Một số dạng bài toán về phương trình hàm

A B f
a ∈ A b ∈ B f A
B f : A −→ B. b a b = f(a).
f : A −→ B
f(A) = {f(a)|a ∈ A} A,
f f
−1
(b) = {a ∈ A|f(a) = b} b
f : A −→ B a
1
, a
2
∈ A
a
1
= a
2
f(a
1
) = f(a
2
).
f : A −→ B f(a
1
) = f(a
2
) a
1

= a
2
.
f : A −→ B b ∈ B
a ∈ A f(a) = b.
f : A −→ B f(A) = B.
f : A −→ B f
f : A −→ B b ∈ B
a ∈ A f(a) = b.
f : A −→ B
y ∈ B x = f
−1
(y) f
f
−1
.
X ⊂ R Y ⊂ R f : X −→ Y
X Y.
f : X −→ Y.
X f.
x
0
∈ X f(x
0
) f x
0
.
f(X) f.
y
0

f f(x) = y
0
f(x) = y
0
y
0
f.
f ⇔ x y = f(x) x ∈ X, y ∈ Y
f ⇔ x y = f (x) x ∈ X, y ∈ Y
f : D −→ R M ⊂ D
M
∀x ∈ M ⇒ −x ∈ M f(−x) = f(x), ∀x ∈ M.
f : D −→ R M ⊂ D M
∀x ∈ M ⇒ −x ∈ M f(−x) = −f(x), ∀x ∈ M.
f : D −→ R a (a > 0)
M M ⊂ D

∀x ∈ M ⇒ x ± a ∈ M
f(x + a) = f (x), ∀x ∈ M
f M. T (T > 0)
f f T
T.
f : D −→ R b (b > 0)
M M ⊂ D

∀x ∈ M ⇒ x ± b ∈ M
f(x + b) = −f(x), ∀x ∈ M
f b
0
M

b
0
M b
0
f M.
f : D −→ R
a (a /∈ {0, 1, −1}) M M ⊂ D

∀x ∈ M ⇒ a
±1
∈ M
f(ax) = f(x), ∀x ∈ M.
f : D −→ R
a (a /∈ {0, 1, −1}) M M ⊂ D

∀x ∈ M ⇒ a
±1
∈ M
f(ax) = −f(x), ∀x ∈ M.
f D ⊂ R x
0
∈ D. f
x
0
lim
x−→x
0
f(x) = f(x
0
).

f(x) (a; b)
(a; b) x ∈ (a; b).
f(x) [a; b]
[a; b] (a; b) lim
x−→a
+
f(x) = f(a), lim
x−→b
−
f(x) = f(b).
f(x) (a; b)
∀x
1
, x
2
∈ (a, b) x
1
≤ x
2
⇒ f(x
1
) ≤ f(x
2
).
f(x) (a; b)
∀x
1
, x
2
∈ (a, b) x

1
≤ x
2
⇒ f(x
1
) ≥ f(x
2
).
(a; b)
(a; b).
f(x)
(a; b)
∀x
1
, x
2
∈ (a, b) x
1
< x
2
⇒ f(x
1
) < f(x
2
).
f(x)
(a; b)
∀x
1
, x

2
∈ (a, b) x
1
< x
2
⇒ f(x
1
) > f(x
2
).
(a, b)
(a; b)
(a; b) (a; b)
f(x) g(x) f(x) + g(x)
f(x) g(x) f(x)g(x)
f(x) (a; b) f(f(x))
f(x)
f(x) = f(−x), ∀x ∈ R.
f(x) =
1
2
[f(x) + f(−x)], ∀x ∈ R.
f(x) =
1
2
[g(x) + g(−x)], ∀x ∈ R
g R f
f f
f(x) =
1

2
[g(x) + g(−x)], ∀x ∈ R
g R
f(x)
f(−x) = −f(x), ∀x ∈ R.
f(x) =
1
2
[f(x) − f (−x)], ∀x ∈ R.
f(x) =
1
2
[g(x) − g(−x)], ∀x ∈ R
g R f
f f
f(x) =
1
2
[g(x) − g(−x)], ∀x ∈ R
g R
x
0
∈ R. f
f(2x
0
− x) = f (x), ∀x ∈ R.
x = x
0
− t(⇔ t = x
0

