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Poland Contests
Poland - Second Round
1993 Poland - Second Round
6
6
Part of
1993 Poland - Second Round
Problems
(1)
f(x)f(f(x)) = 1 , f(1000) = 999, continuous, f(500)?
Source: Polish second round 1993 p6
1/19/2020
A continuous function
f
:
R
ā
R
f : R \to R
f
:
R
ā
R
satisfies the conditions
f
(
1000
)
=
999
f(1000) = 999
f
(
1000
)
=
999
and
f
(
x
)
f
(
f
(
x
)
)
=
1
f(x)f(f(x)) = 1
f
(
x
)
f
(
f
(
x
))
=
1
for all real
x
x
x
. Determine
f
(
500
)
f(500)
f
(
500
)
.
function
continuous function
algebra
functional equation