# Intermediate Value Theorem on [0,1]

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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Please don't dump undigested photocopies here. Tell us why you are interested in the problem; what you know about it; how far you have come toward a solution; where you got stuck; and so on. – Gerry Myerson Nov 7 '12 at 5:57
Consider $f(x)-x^4$. – marty cohen Nov 7 '12 at 6:08
@martycohen: Thank you so much for the hint, I got it this morning, my answer is below. Would you mind tell me how you got that idea in the first place? – Akaichan Nov 8 '12 at 0:57
@GerryMyerson Please don't dump judgemental comments here. I was stuck, I did not know where to start it. If you do not want to answer it, please don't say anything.I would appreciate it so much. Thank you, I'm sure the world knows you are a genius already, and I mean that. So please just be nice on here. – Akaichan Nov 8 '12 at 1:00
I may have said it, but at least 5 other people 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. – Gerry Myerson Nov 8 '12 at 2:50

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 $g(0)<0<g(1)$, 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? – Gerry Myerson Nov 8 '12 at 2:52