− x) 2x
0
− x = x
0
+ t
f(x
0
+ t) = f (x
0
− t), ∀t ∈ R.
g(t) = f(x
0
+ t) g(−t) = f(x
0
− t), f(t) = g(t − x
0
).
g(t) = g(−t), ∀t ∈ R. g(t) R.
f(x) = g(x −x
0
), ∀x ∈ R, g(x) R.
a, b ∈ R. f(x)
f(a − x) + f (x) = 2b, ∀x ∈ R.
a
2
− x = t. x =
a
2
− t a − x =
a

2
+ t.
f(
a
2
+ t) + f(
a
2
− t) = 2b, ∀t ∈ R.
f(
a
2
+ t) − b = g(t), ∀t ∈ R.
g(t) + g(−t) = 0, ∀t ∈ R
⇔ g(t) = −g(−t), ∀t ∈ R.
g(t) R.
f(x) = g(x −
a
2
) + b
g(x) R.
f(x)
f(x) − f(−x) = 2014 sin x, ∀x ∈ R.
f(x) − f (−x) = 1007 sin x − 1007 sin(−x), ∀x ∈ R
⇔ f(x) − 1007 sin x = f(−x) − 1007 sin(−x), ∀x ∈ R.
g(x) = f(x) − 1007 sin x, ∀x ∈ R.
g(x) = g(−x), ∀x ∈ R.
g(x) R.
f(x) = g(x) + 1007 sin x, ∀x ∈ R, g(x) R
f(x)

f(x) + f(−x) =
2 cos x
√
x
2
+1
, ∀x ∈ R.
f(x) + f(−x) =
cos x
√
x
2
+ 1
+
cos x
√
x
2
+ 1
, ∀x ∈ R
⇔ f(x) −
cos x
√
x
2
+1
= −[f(−x) −
cos(−x)
√
(−x)

2
+1
], ∀x ∈ R.
g(x) = f(x) −
cos x
√
x
2
+1
, ∀x ∈ R.
g(x) = −g(−x), ∀x ∈ R.
⇔ g(−x) = −g(x), ∀x ∈ R.
g(x) R.
f(x) = g(x) +
cos x
√
x
2
+ 1
, ∀x ∈ R,
g(x) R.
f
f(x + π) − f (x) = 2 cos x, ∀x ∈ R.
f(x + π) + cos(x + π) = f(x) + cos x, ∀x ∈ R.
g(x) = f(x) + cos x, ∀x ∈ R f(x) = g(x) −cos x, ∀x ∈ R
g(x + π) = g(x), ∀x ∈ R.
g π R.
f(x) = g(x) − cos x, ∀x ∈ R,
g π, R.
f

f(x + 2π) − f(x) = sin x, ∀x ∈ R.
sin x =
(x + 2π) − x
2π
sin(x + 2π) =
x + 2π
2π
sin(x + 2π) −
x
2π
sin x
f(x + 2π) − f (x) =
x + 2π
2π
sin(x + 2π) −
x
2π
sin x, ∀x ∈ R
f(x + 2π) −
x+2π
2π
sin(x + 2π) = f(x) −
x
2π
sin x, ∀x ∈ R.
g(x) = f(x) −
x
2π
sin x, ∀x ∈ R f(x) = g(x) +
x

2π
sin x, ∀x ∈ R
g(x + 2π) = g(x), ∀x ∈ R.
g 2π R.
f(x) = g(x) +
x
2π
sin x, ∀x ∈ R,
g 2π R.
f
f(x + 1) − f(x) = 2x, ∀x ∈ R.
2x = [(x + 1)
2
− (x + 1)] − (x
2
− x)
f(x + 1) − f(x) = [(x + 1)
2
− (x + 1)] − (x
2
− x), ∀x ∈ R
⇔ f(x + 1) − [(x + 1)
2
− (x + 1)] = f(x) − (x
2
− x), ∀x ∈ R.
g(x) = f(x) − (x
2
− x), ∀x ∈ R f(x) = g(x) + (x
2

