MHB Possible Values: $f(8765)-f(4321)$ for Functions Satisfying Inequalities

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The problem involves determining the possible values of the expression $f(8765)-f(4321)$ for functions $f$ that satisfy the inequalities $f(x+14)-14\le f(x) \le f(x+20)-20$. These inequalities impose constraints on the behavior of the function across intervals of real numbers. The discussion notes that no solutions were provided for the previous problem of the week, indicating a lack of engagement. A suggested solution is available for those interested in exploring the problem further. The focus remains on analyzing the implications of the given inequalities on the function's values.
anemone
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Here is this week's POTW:

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Consider functions $f$ defined for all real numbers and taking real numbers as values such that

$f(x+14)-14\le f(x) \le f(x+20)-20$, for all real numbers $x$.

Determine all possible values of $f(8765)-f(4321)$.

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No one answered last week's POTW. (Sadface) However, you can find the suggested solution below:

From $140=7 \times 20 =10 \times 14$, we have the following chains of inequalities:

$f(x)\le f(x+20)-20 \le f(x+40)-40 \le \cdots \le f(x+140)-140\\ f(x)\ge f(x+14)-14 \ge f(x+28)-28 \ge \cdots \ge f(x+140)-140$

So by the squeeze principle, equality holds throughout and we have $f(x)=f(x+20)-20$ and $f(x)=f(x+14)-14$ for all real numbers $x$.

Since $8765-4405=4360$ is a multiple of 20 and $4405-4321=84$ is a multiple of 14, it follows that

$f(8765)-f(4405)=8765-4405$ and $f(4405)-f(4321)=4405-4321$

Adding these equations yields $f(8765)-f(4321)=8765-4321=4444$.

Since $f(x)=x$ satisfies the conditions of the problem, the only possible value of the expression $f(8765)-f(4321)$ is 4444.
 
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