000

250 125000

1. Ba cdc a c ab aa aa?

Cad d 100 a, a d a aa d 100000

d b.

2. Ba cdc aa bd ?

Cad d 100 a a c d 1000 cada a, c d

100000 .

3. C d d a a?

4. C a d a a?

E a ac bca a d a cc db

aa, a cc :Ia d a c.

T d c cacac.

Dac d d.

Gaca d a c.

I . C aa a c a a dcad b

a ca .


24/186

Cac Uaa R Dcc Pdaca




E1:

S 634)( 2 −−== x x x f y

Ha:

−

3

2),0(),2(),1( f f f f

S

Cc d )1( f

Cd d aa )1( f , d da a d y , cd 1= x ; :

Cd 56346)1(346)1(3)1(4)1(,1 2

−=−−=−−=−−=== f y x

5)1( −= f Q dc cd 5,1 −== y x

Cc d )2(− f

2661286)4(386)2(3)2(4)2( 2 −=−−−=−−−=−−−−=− f

26)2( −=− f Q dc cad 26,2 −=−= y x

Cc d )0( f

66)0(306)0(3)0(4)0( 2 −=−−=−−= f

Cc d

3

2 f

3

146

3

4

3

86

9

43

3

86

3

23

3

24

3

2 2

−=−−=−

−=−

−

=

f

E2:

1

4)(,

+

−==

x

x x f ySi

)3(),4(),1(),1(: −− f f f f Halle


25/186

Cac Uaa R Dcc Pdaca




S

2

3

11

41)1( −=

+

−= f

0

5

11

41)1(

−=

+−

−−=− f .

Rcd a d c c a ca c, a

(1) c a d d a c ( = 1 c a d d ()).

050

1444)4( ==

+−= f

2

7

2

7

13

43)3( =

−

−=

+−

−−=− f

E3:

6)(, −== x xg ySi )8(),5(),0(),10(: gggg Halle

S

24610)10( ==−=g

660)0( −=−=g .

Rcd, a a a d a c a a, a,

0= x c a d d d ac.

165)5( −=−=g . 5= x N c a d.

...414213562,1268)8( ==−=g


26/186

Cac Uaa R Dcc Pdaca




Eac aac d c:

://.b.c/ac?=GRSDXA

://.b.c/ac?=BRPI0VcGE&a=ad

://.b.c/ac?=FcS_dQ&a=ad

://.b.c/ac?=FDB19ZG8&a=ad

://.b.c/ac?=DTNX4&a=ad

2.1.3. I: (://../?=1WS4FKA)

U c dd a ca d a c ca cada d .

La cc c ( a) b acd = 0 dad a .

A a d a ca a cc c cd a dd a caca .

La cc c ( a) b acd = 0.

A a d a ca, a cc c dd a ca ca .

NOTA:S, a bca a cc c b , dc, ca

a cc c c

La cac c.

Ua adad aa.

Ra a d a.

D c.

Ca aa.

La d a.

U ca aad a c.

E: D a cc c d cada a d a c.

E1: D c d a c:

67)( 2 +−== x x x f y Solución

C : Hacd = 0


27/186

Cac Uaa R Dcc Pdaca




( ) ( )

101606

160670,0, 2

=→=−=→=−

−⋅−=+−==

x x x x

x x x x ySí

La cdada d c : ( ) ( )0,60,1

C : Hacd = 0

( ) ( ) 66070)0(,0, 2 =+−=== f y xSí

E cdada: ( )6,0

E2: Ha c d a c:

35)( +== x x f y Solución:

C :

x x x ysí =−→=−→+==5

353350,0

Cdada d c:

− 0,

5

3

C :

( ) 3305,0 =+== y xsí Cdada d : ( )3,0

E3: Ha c d a c:

105)( −== t t h y S:

C :

S, = 0, 1050 −= t

Paa d a a cadada a ab ad d a cac a cadad:

( ) 210501050 22 =→−=→−= t t t

C :

S, = 0, 10)0(5 −= y

realeslosenexistenoraízEsta y →−= 10 N a c c . La cc c , cdada (2,0)


28/186

Cac Uaa R Dcc Pdaca




E4: Ha c d a c:

56

26)(

2 ++

+==

x x x

x f y

S:

C .

S, = 0: ( )5 / 2,05

2

5)0(60

2)0(62

→=++

+= y

C :

S, = 0:56

2602 ++

+= x x

x

Mcad a cac a 562 ++ x x da:

( ) ( ) ( )0,3 / 13 / 126056

2656560

2

22 −→−=→+=→++

+++=++ x x

x x

x x x x x

E5: Ha c d a c12

2)(

−== x

x f y

S:

S = 0, c da:

( )2 / 1,02

122 110*2 ⇒=== −− y

S = 0, c daa:

02 12 =− x Ea cac c, a ca a a c.

E6: Ha c d a c:

823log)( −== x xg y

S:

S = 0, da:880 3log3log

2 −− == y

N a d a. N a cc c .


29/186

Cac Uaa R Dcc Pdaca




S = 0, da:

82

3log0

−= x

Acad ca ab ad d a cac

811*)8(13*)8(133 223log23log0 82

−=⇒−=⇒−=⇒= −

x x x x

Q a cac cadca, aad a c da:

3390180 22 −=∨=⇒−=⇒−−= x x x x .

La cdada d c : ( ) ( )0,30,3 ∧−

La dcc cd a a aca.

://.cd.c.ac.c/bMaaca/NSc/ac.

E a a a aa ca cac

http://www.cidse.itcr.ac.cr/cursos-linea/MATEGENERAL/index.htm

2.1.4. Cdad: (://.b.c/ac?=Ad84VSc)

S dc a c ca d d, cad d c da aca aa a a, cad a c ac ca, ad a a a dc a c dca.

La c ( ) x f y = ada a a 2 ca, d c da

ca cdad d aa a a.

La c ( ) xg y = ada a a 3 dca ( ca), a c

ca a aa a a aa ca, a ac ca.

Ea c dca 1.

Paa dca d dcdad db ba a d .


30/186

Cac Uaa R Dcc Pdaca




Fa 2. Fc ca.

(A. E Cba G)

Fa 3. Fc dca.

(Autor. Elkin Ceballos Gómez)

Ia d cc a d dcc d a c:

://.b.c/ac?=_KLO_5Q&a=ad

CRECIMIENTO: S dc a c cc cad a aa a , a ab

aa ( ca).

