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Austrian-Polish
1994 Austrian-Polish Competition
1
1
Part of
1994 Austrian-Polish Competition
Problems
(1)
if f(x +19)>=f (x) + 19 and f (x + 94)<= f (x) + 94 then f (x + 1) = f (x) + 1
Source: Austrian - Polish 1994 APMC
5/3/2020
A function
f
:
R
→
R
f: R \to R
f
:
R
→
R
satisfies the conditions:
f
(
x
+
19
)
≤
f
(
x
)
+
19
f (x + 19) \le f (x) + 19
f
(
x
+
19
)
≤
f
(
x
)
+
19
and
f
(
x
+
94
)
≥
f
(
x
)
+
94
f (x + 94) \ge f (x) + 94
f
(
x
+
94
)
≥
f
(
x
)
+
94
for all
x
∈
R
x \in R
x
∈
R
. Prove that
f
(
x
+
1
)
=
f
(
x
)
+
1
f (x + 1) = f (x) + 1
f
(
x
+
1
)
=
f
(
x
)
+
1
for all
x
∈
R
x \in R
x
∈
R
.
functional
Functional inequality
functional equation
algebra