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Prove that there is a differentiable function $f$ such that $[f(x)]^5 + f(x) + x = 0$. (My textbook offers the following hint: Show that $f$ can be expressed as an inverse function.)

My Progress

$f(x) = -[f(x)]^5 - x$

$x = -[f \ (f^{-1} (x) )]^5 - f^{-1}(x)$

$-x^5 - x = f^{-1}(x)$

However, I do not know where to go next. The only thing that comes to mind is that $f$ is one-one since $f^{-1}$ is certainly a function.

Any hints please?

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  • $\begingroup$ If your inverse function $f^{-1}$ is differentiable/satisfies the criteria for the inverse function theorem to apply, then you apply the theorem and conclude that the inverse function of $f^{-1}$ (namely $f$) is differentiable in a certain neighborhood etc. $\endgroup$ – Moya Apr 24 at 20:22
  • $\begingroup$ @Moya But the inverse function theorem as I learned it (and it was rigorous proof) requires that $f$ be previously known to be differentiable. $\endgroup$ – user_hello1 Apr 24 at 20:25
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    $\begingroup$ you're applying it to $f^{-1}$. Presumably $f^{-1}$ works out to be something obviously differentiable, so you apply the theorem to state $(f^{-1})^{-1}=f$ is differentiable $\endgroup$ – Moya Apr 24 at 20:26
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You’re assuming the inverse is linear, and this is generally not true. Instead, let $g(y)=y^5+y$. $g$ is continuously differentiable and onto all reals which means you can always solve $g(y)=-x$. Moreover $g’(y)=5y^4+1$ is never zero. So just apply the inverse function theorem to conclude.

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  • $\begingroup$ What do you mean by the inverse function theorem here? $\endgroup$ – zhw. Apr 24 at 20:43
  • $\begingroup$ @zhw: en.m.wikipedia.org/wiki/Inverse_function_theorem $\endgroup$ – Alex R. Apr 24 at 20:49
  • $\begingroup$ But that is a local result. $\endgroup$ – zhw. Apr 24 at 20:52
  • $\begingroup$ Since $\forall y~g'(y) \gt 0$, it follows that $g(y)$ is monotonically increasing so the global result follows. $\endgroup$ – Robert Shore Apr 24 at 21:57
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I think the answer of Alex R. is on track, but I want to make things a bit clearer.

Let $g(y) = y^5 + y.$ Argue from one variable results that $g$ is a bijection of $\mathbb R$ onto $\mathbb R,$ with a differentiable inverse $g^{-1}.$ Define $f(x) = g^{-1}(-x).$ Then $f$ is differentiable on $\mathbb R,$ and we have $g(f(x)) = -x.$ This is the desired result.

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