guía 2 funciones

8/18/2019 Guía 2 Funciones

1/12

f : [0, + ∞[→ R f (x ) = x + 1√ x2 + 1 f

x ∈ [0, + ∞[ y > 0 ∀x ∈ Dom (f ) Rec (f )

x

y = x + 1√ x2 + 1

y2 = (x + 1) 2

x2

+ 1x2y2 + y2 = x2 + 2 x + 1

x2(y2 −1) −2x + ( y2 −1) = 0

x = (1 −y2) ± 4 −4(y2 −1)22(y2 −1)

x = (1 −y2) ±2 1 −(y2 −1)22(y2 −1)

y = {−1, 1} 1

−(y2

−1)2

≥ 0

|y2 −1| ≤ 1−1 ≤ y2 −1 ≤ 1

y2 ≥ 0y2 −2 ≤ 0

S 2 = [−√ 2, √ 2] y > 0

Rec (f ) =]0 , √ 2] f (x ) = |x + 1 | −2|x −3| −x.

|x + 1 | −2|x −3| −x ≥ 0|x + 1 | −2|x −3| ≥ x

Dom (f ) = [53

, 72

]


2/12

f : A → B f (x ) = x −13x + 2

A B f (x )

Dom (f ) = R − {−23 } A = R − {−23 } a, b ∈ Dom (f ) f (a ) = f (b) ⇒ a = b

f (a ) = a −13a + 2

= b−13b + 2

= f (b)

(a −1)(3b + 2) = ( b −1)(3a + 2)3ab + 2 a −3b −2 = 3ab + 2 b −3a −2

5a = 5ba = b

∀x ∈ A

B = Rec (f ) xf (x ) = y

y = x −13x + 2 ⇔ y(3x + 2) = x −1 ⇔ 3xy + 2 y = x −1

2y + 1 = x(1 −3y) ⇔ x = 2y + 11 −3y

B = Rec (f ) = R − {13 } f −1(x )

f −1(x ) = 2x + 1

1 −3x

f (x ) = x2 −1 g(x ) =2x + 3 x ≤ 0√ x 0 < x ≤ 8x3 x ≥ 8

g(f (x ))

g(f (x )) =2f (x ) + 3 f (x )

≤ 0

f (x ) 0 < f (x ) ≤ 8(f (x ))3 f (x ) ≥ 8g(f (x )) =

2(x2 −1) + 3 x2 −1 ≤ 0 (1)√ x2 −1 0 < x 2 −1 ≤ 8 (2)(x2 −1)3 x2 −1 ≥ 8 (3)

x2 −1 ≤ 0 ⇔ x ∈ [−1, 1]x2 −1 ≥ 8 ⇔ x2 −9 ≥ 0 ⇔ x ∈]− ∞, −3]∪[3, ∞[0 < x 2 −1 ≤ 8


3/12

0 < x 2 −1 ⇔ x ∈]− ∞, −1[∪]1, ∞[ x2 −1 ≤ 8 ⇔ x2 −9 ≤ 0 ⇔ x ∈ [−3, 3]

S (2) = [−3, −1[∪]1, 3]

g(f (x )) =2(x2 −1) + 3 x ∈ [−1, 1]√ x2 −1 x ∈ [−3, −1[∪]1, 3](x

2

−1)3

x ∈]− ∞, −3]∪[3, ∞[ f (x ) = 2 −√ 2x + 1

Dom (f )

2 −√ 2x + 1 ≥ 0 x

≥ −12

2 ≥√ 2x + 1

4 ≥ 2x + 13 ≥ 2xx ≤

32

Dom (f ) =

−1

2, 3

2 f (x )

a, b ∈ Dom (f ) a < b a < b

2a < 2b2a + 1 < 2b + 1

√ 2a + 1 < √ 2b + 1−√ 2a + 1 > −√ 2b + 1

2 −√ 2a + 1 > 2 −√ 2b + 1

2 −√ 2a + 1 > 2 −√ 2b + 1f (a ) > f (b) f [0, 1[→ R f (x ) =

1√ 1 −x2

Rec (f )


4/12

x

y = 1√ 1 −x2

y2 = 11 −x2

y2 −x2y2 = 1x2 =

y2 −1y2

x = y2 −1y2 x ∈ [0, 1[

0 ≤ y2

−1y 2 < 1

0 ≤ y2

−1y 2 < 10 ≤ y2 −1 < y 2

y2 ≥ 1 y > 0 x

y ≥ 1

Rec (f ) = [1 , + ∞[ f (x ) = 2tan x1+tan 2 x

f (x )