DECRECIMIENTO: S dc a c dcc cad a aa a , a

d ( ca).


31/186

Cac Uaa R Dcc Pdaca




Eda a a aab dd, da a a aab dd.

Fc cc

↓↓

∨

↑↑

Y X

Y X

Fc dcc:

↑↓

∨

↓↑

Y X

Y X

Gca d da c a c cc dcc,

cd a ca d a c d da a dca, a ac b a

ca, dc a a c cc, a ac d baada,

dc a a c dcc.

Paa da a d cc a d dcc a c

d a a d d a d ab cad.

Paa a a 4 a d cc d dcc :

Cc: ( ) ( )∞∧∞−∈ ,21, x x x

Dcc: ( )2,1 x x x ∈

Fa 4. Cc dcc d a c.

(A. E Cba G).


32/186

Cac Uaa R Dcc Pdaca




Eac aa cc dcc d c

://.b.c/ac?=D23EUR

://.b.c/ac?=5HHdM3X6Q

2.1.4.1 O

E a ab cb b d: LGEBRA DE FUNCIONES:

Dada d c g y f d cba aa a a c a:

Sa d c.

( )( ) ( ) ( ) xg x f xg f +=+

Ra d c.

( )( ) ( ) ( ) xg x f xg f −=−

Pdc d c.

( )( ) ( ) ( ) xg x f xg f .. =

Cc d c.

( )( ) ( ) ( ) xg x f xg f / / =

Fc ca cc d c.

( )( ) ( ))( xg f xgo f =

( )( ) ( ))( x f g x f og =

La c ca d da c go f d c: ( ) ))(()( xg f xgo f =

NOTA:

( ) ( ) )()( x f og xgo f ≠


33/186

Cac Uaa R Dcc Pdaca




E1:

S ( ) ( ) 34132 2 −=∧+−= x xg x x x f

Ha ( ) )( xg f +

S

( )( ) ( ) ( ) ( ) 223413234132 222 −+=−++−=−++−=+=+ x x x x x x x x xg x f xg f

( )( ) 22 2 −+=+ x x xg f

E2:

S

( ) ( ) 9713 2 −=−= x xg y x x f

Halle ( ) )( xg f +

S

( ) ( ) ( ) ( ) 107397139713)( 222 −+=−+−=−+−=+=+ x x x x x x xg x f xg f

( ) 1073)( 2 −+=+ x x xg f

E3:

S ( ) ( ) 235 x xg x x f =∧+=

Ha: ( )( ) ( )( ) ( ) x f

g x fg xg f ∧− ,

S

( )( ) ( ) ( ) 3535 22 ++−=−+=−=− x x x x xg x f xg f

( )( ) 352 ++−=− x x xg f

( )( ) ( ) ( ) ( )( ) 232 3535. x x x x xg x f x fg +=+==


34/186

Cac Uaa R Dcc Pdaca




( )( ) 23 35 x x x fg +=

( ) ( )

( ) 35

2

+==

x

x

x f

xg x

f

g

E4:

S 3)()( 2 −=∧= x xg x x f

Ha a c ca: ( ) ( ) )()( x f og xgo f ∧

S

( ) ( )23)3())(()( −=−== x x f xg f xgo f

( ) 33)()())(()( 222 −=−=== x x xg x f g x f og

E5:

://.b.c/ac?=GHITUa4Q

E6:

://.b.c/ac?=LFVDYKPY&a=ad

E7:

://.b.c/ac?=2ICR830CP&a=

E 1

1. E 352 , 1 8.5 . E

,

:

. V .

. V

. D.

. R.


35/186

Cac Uaa R Dcc Pdaca




2. S 3, 5, 7, 8

2

)(2 x

x f = ,

.

3. D :

. ( ) 485 2 −+= x x x f

. ( ) 2520 x x xg −=

. ( ) 12

2 −= x xh

. ( )9

3762

2

+

−−=

x

x x x j

4. P :( ) ( ) x x xg x x x f 72435 22 −=∧+−=

. H:

. ( )( ) xg f −

. ( )

⋅

5

2g f

. ( )

2

3 / g f

. ( )( ) x fog

5. E 5 ( ) x f

F 5. G ( ) x f

(A. E C G. M 12 2011)


36/186

Cac Uaa R Dcc Pdaca




T 5, :

. I .

. L .

. I .

. I , .

2.2. C

S aa a: Idca d c, da d, a

ca a c (ad aa acac a), da a d

dcc a d cc da cdad dcdad d c

(Obad a ca).

2.2.1. F P

Ua c ca da c d a a:

01

1

1)( a xa xa xa x f y n

nn

n ++⋅⋅⋅⋅++== −

−

S dca :

N aab a, aab dad, aab d

d a a, ca, ca.

E d c ca

652)(2107)(37)( 23 +−=+−==−== x x x f x x xh y x xg y

DOMINIO:

E d d a c ca ad c d d a.

CONTINUIDAD: La c ca c d d.

REPRESENTACIN GRFICA: La a d aca a c ca dd d

cada d c.

La c ca caca a :


37/186

Cac Uaa R Dcc Pdaca




2.2.2. F :

E a c d a a:

bmx x f y +== )(

Dd b ym ca a: a d b c d c d

a a ca c .

NOTA:

Cad a d a, a c cc.

Cad a d aa, a c dcc.

La d da ac acca d d cac d a ca c a:

Cad d cac d 90 a d .

Cad d cac a d 90 a d a.

Cad d cac a a 90 a d , aca c d

c, ca a ca ca ca cac = c.

Cad d cac a a c a d ab a a c, ca

a ca a ca cac = b.

GRFICA:Paa aca a c a c c d

L a a :

1. S cca d a d abaa

2. Cada a d ccad aa d aa b a ca .

3. La aa bda bca a caa.

4. U d bd da a a ca.

E1:

Paa a c 53)( −== x x f y .1. Ha d.

2. Rac ca.

3. D a d cc a d cc.

4. Id a c ca dca, ca d dca d

d dcdad.


38/186

Cac Uaa R Dcc Pdaca




S

1. D. lR x ∈ Por ser un polinomio.

2. Gca:

Scc d a d ( d) = 0 = 4, c a b

a ca aad a c.

