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Let $f$ be continuous on $[0,1]$ and such that $f(0)<0$ and $f(1)>1$. Prove that there exists $c\in(0,1)$ such that $f(c)=c^4$

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    I may have said it, but at least 5 other $p$eo$p$le thought it, as shown by the little number next to my comment. It's not just my judgement, but something close to a consensus, that copying out problems from a book or an assignment is not the best way to use this website. I simply let you know what people posting to this site are expected to do. I hope you'll keep my suggestions in mind.2012-11-08

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If we let $g(x)=f(x)-x^4$, then we are given $g(0)=f(0)-0<0$ and $g(1)=f(1)-1>0$. Since $f$ is continuous and $x^4$ is continuous, their difference is, and $g$ is defined on the interval $[0,1]$, so we can apply the Intermediate Value Theorem to $g$. In particular, since $g(0)<0, there exists $c\in (0,1)$ such that $g(c)=0$, so $f(c)-c^4=0$, or $f(c)=c^4$ as we desired.

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consider $g(x)=f(x)-x^4$ $g(x)$ is continuous from $[0,1]$ since both $f(x)$ and $x^4$ continuous on $[0,1]$.

Now, $g(0)=f(0)-0^4=a<0$ (since $f(0)<0$ is given)

$g(1)=f(1)-1^4 = b>0$ (since $f(1)>1$ is given)

$g(x)$ is continuous, so by Intermediate Value Theorem, for every $d$ between $g(0)$ and $g(1)$ $\exists\,c\in(0,1)$ such that $g(c)=d$

Choose $d=0$ so we have $g(c)=0$ or $g(c)=f(c)-c^4=0$ which implies $f(c)=c^4\, for c\in(0,1)$

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    I'm sorry, what was the point of posting this version of the earlier answer of @pi37?2012-11-08