How would I go about showing the existence of a differentiable funciton over the reals such that, for all x,

$\displaystyle {\left( {f\left( x \right)} \right)^5} + f\left( x \right) + x = 0$?

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- Feb 22nd 2009, 09:35 AMFractalMathDifferentiability analysis
How would I go about showing the existence of a differentiable funciton over the reals such that, for all x,

$\displaystyle {\left( {f\left( x \right)} \right)^5} + f\left( x \right) + x = 0$? - Feb 22nd 2009, 05:31 PMHallsofIvy
You don't. You can't. It's not true. If there were a differentiable function satisfying that equation, then, differentiating both sides, $\displaystyle 5f(x)^4f'(x)+ f'(x)= f'(x)(5f(x)^4+ 1)= 0$. So either f'(x)= 0, for all x, or 5f(x)^4+ 1= 0. If f'(x)= 0 for all x, then f(x) is a constant, c, such that $\displaystyle c^5+ c+ x= 0$ or $\displaystyle c^5+ c= -x$ which is impossible: the right side changes as x changes and the left side is a constant. However, all terms in 5f(x)^4+ 1 are non-negative: their sum is 0 only if 5f(x)^4= -1 which again says that f(x) is a constant.

- Feb 23rd 2009, 05:16 AMFractalMath
- Feb 23rd 2009, 09:42 AMHallsofIvy