Paa 55)0(3)0(,0 −=−=== f y x .

E cdada (0, 5).

Paa 75125)4(3)4(,4 =−=−=== f y x .

E cdada (4,7).

Hacd a aba d a da.

X 0 4

Y 5 7

Ub d a caa da a a ca. La ca

a a a 6

Fa 6. Gca d 53)( −== x x f y (A. E Cba G)


39/186

Cac Uaa R Dcc Pdaca




La dcc cd a a a aa a ca d

c.

://.cd.c.ac.c/ca/NUMERICO/d.

P : Paa a a a aa aca

53)( −== x x f y

Db da: 3*5 da a c aca.

://.c./ac/03U004/d.

P : Paa a a a aa aca

53)( −== x x f y

Db da: 35 da a c aca

1. D a d cc a d cc.

S

Obad a ca d a a 7 d a c cc.

Cc: ( )∞∞− , La c dcc.

2. Cdad: La c ca d d.

E2:

Paa a c 104)( +−== x x f y

1. Ha d.

2. Rac ca.

3. D a d cc a d cc.


d dcdad.


40/186

Cac Uaa R Dcc Pdaca




S

1. D: lR x ∈ Gca:

S 210810)2(4)2(2 =+−=+−=== f y x .

S 10102010)5(4)5(5 −=+−=+−=== f y x .

X 2 5

Y 2 10

La ca a a a 7

FIGURA 7. Gca d 104)( +−== x x f y

(A. E Cba G)

1. D a d cc a d cc.

C a d aa, a c dcc, ab d ba a

a 8.

Dcc: ( )∞∞− ,


41/186

Cac Uaa R Dcc Pdaca





d dcdad. La c ca d d.

E3:

Paa a c: 2)( == xg y

1. Ha d.

2. Rac ca.

3. D a d cc a d cc.


d dcdad.

S1. D: ( )∞∞−∈ , x

2. Gca.

S d aa ca a d , a d a. La ca

a a 8

8 8

2 2

FIGURA 8: Gca d 2)( == xg y (A. E Cba G)


42/186

Cac Uaa R Dcc Pdaca




1. Cc d cc.

C a d a a c, a c cc dcc, a c ca.

N cc

N dcc.

2. Cdad. E ca.

L ac aa a c a.

://.b.c/ac?=C3TQ&a=ad

://.b.c/ac?=BVZUHCDM&a=ad

//.b.c/ac?=IG3V0&a=ad

://.b.c/ac?=BVZUHCDM&a=ad

2.2.3. F C F D S G

E a c d a a:

cbxax x f y ++== 2)(

Dd , , ca c d d c ( )0≠a .

DOMINIO:

P a c ca, d cd a d a.

GRFICA:

Ea c cd a a aba ca dd d a d .

S a ( )0>a . La aba ab aca aba, a c d a aba

cd a .

S aa ( )0


43/186

Cac Uaa R Dcc Pdaca




Fa 9. Paba ab aca aba(A. E Cba G. Ma 11 d 2011)

Fa 10. Paba ab aca aba.

(A. E Cba G. Ma 11 d 2011)

Paa aca a aba db ca a:


44/186

Cac Uaa R Dcc Pdaca




1. Idca a d a ca , , :

cc d 2 x cc d x dd, aab.

2. Ddd d d dca aca dd ab a aba.

3. Eca a cdada d c: E c cdada (, ).

)(

2

h f k

a

bh

=

−=

( )

( )

∞

∞−>

,

,0

hCrece

h DecreceaSí ,

( )

( )

−∞

∞−<

,

,0

h Decrece

hCreceaSi

4. S d bca c.

5. E : S da a a add d c aada 3 a a

da 3 a a dca. (E c c). Oba a ca

aad a c.

6. Ub a a caa a da a ca.

NOTA:

U .

E1:

Paa a c: 86)( 2 ++== x x x f y

D: D, a, d a d cc d dcc,

d a c ca dca (d ):

S

D. lR x ∈ P a c ca.

Gca.


45/186

Cac Uaa R Dcc Pdaca




8,6,1 === cba

C a a ( )01,1 >=a . La aba ab aca aba.

( ) ( ) 1181898363)3()(

332

6

)1(2

6

2−=→−=+−=+−+−=−==

−=→−=−=−=

k f h f k

hh

E c cdada ( )1,3 −−

P add d a d c.

A da:

( )

8,6

8836368)6(6)6(6

3,5

3830258)5(6)5(5

0,4

0824168)4(644

2

2

2

=−=

=+−=+−+−=→−=

=−=

=+−=+−+−=→−=

=−=

=+−=+−+−=→−=

y x

y x

y x

y x

y x

y x

A dca:

8,0

88)0(6)0(0

3,1

38618)1(6)1(1

0,2

081248)2(6)2(2

2

2

2

==

=++=→=

=−=

=+−=+−+−=→−=

=−=

=+−=+−+−=→−=

y x

y x

y x

y x

y x

y x

Vc

6 5 4 3 2 1 0

8 3 0 1 0 3 8

Ub a caa da a.

La ca a a a 11.

D a ca d a a 11 a c :


46/186

Cac Uaa R Dcc Pdaca




Cc ( )∞− ,3

Dcc ( )3,−∞− La c ca, a c ca.

FIGURA 11. Gca d: 86)( 2 ++== x x x f y (A. E Cba G)

Na: Ea ca d aa ad a d a a:


E2:

Paa a c 543)( 2 ++−== x x x f y

D: D, a, d a d cc d dcc,

d a c ca dca (d ):

S

D. lR x ∈ Gca:


47/186

Cac Uaa R Dcc Pdaca




( )

3

195

3

8

3

45

3

8

9

4*35

3

24

3

23

3

2)(

3

2

6

4

)3(2

4.033.

5,4,3

2

=++−

=++−=+

+

−=

==

=−

−=

−−=


48/186

Cac Uaa R Dcc Pdaca




Cc: ( )3 / 2,∞−

Dcc: ( )∞,3 / 2

E a c ca.

E3: ://.b.c/ac?=0UHF1FJ2&a=ad

E4: ://.b.c/ac?=IC4ZV4d_J&a=ad

E5: ://.b.c/ac?=PDFZ6L_0&a=

E6: ://.b.c/ac?=VdIKYRF4&a=ad

Eac c cadca

://.b.c/ac?=I654T4c&a=ad

://.b.c/ac?=3M7H0&a=ad

Fc cbca:

E a c d a a:

.)( 23 DCx Bx Ax x f y +++==

C A, B, C, D ca A d d c.