− x), ∀x ∈ R.
g(x) = g(x + 1), ∀x ∈ R.
g(x) 1 R.
f(x) = g(x) + (x
2
− x), ∀x ∈ R,
g(x) 1 R.
f
f(x + 1) − f(x) = 2.3
−x
, ∀x ∈ R.
2.3
−x
= 3.3
−x
− 3
−x
= 3
1−x
− 3
−x
= 3
1−x
− 3
1−(x+1)
.
f(x + 1) − f(x) = 3
1−x
− 3
1−(x+1)

, ∀x ∈ R
⇔ f(x + 1) + 3
1−(x+1)
= f(x) + 3
1−x
, ∀x ∈ R.
g(x) = f(x) + 3
1−x
, ∀x ∈ R f(x) = g(x) − 3
1−x
, ∀x ∈ R.
g(x + 1) = g(x), ∀x ∈ R.
g(x) 1 R.
f(x) = g(x) − 3
1−x
, ∀x ∈ R,
g(x) 1 R.
f
f(3x) = f(x), ∀x ∈ R.
x > 0 x = 3
u
(u = log
3
x)
f(3
u+1
) = f(3
u
), ∀u ∈ R.
g(u) = f(3

u
), ∀u ∈ R.
g(u + 1) = g(u), ∀u ∈ R.
g 1 R.
f(x) = f(3
u
) = g(u) = g(log
3
x), ∀x > 0.
∀x > 0,
f(3x) = g(log
3
(3x)) = g(1 + log
3
x) = g(log
3
x) = f(x)
x > 0 f(x) = g(log
3
x), g 1,
R.
x < 0 −x = 3
u
(u = log
3
(−x)).
f(−3
u+1
) = f(−3
u

), ∀u ∈ R.
h(u) = f(−3
u
), ∀u ∈ R.
h(u + 1) = h(u), ∀u ∈ R.
h 1 R.
f(x) = f(−3
u
) = h(u) = h(log
3
(−x)), ∀x < 0.
∀x < 0
f(3x) = h(log
3
(−x)) = h(1 + log
3
(−x)) = h(log
3
(−x)) = f(x)
x < 0 f(x) = h(log
3
(−x)), h 1
R
f(x) =



g(log
3
x) x > 0

c x = 0
h(log
3
(−x))
x < 0
g, h 1 R,
f
f(−2014x) = f(x), ∀x ∈ R.
f(2014
2
x) = f[(−2014)
2
x] = f[−2014(−2014x)] = f(−2014x) = f(x), ∀x ∈ R.

f(x) =
1
2
[f(x) + f(−2014x)], ∀x ∈ R
f(x) = f(2014
2
x), ∀x ∈ R.
f(x) =
1
2
[g(x) + g(−2014x)], ∀x ∈ R
g 2014
2
R
g(x) = g(2014
2

x), ∀x ∈ R.
f f
f(x) =
1
2
[g(x) + g(−2014x)], ∀x ∈ R
g 2014
2
R.
g(x) =



g
1
(
1
2
log
2014
x) x > 0
c
x = 0
g
2
(
1
2
log
2014

(−x)) x < 0
g
1
, g
2
1 R,
f(ax) = f(x), ∀x ∈ R a = 0, a = ±1 .
f 2 R
f(x + 1) = −2f(x) + 3, ∀x ∈ R.
f 2 R
f(x + 2) = f(x), ∀x ∈ R.
x x + 1
f(x + 2) = −2f(x + 1) + 3, ∀x ∈ R.
f(x) = −2f(x + 1) + 3, ∀x ∈ R.
f(x) = 1, ∀x ∈ R.
f(x) ≡ 1.
h(x) R. f(x)
f(x) + f(x + 1) + f(x + 2) = h(x), ∀x ∈ R.
f(x) 3 R
f(x + 3) = f(x), ∀x ∈ R.
x x + 1, x + 2
h(x) = h(x + 1) = h(x + 2), ∀x ∈ R.
h(x) =
1
3
[h(x) + h(x + 1) + h(x + 2)], ∀x ∈ R.

