Show that there exists a differentiable function f: R to R such that

(f(x))^5+f(x)+x=0 for all x in R.

I'm thinking about using its inverse function g, but not sure if it works. Could anyone please give me some hints? Any help is appreciated!

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- Jun 8th 2010, 03:09 AMnngktrdifferentiable function
Show that there exists a differentiable function f: R to R such that

(f(x))^5+f(x)+x=0 for all x in R.

I'm thinking about using its inverse function g, but not sure if it works. Could anyone please give me some hints? Any help is appreciated! - Jun 8th 2010, 07:57 AMdrjerry
First off, the problem says that, for every real number $\displaystyle x$, the 5-th degree polynomial $\displaystyle y^5 + y + x$ has a real zero. Do you believe this?

You're looking for a function defined by $\displaystyle f(x) = y$, even if you can't write down a formula. I don't know how much background to assume here. You said something about an "inverse function." I would recommend considering the function $\displaystyle g(x,y) = y^5 + y + x$ and applying the implicit function theorem. - Jun 8th 2010, 09:19 AMraulk
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