Paa aca a c c d ad a , a a cca

ada.

P d a a d a a ://.cd.c.ac.c/c

a/NUMERICO/d. aa aa a ca.

C cc aa c a a ca d a c:

( ) 52103 23 ++−= x x x x f

3152)( 234 −+−+= x x x x xg


49/186

Cac Uaa R Dcc Pdaca




2.2.4. F R

E a c d a a:

0)(,)(

)()( ≠== xhcon

xh

xg x f y

S dca a c dad.

La c aca a c cc d d c ca.

DOMINIO:

E ad d a a aa ca / c d ac.

Ua aa ca a d dd dad ac c.

Un hueco es un valor de x donde el numerador y el denominador son iguales a cero.

Paa da a aa ca / c d a c aca ( )

cd d a aa.

1. Ia dad a c.

2. Sca a cac a. S a cac c, dc a

c aca aa ca c.3. L a d bd db a d d d a c.

D d d cada a d a c:

E1:52

5)(

−==

x

x xw y

S

Iaad dad a c.

2

552052 =→=→=− x x x

2

5= x E a aa ca; db c d d.

lR x Dom ∈. M = 5/2. Oa a d dc : D.2

5≠ x


50/186

Cac Uaa R Dcc Pdaca




E2: 62

74)( 2 −−

−

== x x

x x f y

S

( ) ( )

( )( ) ( )( )( )( )

2

332032

202

032202

32)220

2

3242

02

6214062*

2

2062

222

−=→−=→=+

=→=−

=+−→=+−

→=+−

=−−

→=−−→=−−

x x x

x x

x x x x x x

x x x x x x

2

32. −≠∧≠ x x Dom

E3:1

3)(

2 −==

x x f y

S

( )( )

1,1,

101

10101101

2

−≠≠

=→=−

−=→=+

=−+→=−

x x Dom

x x

x x x x x

E4:33

763)(

2

2

++

+−==

x x

x x xk y

S

0332 =++ x x

( ) ( )( )

( ) 233

2

1293

12

31433 2

−±−=

−±−=

−±−= x

Ea cac c R, dc a c aa

ca c, a d d a.


51/186

Cac Uaa R Dcc Pdaca




lR x Dom ∈. E5: ://.b.c/ac?=cIP0GTI&a=ad

E6: ://.b.c/ac?=8AMRXbY

NOTA: U .

Gaca d c aca.

S cda ac a ca ad a a a.

Pcd:

1. D d d a c.

2. A a a c a a da a a dca d cada aa

ca / c. P cada aa ca / c aa, a c

cc a a da cc a dca. T

.

3. Oba a ca aad a c.

4. Ub a caa.

5. Ub a aa ca a caa. La aa ca a

ca ca aa cada a d aa c dad d a

c. La aa ca dd a caa d a aa

c a a dca d a aa ca d c

c a da.

6. Ub a aa a. Paa d ca a d a

a acc dda cc d dad cc d ad,

a bd cd a a a aa a. Tac a ca

a a a dc a.

7. Ua a c a ca; dca a d ca a

aa.

8. S a ca , d bca ab c.

E1: Paa a c:6

5)(+== x x x f y

a. D d.

b. D cc c .

c. Ga.

d. D a d cc a d dcc.

. D a c ca dca, ca d a dca,

d d dcdad.


52/186

Cac Uaa R Dcc Pdaca




S

D:

Iaad dad a c: { }6.06 −∈→=+ lR x Dom x

Paa aca db da a a a a da a a dca d aad

a c b a ca . L a d a aba.

11 10 9 8 7 6 5 4 3 2 1

11 12,5 15 20 35 A. 25 10 5 2,5 1

Paa aa a aa a, a d a ada a a ca 1. E

cc d ad 5 cc d dad 1, dd,c a aa a : 51 / 5 == y

5= y E .

La ca d a c a a d a ca ada a a 13.

FIGURA 13. Gca d:6

5)(

+==

x

x x f y

(A. E Cba G)


53/186

Cac Uaa R Dcc Pdaca




Cc: ( ) ( )∞−∧−∞− ,66,

La c dca = 6.

E2: Paa a c:910

100)(

2 +−==

x x

x x f y

a. D d.

b. D cc c .

c. Ga.

d. D a d cc a d dcc.

. D a c c dc, ca d a dca,

d d dcdad.

S

D:

Iaad dad a c:

( )( )

{ }9,1:min

909

101

09109102

−∈

=→=−

=→=−

=−−→=+−

Rl xiodo

x x

x x

x x x x

Aa a. La d a 2 x . E ad c (

a 2 x ad), dad 1.Aa a : 01 / 0 == y

0= y A

S da a a a da a a dca d = 1, a a da a a dca d = 9

L ad ba a aba.

X 4 3 2 1 0 1 2 3 4 5

Y 6,15 6,25 6,06 5 0 A. 28,57 25 26,66 31,25

X 5 6 7 8 9 10 11 12 13 14

Y 31,25 40 58,33 114,28 A. 111,11 55 36,36 27,08 21,53

La ca a a a 14.


54/186


55/186

Cac Uaa R Dcc Pdaca




Fa 15: Gca d a c

(A. E Cba G)

S

Paa

a. La c ca : ,

b. Aa ca: Aa a:

c. Cc: , Dcc:

E4:

Paa a c

D:

a. D.

b. La cac d a aa.

c. Id a a c ca.

d. Ha:


56/186


57/186

Cac Uaa R Dcc Pdaca




Eac c aca

://.b.c/ac?=adHY0OHM&a=

2.2.5. F I

E a c d a a:

n xg x f y )()( == .

S dca aab d d adca.

Laa c aca a aa a a aab aac ada a

aca .

DOMINIO:

S a d ca:

C #1:

S a a a d cd a d a: D. .lR x ∈ C #2:

S a a a, db aaa a a a aa d

d d a a.

Pcd:

1. Pa a cac 0)( ≥ xg .

2. S ca a cac.

3. La c d a cac d d a c aca.

4. S a cac aada c da d d a a a ,

dc d d a c d a: .lR x Dom ∈

E1: D d d: 152)( 2 −+== x x x f y

S

Ea a cac cadca.