g(x) = g(x + 3)
g(x) + g(x + 1) + g(x + 2) = 0, ∀x ∈ R
g(x) = f(x) −

1
3
h(x), ∀x ∈ R.

g(x) = g(x + 3)
g(x) =
1
3
(2g(x) − g(x + 1) − g(x + 2)), ∀x ∈ R.
g(x) =
1
3
(2q(x) − q(x + 1) − q(x + 2)), ∀x ∈ R.
q(x) 3 R
x ∈ R
g(x + 1) =
1
3
(2q(x + 1) − q(x + 2) − q(x))
g(x + 2) =
1
3
(2q(x + 2) − q(x) − q(x + 1))
g(x + 3) =
1
3
(2q(x) − q(x + 1) − q(x + 2)) = g(x).
g(x) + g(x + 1) + g(x + 2) = 0 g(x + 3) = g(x) g(x)
g(x) q(x) = g(x) q(x) =
q(x + 3), ∀x ∈ R

1
3
(2q(x) − q(x + 1) − q(x + 2)) =
1
3
(2g(x) − g(x + 1) − g(x + 2))
=
1
3
(3g(x) − (g(x) + g(x + 1) + g(x + 2))) = g(x).
h(x) = h(x + 1) = h(x + 2), ∀x ∈ R.
f(x) = g(x) +
1
3
h(x),
g(x) =
1
3
(2q(x) − q(x + 1) − q(x + 2)), ∀x ∈ R.
q(x) 3 R
f(ax + b) = cf(x) + d, a = 0, c = 0.
f(x)
f(x + 1) = f(x) + 3, ∀x ∈ R.
f(x) = 3x + g(x), ∀x ∈ R.
3(x + 1) + g(x + 1) = 3x + g(x) + 3, ∀x ∈ R
⇔ g(x + 1) = g(x), ∀x ∈ R.
g(x) 1 R.
f(x) = 3x + g(x), ∀x ∈ R, g(x) 1 R.
f(x) = 3x + g(x), ∀x ∈ R, g(x) 1
R.

f(x) = 3x + g(x), ∀x ∈ R
f(x + 1) = f(x) + f(1), ∀x ∈ R
f(1) = 3 f(x) = ax, ∀x ∈ R. f(1) = 3 ⇔ a.1 = 3 ⇔ a = 3.
f(x) = 3x. f(x) = 3x + g(x).
f(x)
f(x + 103) = f(x) − 515, ∀x ∈ R.
f(x) = −5x + g(x), ∀x ∈ R.
−5(x + 103) + g(x + 103) = −5x + g(x) − 515, ∀x ∈ R
⇔ g(x + 103) = g(x), ∀x ∈ R.
g(x) 103 R
f(x) = −5x + g(x), ∀x ∈ R, g(x) 103
R.
f(x) = −5x + g(x), ∀x ∈ R, g(x) 103
R
f(x + a) = f(x) + b, ∀x ∈ R.
P (x)
P (1) = 2
P (x + 4) = P (x + 1) + 2, ∀x ∈ R.
x x − 1
P (x + 3) = P (x) + 2, ∀x ∈ R.
P (x) =
2
3
x + Q(x). Q(x)
2
3
(x + 3) + Q(x + 3) =
2
3
x + Q(x) + 2, ∀x ∈ R

⇔ Q(x + 3) = Q(x), ∀x ∈ R
⇔ Q(x) = c, ∀x ∈ R
P (x) =
2
3
x + c, ∀x ∈ R.
P (1) = 2 2 =
2
3
.1 + c ⇔ c =
4
3
.
P (x) =
2
3
x +
4
3
, ∀x ∈ R.
f(x)
f(x − 3) = −f(x) + 2, ∀x ∈ R.
f(x) = 1 + g(x), ∀x ∈ R.
1 + g(x − 3) = −1 −g(x) + 2, ∀x ∈ R
⇔ g(x − 3) = −g(x), ∀x ∈ R.
⇔ g(x) = −g(x + 3), ∀x ∈ R.
g(x) 3 R.
f(x) = 1 + g(x), ∀x ∈ R, g(x) 3
R.
f(x) = 1 + g(x), ∀x ∈ R, g(x) 3

