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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
]
8/18/2019 Guía 2 Funciones
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
8/18/2019 Guía 2 Funciones
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 )
8/18/2019 Guía 2 Funciones
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
8/18/2019 Guía 2 Funciones
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).
8/18/2019 Guía 2 Funciones
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
8/18/2019 Guía 2 Funciones
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
8/18/2019 Guía 2 Funciones
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, + ∞[
8/18/2019 Guía 2 Funciones
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
8/18/2019 Guía 2 Funciones
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
8/18/2019 Guía 2 Funciones
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
8/18/2019 Guía 2 Funciones
12/12
r 1 + r 2 = 4a2 + √ 2
r 1