03 - Resuelto - Guía práctica de Matemática (Rossomando) PDF

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▼❆❚❊▼❆❚■❈❆❘❡s♣✉❡st❛s✭❈át❡❞r❛✿❘♦ss♦♠❛♥❞♦✮❈✐❝❧♦❇ás✐❝♦❈♦♠ú♥✲❯♥✐✈❡rs✐❞❛❞❞❡❇✉❡♥♦s❆✐r❡s❮♥❞✐❝❡ ❣❡♥❡r❛❧✵✳✶✳ Prá❝t✐❝❛✵ ✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳ ✷ ✵✳✷✳ Prá❝t✐❝❛✶✿◆ú♠❡r♦s❘❡❛❧❡s ✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳ ✽ ✵✳✸✳ Prá❝t✐❝❛✷✿❋✉♥❝✐♦♥❡s✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳✳ ✷✵ ✵✳✹✳ Prá❝t✐❝❛✸✿▲✐♠✐t❡s❞❡❋✉♥❝✐♦♥❡s②❆s✐♥t♦t❛s ✳✳✳✳✳✳...

Description

2 2 1 + − 5 5 3 2 2 1 + − 5 5 3 2 1−2 + 5 3 2 1 − 5 3 7 10 − 3 (2)(5) − (1)(3) = = 15 15 (3)(5)  2  2 3 4 1 − . 6 5 2  2  2 1 3 4 − . 6 5 2 9 4 1 − . 4 25 6 6 1 36 1 = − − 4 25 4 150 1 25 − 24 (1)(25) − (6)(4) = = 100 100 (4)(25)  −1 3 1 3 . + 6 2 5  −1 3 3 1 . + 5 6 2   2 3 1 . + 6 3 5 2 1 1 6 + = + 5 6 6 15 17 12 + 5 (2)(6) + (1)(5) = = 30 30 (5)(6)

r

16 3 + 2 9

!

.

2 5 ! 16 3 2 . + 5 2 9   4 3 2 . + 3 2 5     8+9 2 (4)(2) + (3)(3) 2 . . = 6 (3)(2) 5 5   17 34 17 2 . = = 15 30 5 6 r



3 1 . 2 5

2 



1 1 + 2 2

2

. 3+

3 1 . 2 5

2

=



3 10

2

=

9 100

1 4 

1 1 + 2 2

2

. 3+

2

(1) . 3 + 1. 3 + 3+

1 4

1 4

1 4

1 4

13 (3)(4) + (1)(1) = 4 (1)(4)

1 x x + = 3 2 6 x x 1 + = 6 3 2 1 (x)(3) + (x)(2) = (2)(3) 6 1 5x = 6 6 5x (6) = 1 6 5x = 1 ⇒ x =

1 5

2x 1 − +x=2 3 5 2x 1 − +x=2 3 5 (2x)(5)(1) − (1)(3)(1) + (x)(3)(5) =2 (3)(5)(1) 10x − 3 + 15x =2 15 25x − 3 = 2(15) 25x = 30 + 3 x= 

2x 1 + 9 3



.3=

33 25

x 5  

2x 1 + 9 3



.3=

x 5

 x 2x + 1(3) .3= 5 9 2x + 3 x = 5 3

(2x + 3)(5) = x(3) 10x + 15 = 3x 15 = 3x − 10x 15 = −7x ⇒ −

15 =x 7

x+1 3

− 6

2x−3 4

=



 1 x+1 .4 +x− 2 3

  − 2x−3 1 x+1 4 = .4 +x− 2 6 3   (x+1)(4)−(2x−3)(3) )(3)(2)−(x+1)(3)(1) (3)(4) = (1)(1)(2)+(x(3)(1)(2) .4 6 x+1 3

4x+4−(6x−9) 12

2 + 6x − (3x + 3) 6 6   −2x+13 −1 + 3x 12 .4 = 6 6   −2x + 13 −2 + 6x .6 = 3 12 =



−2x + 13 = (−4 + 12x)(12) −2x + 13 = −48 + 144x 61 = 146x 61 =x 146 (x + 1)2 (x + 1)2 (x + 1)(x + 1) x2 + x + x + 1 x2 + 2x + 1 (x − 3)(x + 3)