2tan x1 + tan 2 x

= 2 · sin xcos x1 + sin

2 xcos 2 x

=2sin xcos x

cos 2 x +sin 2 xcos 2 x

= 2sin x

cos x ·cos2 x

= 2 sin x cos x= sin(2 x )

f (x ) = sin(2x ) Dom (f )

sin(2x ) ≥ 0

sin(2x ) ≥ 0 ⇔ 2x ∈ [0, π ]∪[2π, 3π ]∪[4π, 5π ]∪....⇔

x ∈0, π2 ∪

π, 3π

2 ∪2π,

5π2 ∪

....

⇔ x ∈

k∈

Z

kπ, (2k + 1) π

2


5/12

f (x ) = 1+sin x1−cos x

f

cos x = 1

x = 2 kπk ∈ Z

Dom (f ) = R − {2kπ } f

f (x ) = 0 ⇔ 1 + sin x = 0

⇔

sin x =

−1

⇔ x =

3π2

, 7π

2 ,

11π2

,...

⇔ x =

3π2

+ 2 kπ

k ∈ Z |sin x| ≤ 1 |cos x| ≤ 1

1 + sin x ≥ 0 ∧ 1−cos x ≥ 0

1 + sin x

1 −cos x ≥ 0

f

f (x ) = a cos(wx ) + b sin(wx )

f f (x ) = √ a 2 + b2 sin(wx + φ)

√ a2 + b2 f (x )

√ a 2 + b2 = a√ a 2 + b2 cos(wx ) +

b√ a 2 + b2 sin(wx )

φ

cos φ = b√ a 2 + b2

sin φ = a√ a 2 + b2

f (x )√ a2 + b2 = sin φ cos(wx ) + cos φ sin(wx )

f (x ) = a2 + b2 sin(wx + φ).


6/12

f (x ) = sin x −cos x

f (x )√ 2 =

sin x√ 2 −

cos x√ 2

cos φ = 1

√ 2sin φ =

1√ 2

φ = π4

f (x )√ 2 = cos

π4

sin x −cos x sin π4

f (x ) = √ 2sin x − π4

f (x ) = 3sin x cos2 x −sin3 x + 12 cos(3x )

f (x ) = 2 sin x cos2 x + sin x cos2 x −sin3 x + 12

cos(3x )

= 2 sin x cos x cos x + sin x (cos2 x −sin2 x ) + 12

cos(3x )

= sin(2 x )cos x + sin x cos(2x ) + 12

cos(3x )

= sin(3 x ) + 12

cos(3x)

54

f (x )

54=

sin(3x )

54+

cos(3x )

2 54

sin φ = 1

2 54cos φ =

1

54

tan φ = 12 ⇒ φ = arctan

12

f (x ) = 54 sin 3x + arctan 12

log 3 + 2√ 23 −2√ 2 = log 3 + 2√ 2


7/12

log 3 + 2√ 23 −2√ 2 = log 3 + 2√ 23 −2√ 2 ·

3 + 2√ 23 + 2√ 2

= log 3 + 2√ 2 29 −8= log 3 + 2

√ 2

2

= log 3 + 2√ 2

(ex + e−x )2 −(ex −e−x )2

(ex + e−x )2 · 1 − ex −e−xex + e−x2

= 2ex

e2x + 1

(ex + e−x )2

−(ex

−e−x )2

(ex + e−x )2 · 1 − ex −e−xex + e−x2 =

e2x + 2 + e−2x

−e2x + 2

−e−2x

(ex + e−x )2 · (ex + e−x )2 −(ex −e−x )2(ex + e−x )2=

4

(ex + e−x )2 · 2

(ex + e−x )

= 2ex + e−x

= 2e 2 x +1

e x

= 2ex

e2x + 1

f (x ) = e2x −1e2x + 1

f −1(x ) = 12

ln1 + x1 −x

y = f (x ) I ⊆ Dom (f )

y = e2x −1

e2x + 1ye 2x + y = e2x −1

e2x (y

−1) =

−y

−1

e2x = −(1 + y)y −1

2x = ln1 + y1 −y

x = 1

2 ln

1 + y1 −y

f −1(x ) = 12

ln1 + x1 −x


8/12

f : Dom (f ) ⊆ R → R f (x ) = ln2(x ) −1 Dom (f ) x > 0

ln2(x ) −1 ≥ 0 u = ln( x )