R
f(x) = 1 + g(x), ∀x ∈ R
c = −c + 2 f c = 1
f(x) = 1 + g(x), ∀x ∈ R. 2
0
f(x + a) = −f(x) + b, ∀x ∈ R
f
f(x + 2) = 2f(x) + 3, ∀x ∈ R.
f(x) = −3 + g(x), ∀x ∈ R.
−3 + g(x + 3) = −6 + 2g(x) + 3, ∀x ∈ R
⇔ g(x + 2) = 2g(x), ∀x ∈ R.
g(x) = (
√
2)
x
h(x), ∀x ∈ R.
(
√
2)
x+2
h(x + 2) = 2(
√
2)
x
h(x), ∀x ∈ R
⇔ h(x + 2) = h(x), ∀x ∈ R.
h(x) 2 R.
f(x) = −3 + (
√
2)

x
h(x), ∀x ∈ R,
h(x)
2 R.
g(x) = (
√
2)
x
h(x), ∀x ∈ R
g(x + 2) = g(x)g(2), ∀x ∈ R
g(2) = 2. g g(x) = a
x
.
g(2) = 2 ⇒ a
2
= 2 ⇒ a =
√
2.
g(x) = (
√
2)
x
h(x), ∀x ∈ R
2.
f(x + a) = αf(x) + b, ∀x ∈ R
α, a, b α = ±1.
f
f(x + 2) + 3f(x) = x, ∀x ∈ R.
x = [
1

4
(x + 2) −
1
8
] + 3(
1
4
x −
1
8
)
f(x + 2) + 3f(x) = [
1
4
(x + 2) −
1
8
] + 3(
1
4
x −
1
8
), ∀x ∈ R
⇔ f(x + 2) − [
1
4
(x + 2) −
1
8

] = −3[f(x) − 3(
1
4
x −
1
8
)], ∀x ∈ R
⇔ g(x + 2) = −3g(x), ∀x ∈ R
g(x) = f(x) − (
1
4
x −
1
8
)
g(x) = (
√
3)
x
h(x), ∀x ∈ R.
(
√
3)
x+2
h(x + 2) = −3(
√
3)
x
h(x), ∀x ∈ R
⇔ h(x + 2) = −h(x), ∀x ∈ R.

h(x) 2 R.
f(x) = (
1
4
x −
1
8
) + (
√
3)
x
h(x), ∀x ∈ R,
h(x)
2 R.
f
f(3x) = f(x) − 2, ∀x > 0.
f(x) = log
1
√
3
x + g(x), ∀x > 0.
log
1
√
3
(3x) + g(3x) = log
1
√
3
x + g(x) − 2, ∀x > 0

⇔ g(3x) = g(x), ∀x > 0.
x = 3
u
(u = log
3
x).
⇔ g(3
u+1
) = g(3
u
), ∀u ∈ R.
h(u) = g(3
u
), ∀u ∈ R. h(u + 1) = h(u), ∀u ∈ R.
h 1 R.
f(x) = log
1
√
3
x + g(x) = log
1
√
3
x + g(3
u
) = log
1
√
3
x + h(u) = log

1
√
3
x + h(log
3
x), ∀x > 0
f(x) = log
1
√
3
x + h(log
3
x), ∀x > 0.
f(x) = log
1
√
3
x + h(log
3
x), ∀x > 0
h 1
R
f
f(5x) = f(x) + 2, ∀x > 0.
f(x) = log
√
5
x + g(x), ∀x > 0.
log
√

5
(5x) + g(5x) = log
√
5
x + g(x) + 2, ∀x > 0
⇔ g(5x) = g(x), ∀x > 0.
x = 5
u
(u = log
5
x).
⇔ g(5
u+1
) = g(5
u
), ∀u ∈ R.
h(u) = g(5
u
), ∀u ∈ R. h(u + 1) = h(u), ∀u ∈ R.
h 1 R.
f(x) = log
√
5
x + g (x) = log
√
5
x + g (5
u
) = log
√