(x − 2)2

(x − 3)(x + 3) x2 + 3x − 3x − 9 x2 − 9

(x − 2)2 (x − 2)(x − 2) x2 − 2x − 2x + 4 x2 − 4x + 4



.4

(x − 2)2 (x + 2) (x − 2)2 (x + 2) (x − 2)(x − 2)(x + 2) (x2 − 4x + 4)(x + 2) x3 + 2x2 − 4x2 − 8x + 4x + 8 x3 − 2x2 − 4x + 8 (x + 1)2 (x + 1)2 (x + 1)(x + 1) x2 + x + x + 1 x2 + 2x + 1 (x − 3)(x + 3)

(x − 2)2

(x − 2)2 (x + 2)

(x − 3)(x + 3) x2 + 3x − 3x − 9 x2 − 9

(x − 2)2 (x − 2)(x − 2) x2 − 2x − 2x + 4 x2 − 4x + 4

(x − 2)2 (x + 2) (x − 2)(x − 2)(x + 2) (x2 − 4x + 4)(x + 2) x3 + 2x2 − 4x2 − 8x + 4x + 8 x3 − 2x2 − 4x + 8

(a + b)2 a2 + 2ab + b2 (a − b)(a + b) a2 − b2

(a + b)2 − 2ab

a2 + b2 a(2ab + a) 2a2 b + a2 a 6= 0 :

a+b a 1+

b 6= 0 :

b a

a+b b a +1 b

7

a5

√ 5 a7 2 √ 2 √ 2 a b c d

ad : b= 6 0, c = 6 0 bc b 6= 0 :

 a −1 b

a 6= 0 :

b a

A = {x ∈ ℜ/x < 2} (−∞; 2) B = {x ∈ ℜ/x > −2} (−2; +∞) C = {x ∈ ℜ/x ≤ 2} (−∞; 2] D = {x ∈ ℜ/x ≥ 2} [2; +∞) E = {x ∈ ℜ/1 ≤ x ≤ 3} [1; 3] F = {x ∈ ℜ/x < 1σx > 3} (−∞; 1) ∪ (3; +∞) G = {x ∈ ℜ/0 ≤ x ≤ 6} ∪ {x ∈ ℜ/1 ≤ x ≤ 8} [1; 6] H = {x ∈ ℜ/0 ≤ x ≤ 6} ∩ {x ∈ ℜ/1 ≤ x ≤ 8} [0; 8]

A = {x ∈ ℜ/2x + 3 ≤ 0} 2x + 3 ≤ 0 2 x ≤ −3 x ≤ − 32 (−∞; − 32 ]

B = {x ∈ ℜ/ − 1 < 2x + 4 ≤ 0} −1 < 2x + 4 ≤ 0 −1 − 4 < 2x ≤ 0 − 4 −5 < 2x ≤ −4 −25 < x ≤ − 42 − 52 < x ≤ −2 (− 52 ; −2] C = {x ∈ ℜ/x ≤ 1 + 2x} x ≤ 1 + 2x −1 ≤ 2x − x −1 ≤ x [−1; +∞) D = {x ∈ ℜ/2x + 1 ≥ x − 3} 2x + 1 ≥ x − 3 2x − x ≥ −3 − 1 x ≥ −4 [−4; +∞) E = {x ∈

ℜ/ x+3 2

≥

2x−6 } 4

x+3 2x − 6 ≥ 4 2 (x + 3)(4) ≥ (2x − 6)(2) 4x + 12 ≥ 4x − 12 0 ≥ −24 F = {x ∈ ℜ/ x2 +

x 3

<

x+9 } 3

x x+9 x + < 2 3 3 x+9 (x)(3) + (x)(2) < (2)(3) 3 5x x + 9 < 3 6 (5x)(3) < (x + 9)(6) 15x < 6x + 54 9x < 54 → x < 6 (−∞; 6)

A = {x ∈ ℜ/x(x + 1) ≥ 0} (−∞; −1) (−1; 0)

x = −1 (0; +∞)

f (−2) = (−2)((−2) + 1) f (−0, 5) = (−0, 5)((−0, 5) + 1) f (1) = (1)((1) + 1)

x=0

=2 = −0, 25 =2

(−∞; −1] ∪ [0; +∞) B = {x ∈ ℜ/x(x − 1) > 0}

x=1

(−∞; 0) (0; 1) f (−1) = (−1)((−1) − 1) f (0, 5) = (0, 5)((0, 5) − 1) f (2) = (2)((2) − 1)