u 2

−1

≥ 0

u ∈]− ∞, −1]∪[1, + ∞[

u ≤ −1∨

u ≥ 1

ln(x ) ≤ −1∨

ln(x ) ≥ 1

x ≤ e−1∨

x ≥ e

x ∈]− ∞, e−1

]∪[e, + ∞[

x > 0

Dom (f ) =]0 , e−1]∪[e, + ∞[


9/12

f (x ) = √ 3x −√ 7 f (x ) =

x2 −6x + 9x2 + 9

f (x ) =

1 −2x8 + 3 x

f (x ) = 1

3|x| − |x −2| f

f (x ) = |x|+ |x −1| g(x ) = |2x −3|+ |3x −4| − |3x −1| h (x ) =

1

3|x| − |x −2| f (x ) f (x ) |f (x )| max {f (x), 0} max {−f (x ), 0}

f (x ) = 3 x −2 f (x ) = x3

f (x ) = x2 −3x −1 f (x ) = 3x −2, x 11 −4x, x > 1

f ]−2, 1]→ M f (x) = 1 −xx + 2 f

M

f

f

f (x ) = x

x2 −9 f (x ) = x√ 3 −x f (x ) =

x2 −4x2 −9

f (x ) = x−1x 2 +1 g(x ) = 2x −13x +1 (fog )(x ) (gof )(x )

f (x ) = 6

− |2x + 1

| |x + 3 | −4

g(x ) = √ 5 −x2

f

(f + g)(2) 1 g

f

f : [0, + ∞[ → Bx → f (x ) = 1 −√ x + 1

B f


10/12

f : [0, + ∞[→ R f (x ) = x

x2 + 1

Rec (f ) f

f (x ) = 2x1 − |x|

∀y > 0 x ∈ [0, 1] y = f (x ) ]0, 1[ f

f f (x ) = x

x2 − |x|

f

g :]1, + ∞[→ g(]1 ∞[) g(x ) = f (x ) g g−1(x )

f (x ) = xx −1

g(x ) = x2

−1 (( f + g) ◦g)(x )

f (x ) = 1

x2 + 1 g(x ) =

x1 + x

A(h )

A(h ) = f (x + h ) −(f ◦g)(x + h )

(f ·g)(x + h ) − f (x + h )g (x + h )

A(h ) h x A(1)

A(−1)

.

f (x ) = x −1x −2 −1 Dom (f ) f (x ) =

x + 1x −1

, x < 0x −1x + 1

, x ≥ 0 f Dom (f )

f (g(x )) g(x ) = x2 −5x + 6

f (x ) = x −5 x < 5−x −5 x ≥ 5 (f ◦f )(x )

f : R → R g : R −{1} → R f (x ) =2x + 3 , 9 < xx − |x| , −9 ≤ x ≤ 9x −4 , x < −9

g(x ) = x + 1x −1

g−1of f (g(x))

tan( θ) = x sin(θ) cos(2θ) x


11/12

f (x ) = sin( x ) g(x ) = 2 arctan( x ) (f ◦g)(x ) = 2xx2 + 1

f (x ) = 1 2+ 2+ √ 2+2cos(8 x )

f f (x ) = sec x2 f f

f (x ) = 1

√ cos x

f

g(x ) = f (x) ·cos x (g(x ))2 = −sin x h (x ) = |sin x|

f (x ) = cos(2 x ) + √ 3 sin(2x ) f (x ) = A sin(wx + φ) A w φ

f (x ) = 2 2 + √ 2 sin(3x ) + 2 2 −√ 2cos(3x ). f (x ) = A sin(wx + φ) A w φ f (x ) = cos( π

−2x )

−√ 3cos 2x + π2

f

OX 0, 49π12

f (x ) = ex −1ex + 1

Rec (f )

w = W 1 r ·(1 + r )s

(1 + r )s −1s =

log(w) −log(w −W 1 r )log(1 + r )

x ∈ R log2 + log(4 x −2 + 9) = 1 + log(2 x −2 + 1)

log2 x(log2 a )2 −

2loga xlog 1

2a

= log 3√ a x ·(loga x ) a = 1 a ∈ R +

f (x ) = log 3(x2 −3x −4) x ∈ Dom (f ) f (x ) < 1 x ABC b h P

S x

x r x

2 p = 12 V (x ) V (x ) x

2a r1 r2


12/12

r 1 + r 2 = 4a2 + √ 2

r 1

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