5
x + h(u) = log
√
5
x + h(log
5
x), ∀x > 0
f(x) = log
√
5
x + h(log
5
x), ∀x > 0.
f(x) = log
√
5
x + h(log
5
x), ∀x > 0
h 1
R
f(x) = log
√
5
x+g(x)
f(5x) = f(5) + f(x), ∀x > 0.
f(x) = log
a
x, ∀x > 0 f(5) = 2.
log

a
5 = 2 ⇒ a
2
= 5 ⇒ a =
√
5.
f(x) = log
√
5
x + g(x), ∀x > 0.
f(ax) = f(x) + b, ∀x > 0 a > 0, a = 1
f
f(2x − 1) = −f(x) + 2.∀x ∈ R.
f(x) = 1 + g(x), ∀x ∈ R.
g(2x − 1) + 1 = −g(x) − 1 + 2, ∀x ∈ R
⇔ g(2x − 1) = −g(x), ∀x ∈ R.
x = 1 + t 2x = 2t + 1 t = x − 1
g(2t + 1) = −g(t + 1), ∀t ∈ R.
h(t) = g(t + 1), ∀t ∈ R
h(2t) = −h(t), ∀t ∈ R.
f(x) = 1 + g(x) = 1 + g(1 + t) = 1 + h(t) = 1 + h(x − 1), ∀x ∈ R.
f(x) = 1 + h(x − 1), ∀x ∈ R
h h(2t) = −h(t), ∀t ∈ R.
f
f(2x + 1) = 3f(x) + 2, ∀x ∈ R.
f(x) = −1 + g(x), ∀x ∈ R.
−1 + g(2x + 1) = 3.(−1 + g(x)) + 2, ∀x ∈ R.
⇔ g(2x + 1) = 3g(x), ∀x ∈ R.
x = −1 + t 2x + 1 = 2t − 1 t = x + 1
g(2t − 1) = 3g(t − 1), ∀t ∈ R.

h(t) = g(t − 1), ∀t ∈ R.
h(2t) = 3h(t), ∀t ∈ R.
t = 0 h(0) = 0.
t = 0, h(t) = |t|
log
2
3
.k(t).
|2t|
log
2
3
.k(2t) = 3|t|
log
2
3
.k(t) ⇔ k(2t) = k(t)
f(x) =

2
3
x = −1
2
3
|x + 1|
log
2
3
k(x + 1) x = −1
k k(2t) = k(t), ∀t = 0.

f
f(−2x + 3) = 3f(x) − 5, ∀x ∈ R.
f(x) =
5
2
+ g(x), ∀x ∈ R.
5
2
+ g(−2x + 3) = 3.(
5
2
+ g(x)) − 5, ∀x ∈ R
⇔ g(−2x + 3) = 3g(x), ∀x ∈ R.
x = 1 + t −2x + 3 = −2t + 1, t = x − 1
g(−2t + 1) = 3g(t + 1), ∀t ∈ R.
h(t) = g(t + 1), ∀t ∈ R.
h(−2t) = 3h(t), ∀t ∈ R
⇔ h(4t) = 9h(t), ∀t ∈ R.
t = 0 h(0) = 0.
t = 0, h(t) = |t|
log
2
3
.k(t).
|4t|
log
2
3
.k(4t) = 9|t|
log

2
3
.k(t) ⇔ k(4t) = k(t)
f(x) =

5
2
x = 1
5
2
|x − 1|
log
2
3
k(x − 1)
x = 1
k k(4t) = k(t), ∀t = 0.
x = 1 + t
−2x + 3 = x ⇔ x = 1. x = 1
t = 0, x = 1 + t x − 1 = t.
h(4t) = h(4)h(t), ∀t ∈ R h(4) = 9. h(t) = t
a
. h(4t) = 9h(t), ∀t ∈ R
(4t)
a
= 9t
a
⇔ 4
a
= 9 ⇔ a = log