PASOS:

1. Td d d a a db a a c, db aa a

cac: 01522 ≥−+ x x


58/186

Cac Uaa R Dcc Pdaca




2. Ec cc d a a. E a a c a

a cac a, a bd cc d a . U

cc dd a ac c, dc, dd

b a cab d a .

Paa : 3501522 =∨−=→=−+ x x x x E cc.3. Ub N cc a ca ca a a a 16.

4. Ea d a bda a . Paa a

c a a da d cc, a

c ab cc a c a a

dca d d cc. E aa a

bda a d ad cca a ca ca, a aa 16.

5. La a c d a cac, da ad a ca c

d d a dadad. Paa a d d a dadad d a

bda a d.

S dc: 0> S a ++++, c cc.

S dc: 0≥ S a ++++, cd cc.

S dc: 0< S a , c cc.

S dc: 0≤ S a , cd cc.

La c d a cac : ( ] [ )∞∪−∞−∈ ,35, x E d d a c: D. ( ] [ )∞∪−∞−∈ ,35, x

Fa 16. D d 152)( 2 −+== x x x f y

(A. E Cba G)


59/186


60/186


61/186

Cac Uaa R Dcc Pdaca




( ) ( )( ) ( ) 0130320133206

132323

=+∨=−⇒=+−⇒=

+−

x x x x

x x

3

113013

2

332032 −=⇒−=⇒=+∨=⇒=⇒=− x x x x x x

La c d a cac :

3

1

2

3−=∨= x x

Ubca d a ca ca da a a da a a

dca d cada d .

Fa 19. D d a c 376)( 2 −−== x x xg y

(A. E Cba G)

E d d a c :

∞∪

−∞−∈= ,

2

3

3

1,.. x Dom

E5: D d d: 7 75)( −== x xg y

S

C a a a d d a

lR x Dom ∈.


62/186

Cac Uaa R Dcc Pdaca




E6: ://.b.c/ac?=BCQBEBGdU

E7: ://.b.c/ac?=QUCCVAE4TM

E8: ://.b.c/ac?=GbPJHD&a=ad

GRFICA DE LA FUNCIN IRRACIONAL:

El procedimiento a seguir es:

1. D d d a c.

2. A a a d d cada a ad . P

cada a a aada cc a.

3. Paa cada a b a ca aad a c.

4. S bca a caa.5. S da a.

6. L d a d d .

E1: Paa a c: 56)( 2 ++== x x x f y

a. D d.

b. Ga.

c. D a d cc d dcc.

d. D c dc.

S

S db da d.

0562 ≥++ x x

( )( )

51

0501

051

0562

−=∨−=

=+∨=+

=++

=++

x x

x x

x x

x x

Ubcad a ca ca


63/186

Cac Uaa R Dcc Pdaca




Fa 20. D d a c 56)( 2 ++== x x x f y

E d da.

]( [ )∞−∪−∞−∈ ,15,. x Dom

L a d a da : E cc ca a a a da d = 5

ca a a a dca d = 1.

La ca a ada a a 21

9 8 7 6 5 1 0 1 2 3

5,65 4,58 3,46 2,23 0 0 2,23 3,46 4,58 5,65


64/186

Cac Uaa R Dcc Pdaca




Fa 21. Gca d 56)( 2 ++== x x x f y (A. E Cba G)

Cc: ( )∞− ,1

Dcc: ( )5,−∞−

E2: Paa a c x x f y −== 2)(

a. D d.

b. Ga.

c. D a d cc d dcc.

d. D c dc.

Sc

D:

( ]2,.

2202

∞−∈

≤→−≥−→≥−

x Dom

x x x

S da a ad 2 , d a aba.


65/186


66/186

Cac Uaa R Dcc Pdaca




a. D.

b.

Ha:

S

A. D

D:

B.

F

S aa c abaca a a c d a c cbac d a,

a, d, ca ac d c ca.

Ec da a c ca, aca aca abaca, ab

ca cbac d a.


67/186

Cac Uaa R Dcc Pdaca




E1:

1532)(

+

−== x x xg y

E2:

25)(

+=

x

x x f

Paa da d d a c db ca d c

cbad.

D d d a c

E1:1

)(−

== x

x x f y

S

S db c 010 ≠−∧≥ x x E dc: 10 ≠∧≥ x x

Ubcad a ca ca :

Fa 23. D d a c1

)(

−

==

x

x x f y

(A. E Cba G)

D: [ ) ( ]∞∪∈ ,11,0 x

E2:

( )12

92

−

−==

x

x x f y


68/186


69/186

Cac Uaa R Dcc Pdaca




Ubcad a ca ca:

Fa 25. D d a c x

x xg y

35

720)(

−

−==

(A. E Cba G)

D:

∞−∈

3

5, x

NOTA:

Ua c abaca ca d d.

Paa aca a c ada a cca aada ad a

dada. P d ac a ca d a a d a a:


P d ac ad a aa ca c Ec D.

Eac d c abaca.

://.b.c/ac?=N5UZW&a=

://.b.c/ac?=IOY192EU&a=ad

://.b.c/ac?=7PU0A&a=

FUNCIONES TRASCENDENTES

S c abaca, a c a c ca, a

c ca, a c aca, a c ca a a

c bca a.


70/186


71/186


72/186

Cac Uaa R Dcc Pdaca




X 3 2 1 0 1 2 3

Y 0,008 0,04 0,2 1 5 25 125

4. D a d cc a d dcc. Cc ( )∞∞− ,

FIGURA 26. Gca d: x x f y 5)( ==

(A. E Cba G)

E2:

Paa a c: 153)( +== x x f y

1. D d: E d d a, a a

. D. lR x ∈ .

2. D c:

)3,0(3330 11)0(5 ⇒===⇒= + y x Ea a cc c .

C cc, a = 0, a cac 1530 += x , c.

3. Ga.


73/186

Cac Uaa R Dcc Pdaca




Paa ac a ca, db aa a : E c cc a a da cc a

dca, c a a aba.

X 5 4 3 2 1 0 1 2 3 4 5

0.0 0.0 0.0 0.0 0.01 3 729 17714 43046721 10460353203 2541865828329

La aca a a a 27.