C = {x ∈

x ℜ/ x+1

≥0

(1; +∞)

=2 = −0, 25 =2 >0

(−∞; 0) ∪ (1; +∞) ≤ 0} x ≤0 x+1

A≤0∧B>0:

x≤0 ∧ x≤0 ∧

x+1>0 x > −1

x≥0 ∧ x≥0 ∧

x+10 x > −1

(−1; +∞) x+2≤0 ∧ x ≤ −2 ∧

A≤0∧B1 x+1 x −1>0 x+1 x − 1(x + 1) >0 x+1 −1 >0 x+1 A< 0

A1 ∧ x>1

[4; +∞)

A≥0∧B0 x+1 −x − 7 >0 x+1

A>0∧B>0: ∅

−x − 7 > 0 −x > 7 x < −7

∧ x+1>0 ∧ x > −1 ∧ x > −1

−x − 7 < 0 −x < 7 x > −7

A 2x + 1 > 3 −4 > 2x > 2 −2 > x > 1 (−∞; −2) ∪ (1; +∞)

F = {x ∈ ℜ/|2x + 1| < 3}

−3 < 2x + 1 < 3 −4 < 2x < 2 −2 < x < 1 (−2; 1) G = {x ∈ ℜ/3 − |2x + 1| > 0} 3 − |2x + 1| > 0 −|2x + 1| > −3 |2x + 1| < 3 −3 < 2x + 1 < 3 −4 < 2x < 2 −2 < x < 1 (−2; 1)

A = (1, 1) B = (−1, 0) C = (2, 3) D = (0, 1) E = (−1, −2)

A = (1, 1)

B = (−1, 1) d 2 = (1 − (−1))2 + (1 − 1)2 d 2 = (2)2 + (0)2 d 2 = 4√+ 0 d= 4 d=2

A = (2, 3)

B = (−1, 2) d 2 = (2 − (−1))2 + (3 − 2)2 d 2 = (3)2 + (1)2 d 2 = 9√+ 1 d = 10 √ 30

a∈ℜ A = (1, a)

B = (2, 7) √ ( 30)2 = (1 − 2)2 + (a − 7)2 30 = (−1)2 + (a − 7)2 30 = 1 + a2 − 7a − 7a + 49 30 = a2 − 14a + 50 0 = a2 − 14a + 20

7+

√ 29

a=7−

√ 29

a =

√ 29

a∈ℜ A = (−2, a)

B = (a, 5) √ ( 29)2 = (−2 − a)2 + (a − 5)2 29 = (a2 + 4a + 4) + (a2 − 10a + 25) 29 = 2a2 − 6a + 29 0 = 2a2 − 6a 0 = a(2a − 6) a=0

a=3

A = (3; 7) d(A; P ) = 5

P = (−1; y)

y∈ℜ d(A; P ) =

5 (x1 , y1 )

(3; 7)

(x2 , y2 )

(−1; y )

2

d = (x2 − x1 )2 + (y2 − y1 )2 52 = (−1 − 3)2 + (y − 7)2 25 = (−4)2 + (y − 7)(y − 7) 25 = 16 + (y 2 − 7y − 7y + 49) 9 = y 2 − 14y + 49 0 = y 2 − 14y + 40 a = 1 b = −14

c = 40

√ −b b2 − 4ac 2a p −(−14) (−14)2 − 4(1)(40) 2(1) √ 14 36 2

14 + 6 = 10 2 14 − 6 =4 2

f (x) = −x + 2 B = (−4; f (−4))

A = (0; f (0))

f (0) = −(0) + 2 ⇒ f (0) = 2 f (−4) = −(−4) + 2 ⇒ f (−4) = 6 A = (0, 2)

B = (−4, 6)

(x1 , y1 ) (x2 , y2 )

(0, 2)

(−4, 6) d 2 = (x2 − x1 )2 + (y2 − y1 )2 d 2 = (−4 − 0)2 + (6 − 2)2 d 2 = (−4)2 + (4)2 d 2 = 16 + 16 √ d = 32 √ 32

5, 6568

f (x) = 2x + 6

0 = 2x + 6 −6 = 2x −3 = x P = (−3, 0)

f (0) = 2(0) + 6 f (0) = 6 Q = (0, 6)