2
3.
h(t) = |t|
log
2
3
.k(t), ∀t = 0.
f(ax + b) = αf(x) + c, ∀x ∈ R, a = ±1, α = 1.
f
f(0) = 0 f(2x + 1) = 3f(x) + 5, ∀x ≥ 0.
f(x) =
−5
2
+ g(x), ∀x ≥ 0.
−5
2
+ g(2x + 1) = 3.(
−5
2
+ g(x)) + 5, ∀x ≥ 0
⇔ g(2x + 1) = 3g(x), ∀x ≥ 0.
x = −1 + t 2x + 1 = 2t − 1, t = x + 1
g(2t − 1) = 3g(t − 1), ∀t ≥ 1.
h(t) = g(t − 1), ∀t ≥ 1.
h(2t) = 3h(t), ∀t ≥ 1.
h(t) = t
log
2
3
.k(t), ∀t ≥ 1.

t
log
2
3
.k(2t) = 3|t|
log
2
3
.k(t) ⇔ k(2t) = k(t).
x ≥ 0
f(x) =
−5
2
+g(x) =
−5
2
+g(t−1) =
−5
2
+h(t) =
−5
2
+t
log
2
3
.k(t) =
−5
2
+(x+1)

log
2
3
.k(x+1)
k k(2t) = k(t), ∀t ≥ 1.
f(0) = 0 ⇔ k(1) =
5
2
.
k(t) =
5
2
, ∀t ≥ 1.
f(x) =
−5
2
+
5
2
.(x + 1)
log
2
3
, ∀x ≥ 0.
f
f(x) =
−5
2
+
5

2
.(x + 1)
log
2
3
, ∀x ≥ 0.
ω(x) =
ax + b
cx + d
, c = 0, ad − bc = 0.
f(ω(x)) = αf(x) + β.
f R\{−2}
f(
−1
x + 2
) = 3f(x) − 4, ∀x = −2.
−1
x + 2
= x ⇔ x
2
+ 2x + 1 = 0 ⇔ x = −1.
x = −1 f(−1) = 2.
x = −1, x = −2. f (x) = 2 + g(x), ∀x = −1, x = −2.
2 + g(
−1
x + 2
) = 3.[2 + g(x)] − 4, ∀x = −1, x = −2
⇔ (
−1
x + 2

) = 3g(x), ∀x = −1, x = −2.
x = −1, x = −2,
1
x+1
= t
x = −1 +
1
t
−1
x + 2
=
−1
1 +
1
t
=
−t
t − 1
= −1 +
1
t + 1
.
g(−1 +
1
t + 1
) = 3g(−1 +
1
t
), ∀t = −1, t = 0.
g(−1 +

1
t
) = h(t), ∀t = −1, t = 0.
h(t + 1) = 3h(t), ∀t = −1, t = 0.
h(t) = 3
t
k(t), ∀t = −1, t = 0.
3
t+1
k(t + 1) = 3.[3
t
k(t)], ∀t = −1, t = 0
⇔ k(t + 1) = k(t), ∀t = −1, t = 0.
f(x) =

2
x = −1
2 + 3
1
x+1
k(
1
x+1
)
x = −1, x = −2
k
k(t + 1) = k(t), ∀t = −1, t = 0.
1
x+1
= t, ∀x = −1, x = −2

−1
x+2
= x
x = −1. x = −1 0
x = −1 ∞ x = −1 ∞
1
x+1
= t,
x −→ −1 t −→ ∞. x = α +
β
at+b
−1
x+2
= α +
β
ct+d
.
f R\{2}
f(
−1
x − 2
) = 2f(x) + 3, ∀x = 2.
−1
x − 2
= x ⇔ x
2
− 2x + 1 = 0 ⇔ x = 1.
x = 1 f(1) = −3.
x = 1, x = 2. f(x) = −3 + g(x), ∀x = 1, x = 2.
−3 + g(

−1
x − 2
) = 2.[−3 + g(x)] + 3, ∀x = 1, x = 2
⇔ g(
1
2 − x
) = 2g(x), ∀x = 1, x = 2.
x = 1, x = 2,
1
x−1
= t x = 1 +
1
t
1
2 − x
=
1
1 −
1
t
=
t
t − 1
= 1 +
1
t − 1
.