3. D a d cc a d dcc.

La c cc ( )∞∞− ,

Fa 27: Gaca d 153)( +== x x f y

(A. E Cba G)

E3:

Paa a c: 412

2

)( −−

== x x

e x f y

4. D d:

C a a c aca aa da d ac dad d a a c

22

02020)2)(2(042

−=∨=

=+∨=−⇔=+−⇔=−

x x

x x x x x


74/186

Cac Uaa R Dcc Pdaca




D: 22 −≠∧≠ x x

1. D c:

...284,10 4 / 141

4)0(

1)0(22

=⇒=⇒=⇒=⇒= −−

−

−

ye ye ye y x ...)284,1,0( E c c .

C c.

2. Ga.

Paa a ca a a cc a a a da d 2, cc a 2 2 cca a a dca d 2. A cac a a a aba.

X 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7

Y 0.7 0.6 0.5 0.4 0.2 ada 2.7 1.2 0.7 ada 2.7 1.7 1.5 1.4 1.3

La aca a a a 28.

3. D a d cc a d dcc.

La c cc ( )∞∞− ,

Fa 24. Gaca d 412

2

)( −−

== x x

e x f y

(A. E Cba G)

E4: ://.b.c/ac?=U7HC4


75/186

Cac Uaa R Dcc Pdaca




E5: ://.b.c/ac?=0SSbEE6: Sc d cac ca.

://.b.c/ac?=aPVTZAOY&a=PaL&=77286F52BE3722A4&d=0&a=1E7: Sc d cac ca.://.b.c/ac?=W7UQGA_A&a=PaL&=77286F52BE3722A4&a=1&a_=PL&d=2

NOTA:Ua c ca ca d d.

FUNCIN LOGARTMICA

://.b.c/ac?=FGP3F5Q0

://.b.c/ac?=MMXOESYY

://.b.c/ac?=aU9GN_c

E a c d a a:

)(log)( xg x f y b== .1,0 ≠> bbCon

DOMINIO:

Paa aa d db a cac: 0)( > xg

D d c d a c:

E1: x x f y 3log)( ==

S

DOMINIO: S db aa ca a cac: 0> x D. ( )∞∈ ,0 x .

INTERCEPTOS:C = 0, 0log3= y N , a cc c .

C : 0log00 3 ==⇒= x y 130 =⇒=⇒ x x ⇒ (1, 0).

E2: )32(log)( 2 −== x x f y


76/186

Cac Uaa R Dcc Pdaca




S

DOMINIO:2

332032032 =⇒=⇒=−⇒>− x x x x

Ubcad a ca ca:

Fa 29. D d )32(log)( 2 −== x x f y

(A. E Cba G)

INTERCEPTOS:

S existe No y x )3(log)3)0(2(log0 22 −=−=⇒= . N a cc c .

S0

2 232)32(log00 =−⇒−=⇒= x x y

xejeconersescción

x x x x x

int)0,2(

22 / 442312132 =⇒=⇒=⇒+=⇒=−

E3: ( )4log)( 2 −== x x f y

S

DOMINIO: 042 >− x ( )( )

22

0202022042

=∨−=

=−∨=+⇒=−+⇒=−

x x

x x x x x

Ubcad a ca ca:


77/186

Cac Uaa R Dcc Pdaca




Fa 30. D d ( )4log)( 2 −== x x f y

(A. E Cba G)

( ) ( )∞∪−∞−∈ ,22,. x Dom

INTERCEPTOS:

S existeno y x )4log()4)0log((0 2 −=−=⇒= . N a cc c .

S 022 104)4log(00 =−⇒−=⇒= x x y

55

0501414 222

−=∨=

=−⇒=−−⇒=−

x x

x x x

0,5)0,5( −∧ Icc c .

E4: )105ln( += x y

S

DOMINIO: 0105 >+ x

25 / 101050105 −=⇒−=⇒−=⇒=+ x x x x

( )∞−∈ ,2. x Dom

INTERCEPTOS:

S 302,210ln)10)0(5ln(0 =⇒=⇒+=⇒= y y y x ...)302,2,0(⇒ c .

S0105)105ln(00 e x x y =+⇒+=⇒=

5 / 99510151105 −=⇒−=⇒−=⇒=+ x x x x


78/186

Cac Uaa R Dcc Pdaca




)0,5 / 9(− C .

E5: ://.b.c/ac?=6XK99cV

E6: ://.b.c/ac?=8cI&a=ad

ASPECTOS A TENER EN CUENTA PARA OPERAR CON LOGARITMOS:

La cacada d ca aa abaa a:

La ca log ca 10log

La ca ln ca elog

Ea c a cacada c.

E a cacada Ca MS fx 82− a, dca a a a

c cb, cd :

E1:Oba:

S

20log . D a ca log , d 20, d a ca a ad

bd : 1.30109996

La cca :

E2: 135ln

Oba:

S

135ln . D a ca ln , d 135 , d a ca a, ad bd :905274778.4

Cad a ba d a d a 10 d , db ac cab d ba,

aa aca a a:

A

B

A

B B A

log

log

ln

lnlog ==


79/186


80/186

Cac Uaa R Dcc Pdaca




X 0,2 0,5 1 2 3 4

Y 2,32... 1 0 1 1,58... 2

FIGURA 31. Gca d x x f y 2log)( ==

(A. E Cba G)

E2:

Paa a c: )102(log)( 5 −==

x x f y 1. D d:

S

52 / 101020102

0102

=⇒=⇒=⇒=−

>−

x x x x

x

D. ( )∞∈ ,5 x .

2. D c:

S

S = 0, ( ) ( ) existe No y 10log)1002(log 55 −=−= N c .

C S = 0,


81/186

Cac Uaa R Dcc Pdaca




x x x x 210110211025)102(log0 05 =+⇒−=⇒−=⇒−=

x x =⇒=2

11211

Ic c : (11/2,0).

3. Ga:

L a a a aba, a ca a a a 32.

= 5 a aa ca.

X 5 6 7 8 9 10 11

Aa

ca.

0.43... 0.86... 1.11... 1.29... 1.463... 1.54...

4. D a d cc a d dcc. Cc

Fa 26. Gaca d )102(log)( 5 −== x x f y

(A. E Cba G)

NOTA:Ua c aca ca d d.

E ac a a d a c cadca, aca

ca.