(x1 , y1 ) (x2 , y2 )

(0, 6) d 2 = (x2 − x1 )2 + (y2 − y1 )2 d 2 = (0 − (−3))2 + (6 − 0)2 d 2 = (3)2 + (6)2 d 2 = 9 + 36 √ d = 45

(−3, 0)

√ 45

6, 7082

x+5 A = {x ∈ ℜ/ 2x+1 < 1}

A 0∧B 0 ∧ −x > −4 ∧ x 0} x2 + 4x − 5 > 0

a=1 b=4

c = −5

p √ −(4) (4)2 − 4(1)(−5) −4 36 ⇒ 2 2(1)

−4 + 6 =1 2

x = −6 x = 0 x = −6 x=0 x=2

−4 − 6 = −5 2

x=2

⇒ (−6)2 + 4(−6) − 5 > 0 ⇒ (0)2 + 4(0) − 5 > 0 ⇒ (2)2 + 4(2) − 5 > 0

(−∞; −5) ∪ (1; +∞)

⇒ ⇒ ⇒

7>0 −5 > 0 7>0

f (x) = x2 + 1 f (0) = (0)2 + 1 f (1) = (1)2 + 1 f (−1) = (−1)2 + 1 f (2) = (2)2 + 1

f (0) f (1) f (−1) =1 =2 =2 =5

1 ∈ Domf f (x) = x + 10

f (x) = x2 − 4

∀ℜ

∀ℜ

∀ℜ

∀ℜ

f (x) = x3 + 2x2 + x − 4

f (x) =

x+3 x−1

∀ℜ

∀ℜ

x − 1 6= 0 x 6= 1 ∀ℜ − {1} ∀ℜ − {1} f (x) =

x x2 − 9

x2 − 9 6= 0 x2 6=√9 |x| 6= 9 |x| 6= 3 ∀ℜ − {−3; 3}

f (2)

f (x) =

1 3 − x−1 x

x

0

x − 1 6= 0 x 6= 1 ∀ℜ − {0; 1} ∀ℜ − {0; 1} f (x) =

√ x−1

x−1≥0 x≥1 [1; +∞)

[1; +∞)

f (x) = 2x ∀ℜ ∀ℜ

f (x) = 2x

f (x) = 2x − 1

f (x) = 2x + 1

f (x) = −x

f (x) = −x + 6

f (x) = −x − 6

f (x) = 6

f (x) = −6

f (1) = 2

f (−1) = −2

m=

2 − (−2) 4 = =2 1 − (−1) 2 y = 2x + b

2 = 2(1) + b 2=2+b 0=b

y = 2x + 0 f (2) = 3

f (4) = 0

m=

3 3−0 = 2−4 −2

y = − 32 x + b 3 = − 23 (2) + b 3 = −3 + b 6=b y = − 32 x + 6

f (1) = 2

f (2) = 5

m=

−3 2−5 =3 = 1 − 2 −1 y = 3x + b

2 = 3(1) + b 2=3+b −1 = b

f (0) = 1

y = 3x − 1

f (2) = 1

m=

0 1−1 = =0 0 − 2 −2 y = 0x + b y=b

1=b

y=1

P = (−1, 1)

Q = (2, 4)

m=

1−4 −3 =1 = −3 −1 − 2 y = 1x + b

1 = 1(−1) + b 1 = −1 + b 2=b

y = 1x + 2

P = (0, 1)

Q = (1, 3)

m=

1−3 −2 =2 = 0 − 1 −1 y = 2x + b

1 = 2(0) + b 1=0+b 2=b

y = 2x + 1 P = (−2, 3)

Q = (−3, 2)

m=

1 3−2 = =1 −2 − (−3) 1 y = 1x + b

3 = 1(−2) + b 3 = −2 + b 5=b

y = 1x + 5 P = (1, 3)

Q = (2, 3)

m=

0 3−3 = =0 1 − 2 −1 y = 0x + b y=b

3=b

y=3

m P = (1, 3)

m=2

y = 2x + b 3 = 2(1) + b 3=2+b 1=b y = 2x + 1 P = (−1, 4)

m = −2 y = −2x + b 4 = −2(−1) + b 4=2+b 2=b

P = (1, 1)

m=0

y = −2x + 2

y = 0x + b y=b 1=b y=1

(0, 1)

m=

(3, 6)

