# Real Variable. Function. [closed]

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 MummertJun 10 at 15:54

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Solid Snake, Davide Giraudo, Yiorgos S. Smyrlis, Carl Mummert
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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

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?)