://.b.c/ac?=bSSWRVY&a=ad


82/186

Cac Uaa R Dcc Pdaca




2.2.6. F D P T

E d c ab cb b d:

FUNCIN DEFINIDA POR PARTES O FUNCIN DEFINIDA POR TRAMOS O FUNCIN

SECCIONALMENTE DEFINIDA.

E1: Ga a c:

>

≤−==

1

11)(

2 xsi x

xsi x x f y

S

Cad 1≤ x S db a x x f −= 1)( , c a a ca aa a d a a ,

1 d aa ac ca.

x 1 2 x y −= 1 0 3

E a cd a 1= x ac a d a ca, cca dc .

Cad 1> x , db a 2)( x x f = , a aba, da a a d 1 ada:

x 1 2 3 4 52 x y = 1 4 9 16 25

Ta ca aa a a 1= x ac a d a a, aa dca a ca ac c c.

La ca a a a 33.


83/186

Cac Uaa R Dcc Pdaca




Fa 33. Gaca d a c:

>

≤−==

1

11)(

2 xsi x

xsi x x f y

(A. E Cba G)

E2: Paa a c:

−>+

−≤+==

21

223)(

2 xsi x

xsi x x f y

Ha:

( )2),5(),10(),2(),1(),0( −−−− f f f f f f

S

C 0= x a 2, db aa a: 12 + x

( ) 110)0( 2 =+= f

C 1−= x a 2, db aa a: 12 + x

( ) 21111)1( 2 =+=+−=− f


84/186

Cac Uaa R Dcc Pdaca




C 10= x a 2, db aa a: 12 + x

( ) 1011100110)10( 2 =+=+= f

C 5−= x 2, db aa a: 23 + x

132152)5(3)5( −=+−=+−=− f

C 2−= x a a 2, db aa a: 23 + x

4262)2(3)2( −=+−=+−=− f

Ga:

S

Paa aca a c da 5 a a a a 2 (cd 2)

aa 23 + x . Y da 5 a a a 2 ( c 2) aa

12 + x .

Paa ca a aba d a:

x 6− 5− 4− 3− 2− 23 += x y 16− 13− 10− 7− 4−

x 1− 0 1 2 3

12 += x y 2 1 2 5 10

La aca a ada a a 34.


85/186


86/186

Cac Uaa R Dcc Pdaca





87/186

Cac Uaa R Dcc Pdaca




Fc d :

)()( xsigno x f = S d c:

=

0,1

0,0

0,1

)(

xsi

xsi

xsi

x f

Paa a c a:

1. 1)10( = f

2. 1)2( −=− f

3. 0)0( = f

4. 1)3( = f

5. 1)3 / 2( −=− f

E 2.

1. D , :

a. ( ) 4116 2 +−= x x x f

b. ( ) 252 − x x f

c. ( )

45

372

−+

−=

x x

x x f

P

. D .

2. G: 12)( +== x x f y

3. G: 45)( 2 +−== x x x f y

4. P : [ ][ ]32)( −== x x f y D:

1. )0( f

2. )3 / 2( f

3. )5( f

4. )3( f

5. )(e f

6. )(π f


88/186

Cac Uaa R Dcc Pdaca




2.3. A

E1:

S a a caa d ca d d a b ccda, a a a 35.

Fa 35.

(A. E Cba G)

C a a d ca d abca caa aa d aa . Paa cada a

d a caa ca cadad dc d ad . Va a 36.

Fa 36

(A. E Cba G)

L dba ad aca aba. Va a a 37

Fa 37

(A. E Cba G)


89/186

Cac Uaa R Dcc Pdaca




E d a caa d ba caa b c:

V = aa cada a cad ac, dad cbca.

Paa a caa :

Aa =

La = a 2

Ac = b 2

P a , bad c a a:

( )( ) xb xa xv 22 −−= Ecad a cac:

( ) ( )( ) ( ) abxbxax x xbxaxabx xb xax xv ++−⇒+−−=−−= 2233222 22442222

La aa :

( ) abxba x xv ++−= 23 24

TENIENDO EN CUENTA LA SITUACIN ANTERIOR RESUELVA EL SIGUIENTE PROBLEMA PARTICULAR:

S da c a caa d a caa aa a a d a a d ca d 20

c 15 c. Paa ca cadad dc a ca a dba

ad aca aba.

1. Ecba a c aa .

S

S a c aa :

( ) abxba x xv ++−= 23 24 C: cmbcma 1520 =∧=

Raad a c d :

( ) ( )( ) x x xv 1520152024 23 ++−= 323 300704 cm x x xv +−=

2. D d ac aa a c.


90/186

Cac Uaa R Dcc Pdaca




S

P a c ca : lR x ∈

3. D d dd d a d a ac a aa a c d

.

S

S db c a a c.

( )( ) 0220 >−−

⇒>

xb xa xv

Raad a d a d b a dadad:

( )( ) 0215220 >−− x x x

021502200 >−∨>−∨> x x x

S :2

15100 =∨=∨= x x x

La c d a dadad :

Fa 38. Sc d a dadad.

(A. E Cba G)

E d : ( )∞+∪

∈ ,102

15,0 x

P c a d a

( )∞+,10

S da d aa, ca, b ba a.

P a, d dd d a d a ac a :


91/186

Cac Uaa R Dcc Pdaca




∈

2

15,0 x

L a2

150 =∧= x x c, a c a a a a c,

dc, aba caa.

1. D a d d a caa d a aa a d 378 3cm . D a c a c d dca.

S

S bb aa ca a cac:

378300704 23 =+− x x x

SOLUCIN DE LA ECUACIN

0378300704 23 =−+− x x x

( ) 01891503522 23 =−+− x x x 0189150352 23 =−+− x x x

La cac ca Facad d ad 3 d a cac d

aac.

Rcd b ac d 189150352 23 −+− x x x d d

dd:

L b ac :

( ) ( ) ( ) 376189115013511 23 −=−−+−−−⇒−= x N ac

( ) ( ) ( ) 72189115013511 23 −=−+−⇒= x N ac.

( ) ( ) ( ) 1008189315033533 23 −=−−+−−−⇒−= x N ac.