1−6 5 −5 = = −3 3 0−3 y = 35 x + b

1 = 53 (0) + b 1=b

y = 35 x + 1

(0, 7)

m=

(4, 0)

7−0 7 7 =− = 4 0 − 4 −4 7 y =− x+b 4

7 7 = − (0) + b 4 7=b

7 y =− x+7 4

6 y=6

f (x) f (x) = x + 1

g(x)

g(x) = −2x + 4 x + 1 = −2x + 4 x + 2x = 4 − 1 3x = 3 x=1

f (1) = (1) + 1 f (1) = 2 (1, 2) f (x) = 6

g(x) = 2x − 4 6 = 2x − 4 6 + 4 = 2x 10 = 2x 5=x

g(5) = 2(5) − 4 f (1) = 6 (5, 6) f (x) = 6 y

(x, 6)

A = {x ∈ ℜ/x + 1 ≤ 2x − 1} x + 1 ≤ 2x − 1 1 + 1 ≤ 2x − x 2≤x

B = {x ∈ ℜ/ − 2x + 3 > x + 3} −2x + 3 > x + 3 −2x − x > 3 − 3 −3x > 0 x0 (0; +∞)



Dominio ∀ℜ Imagen (0; +∞ )  Dominio (0; +∞) −1 f (x) : Imagen ∀ℜ f (x) :

f (x) = ln(x) f (x) x>0 (0; +∞) y = ln(x) ey = eln(x) ey = x y −1 = ex

f (x)

∀ℜ



Dominio (0; +∞) ∀ℜ  Imagen Dominio ∀ℜ −1 f (x) : Imagen (0; +∞) f (x) :

f (x) = ex + 15 f (x)

f (x)

∀ℜ

y = ex + 15 y − 15 = ex ln(y − 15) = ln(e)x ln(y − 15) = x y −1 = ln(x − 15)

x − 15 > 0 x > 15 (15; +∞)



Dominio ∀ℜ Imagen (15; +∞ )  Dominio (15; +∞) −1 f (x) : Imagen ∀ℜ f (x) :

f (x) = 15 + ln(x) f (x) x>0 (0; +∞) y = 15 + ln(x) y − 15 = ln(x) ey−15 = eln(x) ey−15 = x

y −1 = ex−15

f (x)

∀ℜ



Dominio (0; +∞) ∀ℜ  Imagen Dominio ∀ℜ −1 f (x) : Imagen (0; +∞) f (x) :

f (x) = 4ex f (x)

f (x)

∀ℜ

y = 4ex y x 4 =e y ln( 4 ) = ln(e)x ln( y4 ) = x y −1 = ln( x4 )

x >0 4 x>0

(0; +∞)



Dominio ∀ℜ Imagen (0; +∞ )  Dominio (0; +∞) −1 f (x) : Imagen ∀ℜ f (x) :

f (x) = ln(x)(4) f (x) x>0 (0; +∞)

y = ln(x)(4) y = ln(x) 4y e 4 = eln(x) y e4 = x

x

y −1 = e 4

f (x)

∀ℜ



Dominio (0; +∞) ∀ℜ  Imagen Dominio ∀ℜ f −1 (x) : Imagen (0; +∞) f (x) :

f (x) = 2ex + 1 f (x)

f (x)

∀ℜ

y = 2ex + 1 y−1 x 2 =e y−1 ln( 2 ) = ln(e)x )=x ln( y−1 2 y −1 = ln( x−1 ) 2

x−1 2

>0 x−1>0 x>1 (1; +∞)



Dominio ∀ℜ Imagen (1; +∞ )  Dominio (1; +∞ ) f −1 (x) : Imagen ∀ℜ f (x) :

f (x) = 1 + 4ln(x) f (x) x>0 (0; +∞) y = 1 + 4ln(x) y−1 = ln(x) 4 y−1 e 4 = eln(x) y−1 e 4 =x

y −1 = e

f (x)

x−1 4

∀ℜ

Dominio (0; +∞) ∀ℜ  Imagen Dominio ∀ℜ f −1 (x) : Imagen (0; +∞) f (x) :

f (x) = e2x−3



f (x)

f (x) y = e2x−3 ln(y) = ln(e)(2x − 3) ln(y) = 2x − 3 ln(y) + 3 =x 2

y −1 =

ln(x) + 3 2

x>0

∀ℜ

(0; +∞)



