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Let $f:[-1,1]\rightarrow R$ a continuous function: $-1\leq f(x)\leq 1 \forall x\in [-1,1]$. Prove that exist some $c\in[-1,1]: f(c)=c^3.$

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closed as off-topic by Jonas Meyer, Solid Snake, Davide Giraudo, Yiorgos S. Smyrlis, Carl Mummert Jun 10 at 15:54

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Homework? What did you try? –  Did Jan 21 '13 at 15:47
@RafaelJiménezGuerra your above comments means: "Hey seriously if you do not have help, do not help, but do not let unnecessary comments"...dont be rude here in MSE.no one will give answer to you, -1 for you –  La Belle Noiseuse Jan 21 '13 at 16:01
If really you are not looking for answers but for hints, why do you never indicate what you tried and do you always reproduce verbatim the text of the exercise? This conveys exactly the opposite message. –  Did Jan 21 '13 at 16:15

1 Answer 1

consider the continous(why?) function $g(x)=f(x)-x^3$ , $g(1)=f(1)-1\le 0$(why?) and $g(-1)=f(-1)+1\ge 0$(why?) hence there exist $c\in [-1,1]$ such that $g(c)=0\Rightarrow f(c)-c^3=0\Rightarrow f(c)=c^3$(why?)

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+1 for the four why? –  Did Jan 21 '13 at 15:52
@RafaelJiménezGuerra Why? The answer is excellent. Gives HINTS, exactly what you say you want. –  Did Jan 21 '13 at 16:16
Erasing your tracks? (By the way, since you come to MSE to see your homework solved, surely you will mention the fact to your teacher, won't you?) –  Did Jan 21 '13 at 16:28

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