Home Competitions CIIM 2009 Problem 5
Let f : R → R f:\mathbb{R} \to \mathbb{R} f : R → R , such that
i) For all a ∈ R a \in \mathbb{R} a ∈ R and all ϵ > 0 \epsilon > 0 ϵ > 0 , exists δ > 0 \delta > 0 δ > 0 such that ∣ x − a ∣ < δ ⇒ f ( x ) < f ( a ) + ϵ . |x-a| < \delta \Rightarrow f(x) < f(a) + \epsilon. ∣ x − a ∣ < δ ⇒ f ( x ) < f ( a ) + ϵ . ii) For all b ∈ R b\in \mathbb{R} b ∈ R and all ϵ > 0 \epsilon > 0 ϵ > 0 , exists x , y ∈ R x,y \in \mathbb{R} x , y ∈ R with b − ϵ < x < b < y < b + ϵ b - \epsilon < x < b < y < b + \epsilon b − ϵ < x < b < y < b + ϵ , such that ∣ f ( x ) − f ( b ) ∣ < ϵ |f(x)-f(b)|< \epsilon ∣ f ( x ) − f ( b ) ∣ < ϵ and ∣ f ( y ) − f ( b ) ∣ < ϵ . |f(y)-f(b)| < \epsilon. ∣ f ( y ) − f ( b ) ∣ < ϵ . Prove that if f ( a ) < d < f ( d ) f(a) < d < f(d) f ( a ) < d < f ( d ) there exists c c c with a < c < b a < c < b a < c < b or b < c < a b < c < a b < c < a such that f ( c ) = d f(c) = d f ( c ) = d .
Dataset text AI-MO AoPS . Quoted with source.
Solutions
No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.
Add a solution
Similar problems
These problems have the most words in common with this statement.
Locator ciim/2009/p5 · Tag ZVHXCK1SRF · Edit this page on GitHub · Keys: j next, k previous