Dominio ∀ℜ Imagen (0; +∞ )  Dominio (0; +∞ ) f −1 (x) : Imagen ∀ℜ f (x) :

f (x) = ln(3x + 5) f (x) 3x + 5 > 0 3x > −5 x > − 53

(− 53 ; +∞)

y = ln(3x + 5) ey = eln(3x+5) ey = 3x + 5 ey − 5 =x 3

y −1 =

f (x)

ex − 5 3

∀ℜ

Dominio (− 53 ; +∞) ∀ℜ  Imagen Dominio ∀ℜ f −1 (x) : Imagen (− 35 ; +∞) f (x) :



f (x) = 4e2x−2 + 8 f (x)

f (x) y = 4e2x−2 + 8

∀ℜ

y−8 = e2x−2 4   y−8 ln = ln(e)(2x − 2) 4   y−8 = 2x − 2 ln 4   +2 ln y−8 4 =x 2

y −1

=

ln

 x−8  4

+2

2

x−8 >0 4 x−8>0 x>8 (8; +∞)



Dominio ∀ℜ Imagen (8; +∞ )  Dominio (8; +∞) −1 f (x) : Imagen ∀ℜ f (x) :

f (x) = 6ln(3x + 3) − 6 f (x) 3x + 3 > 0 3x > −3 x > −1

(−1; +∞)

y = 6ln(3x + 3) − 6 y+6 = ln(3x + 3) 6 e

y+6 6

e

= eln(3x+3)

y+6 6

e

= 3x + 3

y+6 6

3

−2

=x

y −1 =

f (x)

x+6 6

3

−2

∀ℜ

x x+1



Dominio (−1; +∞) ∀ℜ  Imagen Dominio ∀ℜ −1 f (x) : Imagen (−1; +∞) f (x) :

f (x) =

e

f (x) x + 1 6= 0 x 6= −1

∀ℜ − {−1}

y=

x x+1

y(x + 1) = x xy + y = x y = x − xy y = x(1 − y) y =x 1−y y −1 =

x 1−x

1 − x 6= 0 −x = 6 −1 x = 6 1

∀ℜ − {1}



Dominio ∀ℜ − {−1} ∀ℜ − {1}  Imagen Dominio ∀ℜ − {1} f −1 (x) : Imagen ∀ℜ − {−1} f (x) :

f (x) =

x−2 3x + 5 f (x) 3x + 5 6= 0 3x 6= −5 x 6= − 53

∀ℜ − {− 53 }

x−2 3x + 5

y=

y(3x + 5) = x − 2 3xy + 5y = x − 2 3xy − x = −2 − 5y x(3y − 1) = −2 − 5y x=

−2 − 5y 3y − 1

y −1 =

−2 − 5x 3x − 1

3x − 1 6= 0 3x 6= 1 x 6= 31

∀ℜ − { 31 }

Dominio ∀ℜ − {− 53 } ∀ℜ − { 13 }  Imagen Dominio ∀ℜ − { 31 } f −1 (x) : Imagen ∀ℜ − {− 35 } f (x) :

f (x) =

√ 2x + 1



f (x) 2x + 1 ≥ 0 2x ≥ −1 x ≥ − 12

[− 12 ; +∞)

√ y = 2x + 1 2 y = 2x + 1 y 2 − 1 = 2x y2 − 1 =x 2

y −1 =

x2 − 1 2 ∀ℜ

Dominio [− 12 ; +∞) ∀ℜ  Imagen Dominio ∀ℜ −1 f (x) : Imagen [− 21 ; +∞) f (x) :



f (x) = ex

ex = 0 x = ln(0)

f (0) = e0 = 1

C + : ∀ℜ C− : ∅

ex = y x = ln(y) y −1 = ln(x) x>0 (0; +∞) ∀ℜ l´ım ex = ∞

x→∞

l´ım ex = e−∞ =

x→−∞

y=0 f (x) = e

1 1 = =0 ∞ e∞

−x

e−x = 0 −x = ln(0) x = −ln(0)