( ) ( ) ( ) 0189315033533 23 =−+−⇒= x E dc : 3− x ac. Paa

ca ac db ca a d:


92/186

Cac Uaa R Dcc Pdaca




3

189150352 23

−

−+−

x

x x x

Uad d ca:

2 35 150 189 3

6 87 189

2 29 63 0

La cac da:

( )( ) 0632923 2 =+−− x x x 06329203 2 =+−∨=− x x x

303 =⇒=− x x 063292 2 =+− x x

Ea a cac a ad a a.

( ) ( ) ( )( )

( ) 450484129

22

63242929063292

2

2 −±=⇒−−±−−

=⇒=+− x x x x

661.24

37729839.114

37729

4

37729≈

−=∨≈

+=⇒

±= x x x

C 3= x , a d d a caa :

Aa: cm x 3⇒

La: ( ) cm x 143220220 ⇒−⇒−

Ac: ( ) cm x 93215215 ⇒−⇒−

C 661.2= x , a d d a caa :Aa: cm x 661.2⇒

La: ( ) cm x 678.14661.2220220 ⇒−⇒−

Ac: ( ) cm x 678.9661.2215215 ⇒−⇒−

C 839.11= x , b, a :Aa: cm x 661.2⇒

La: ( ) cm x 678.3839.11220220 −⇒−⇒− Daa a d aa, b.


93/186

Cac Uaa R Dcc Pdaca




1. D a cadad d aa ad.

S

C 3= x a cadad d aa ad :

Cadad d aa ad = 22363*9*23*14*29*14 cm=++

C 661.2= x a cadad d aa ad :Cadad d aa ad:

2676316.271661.2*678.9*2661.2*678.14*2678.9*678.14 cm=++

1. Uc a d aada d a aa ddc d aa a ba d d b aa d 378 c3.

S

Obad ad a, ddc a cad a

d d a caa :

Aa: cm661.2

La: cm678.14

Ac: cm678.9

E2: S a a cadada d ca d ad , a a a 39.

Fa 39.

(A. E Cba G)

C a a d ca d abca caa aa d aa . Paa cada a

d a caa ca cadad dc d ad . Va a 40.


94/186

Cac Uaa R Dcc Pdaca




Fa 40

(A. E Cba G)

L dba ad aca aba. Va a a 41.

Fa 41.

(A. E Cba G)

E d a caa d ba caa b c:

V = aa cada a cad ac dad cbca.

Paa a caa d :

Aa =

La = 2

Ac = 2

P a , d ba c a a:

Ecad a cac a aa :

Td ca a ac bca a a:

S da c a caa aa. Paa a a a cadada d ca

ca a ca a cadad dc d 5 c d ad dba aca aba.

D d d a c d .

1. Ecba a c aa .

2. D d ac aa a c.


95/186

Cac Uaa R Dcc Pdaca




3. D d dd d a a aa d d .

4. D a d d a a d ca a a, S a caa ca aa

c d 2000 c3.

5. D a d d a caa.

6. D a cadad d aa ad.

S:

1. Ecba a c d d a caa.

Ua a aad a

S :

Oa a d d a aa:

U cadad c ca ad a, a a a 42.

Fa 42.

(A. E Cba G)

Sa ad d cadad; a a a a ca a 5 c a cada ad d a a.

Va a a 43.

Fa. 43

(A. E Cba G)


96/186

Cac Uaa R Dcc Pdaca




Qad 5 c cada a ad d a caa 5 5 = 10. Va a a 44.

Fa 44.

(A. E Cba G)

Dbad ad aca aba a caa da:

Fa 45

La c aa : ( ) )10)(10(5 −−= x x xv Scad da:

( ) 32 5001005 cm x x xv +−=

1. D ac:

S

P d :

2. D d dd d a a aa d d :

S

S db c :

( ) 0> xv E ca :

1010)10)(10(5 >⇒>−− x x x


97/186

Cac Uaa R Dcc Pdaca




D:

3. D a d d a a d ca a a, S a caa ca aa

c d 2000 c3.

S

S db aa ca:

E ca :

20005001005 2

=+− x x

( )( )

3010

030010

03010

030020

040010020

400100205

200010020

2000)10020(5

2

2

22

2

=∨−=

=−∨=+

=−+

=−−

=−+−

=+−→=+−

=+−

x NO x

x x

x x

x x

x x

x x x x

x x

La d d a a d ca db : 30 c 30 c.

1. D a d d a caa.

S

Obad a a 45, :

La:

Ac:

A:

2. D a cadad d aa ad:

S

Obad a a 43 44 d ddc a cadad d aa ad :

Cadad d aa ad a a: E a d a ba 4 c a d cad.


98/186

Cac Uaa R Dcc Pdaca




Cadad d aa = 20 c * 20 c + 4 * 20 c * 5 c

E3: S ca c 1200 c2 d aa aa ac a caa c ba cadada aa.

3. Ec a aa d a caa d a a aab.

S

Sa:

: x Cada ad d a ba cadada. : y La aa d a caa Va a a 46

Fa 46

(A. E Cba G)

E d a caa :

2 3* *v x x y v x y cm= ⇒ =

Paa cb d , a a cdc aa a cadad d aa.

Cadad d aa : 1200 c2

Paa a caa d a a 46 a cadad d aa a a:

La cadad d aa d a ba cadada 4 c a cadad d aa d ca

cad, :

21200 * 4 * 1200 4 x x x y x xy= + ⇒ = +


99/186

Cac Uaa R Dcc Pdaca




Dad :

2

2 2 12001200 4 1200 44

x x xy x xy y x

−= + ⇒ − = ⇒ =

Raad a a 2 3v x y cm=

2 32 2 31200 1200

4 4

x x xv x y v x v cm

x

− −= ⇒ = ⇒ =

La aa d a caa :

( )3

31200

4

x xv x cm

−=

1. D d d a a.

S

P a c ca d : x lR∈ P a ac bca aa da cda a caa, d db

a a a ca :

( )31200

04

x xv x

−= >

SOLUCIN DE LA DESIGUALDAD

( ) ( )( )3

3 21200 0 1200 0 1200 0 1200 1200 04 x x x x x x x x x− > ⇒ − > ⇒ − > ⇒ + − >

La ac :

0, 1200, 1200 x x x= = = −

La c d a dadad, d d a c :


100/186

Cac Uaa R Dcc Pdaca


Corporación Universitaria Remington - Calle 51 51-27 Conmutador 5111000 Ext. 2701 Fax: 5137892. Edif

Documents

Calculo II C Difrerencial