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Problems
Contests
International Contests
Nordic
2000 Nordic
4
4
Part of
2000 Nordic
Problems
(1)
1/3 <= f(1/3) <= 4/7
Source: Nordic Mathematical Contest 2000 #4
10/3/2017
The real-valued function
f
f
f
is defined for
0
≤
x
≤
1
,
f
(
0
)
=
0
,
f
(
1
)
=
1
0 \le x \le 1, f(0) = 0, f(1) = 1
0
≤
x
≤
1
,
f
(
0
)
=
0
,
f
(
1
)
=
1
, and
1
2
≤
f
(
z
)
−
f
(
y
)
f
(
y
)
−
f
(
x
)
≤
2
\frac{1}{2} \le \frac{ f(z) - f(y)}{f(y) - f(x)} \le 2
2
1
≤
f
(
y
)
−
f
(
x
)
f
(
z
)
−
f
(
y
)
≤
2
for all
0
≤
x
<
y
<
z
≤
1
0 \le x < y < z \le 1
0
≤
x
<
y
<
z
≤
1
with
z
−
y
=
y
−
x
z - y = y -x
z
−
y
=
y
−
x
. Prove that
1
7
≤
f
(
1
3
)
≤
4
7
\frac{1}{7} \le f (\frac{1}{3} ) \le \frac{4}{7}
7
1
≤
f
(
3
1
)
≤
7
4
.
inequalities
Functional inequality
calculus