f (0) = e−0 = 1

C + : ∀ℜ C− : ∅

e−x = y −x = ln(y) y −1 = −ln(x) x>0 (0; +∞) ∀ℜ

l´ım e−x = e−∞ =

x→∞

1 1 = =0 ∞ e∞

y=0 f (x) = 2

x

2x = 0 x = log2 (0)

f (0) = 20 = 1

C + : ∀ℜ C− : ∅

2x = y x = log2 (y) y −1 = log2 (x)

x>0 (0; +∞) ∀ℜ l´ım 2x = ∞

x→∞

l´ım 2x = 2−∞ =

x→−∞

y=0

1 1 = =0 2∞ ∞

f (x) = 10x

10x = 0 x = log10 (0)

f (0) = 100 = 1

C + : ∀ℜ C− : ∅

10x = y x = log10 (y) y −1 = log10 (x) x>0 (0; +∞) ∀ℜ

l´ım 10x = ∞

x→∞

l´ım 10x = 10−∞ =

x→−∞

y=0

1 1 = =0 10∞ ∞

f (x) = 10−x

10−x = 0 −x = log10 (0) x = −log10 (0)

f (0) = 10−0 = 1

C + : ∀ℜ C− : ∅

10−x = y −x = log10 (y) y −1 = −log10 (x) x>0

(0; +∞) ∀ℜ l´ım 10x = 10−∞ =

x→∞

y=0 f (x) =



1 10

1 1 = =0 10∞ ∞

x

10−x = f (x) = ex + 1



1 10

x

ex + 1 = 0 ex = −1 x = ln(−1)

f (0) = e0 + 1 = 2

C + : ∀ℜ C− : ∅

ex + 1 = y ex = y − 1 x = ln(y − 1) y −1 = ln(x − 1) x−1>0 x>1 (1; +∞) ∀ℜ l´ım ex + 1 = ∞

x→∞

l´ım ex + 1 = e−∞ + 1 =

x→−∞

y=1

1 1 +1= +1=0+1=1 ∞ e∞

f (x) = 2ex

2ex = 0 ex = 0 x = ln(0)

f (0) = 2e0 = 2

C + : ∀ℜ C− : ∅

2ex = y ex = y2 x = ln( y2 ) y −1 = ln( x2 ) x 2

>0 x>0 (0; +∞) ∀ℜ l´ım 2ex = ∞

x→∞

l´ım 2ex = 2e−∞ = 2

x→−∞

y=0 f (x) = −2e



1 e∞



=2

x

−2ex = 0 ex = 0 x = ln(0)



1 ∞



= 2(0) = 0

f (0) = −2e0 = −2

C+ : ∅ C − : ∀ℜ

−2ex = y y ex = −2 y ) x = ln( −2 x −1 ) y = ln( −2 x −2

>0 x0 x−1>0 x>1 (1; +∞) ∀ℜ l´ım 2ex + 1 = ∞

x→∞

x

l´ım 2e +1 = 2e

x→−∞

−∞

+1 = 2



y=1 x

f (x) = −2e + 1

1 e∞



+1 = 2



 1 +1 = 2(0)+1 = 1 ∞

−2ex + 1 = 0 −2ex = −1 ex = 0, 5 x = ln(0, 5) ln(0, 5)

f (−1) = −2e−1 + 1 = 0, 2642 f (0) = −2e0 + 1 = −1 C + : (−∞; ln(0, 5)) C − : (ln(0, 5); +∞)

−2ex + 1 = y −2ex = y − 1 ex = y−1 −2 x = ln( y−1 ) −2 ) y −1 = ln( x−1 −2 x−1 > 0 −2 x−1 0 x > −2

(−2; +∞) ∀ℜ

l´ım 2ex − 2 = ∞

x→∞

x

l´ım 2e − 2 = 2e

x→−∞

−∞

−2 = 2



y = −2

1 e∞



−2 = 2

l´ım f (x)

x→+∞



1 ∞



l´ım f (x)

x→−∞

l´ım ex

x→+∞

l´ım ex = e∞ = ∞

x→+∞

l´ım ex

x→−∞

l´ım ex = e−∞ =

x→−∞

l´ım e−x

x→+∞

= 2(0) − 2 = −2

1 1 = =0 ∞ e∞

l´ım e−x = e−∞ =

x→∞

l´ım e

x→−∞

1 1 = =0 ∞ ∞ e

−x