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Let be a function satisfying and
MATHEMATICS · Single correct · AoPS · Mathematics · Functions · Functions
Problem
Let
f
:
R
→
R
f: R \to R
f
:
R
→
R
be a function satisfying
f
(
x
y
2
)
=
f
(
x
)
f
(
y
)
2
,
∀
x
,
y
∈
R
f\left(\frac{xy}{2}\right) = \frac{f(x)f(y)}{2}, \forall x, y \in R
f
(
2
x
y
)
=
2
f
(
x
)
f
(
y
)
,
∀
x
,
y
∈
R
and
f
(
1
)
=
f
′
(
1
)
≠
0
f(1) = f'(1) \neq 0
f
(
1
)
=
f
′
(
1
)
=
0
. Then,
f
(
x
)
+
f
(
−
x
)
f(x) + f(-x)
f
(
x
)
+
f
(
−
x
)
is (for all non-zero real values of
x
x
x
) (1) constant (2) can't be discussed (3)
1
x
\frac{1}{x}
x
1
(4)
1
x
2
\frac{1}{x^2}
x
2
1
Related theory
Functions
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