Show that there exists a differentiable function $f$ s.t. $(f(x))^5+f(x)+x=0$

Question

Show that there exists a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ s.t.

$(f(x))^5+f(x)+x=0$ for all $x \in \mathbb{R}$

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I am meant to use the Inverse Function Theorem for differentiability. However, I have no idea how to use it. I am sorry for the lack of any work. I genuinely have no idea.

Answer 1

Hint: The graph of your function would need to be a subset of the solution set of $y^5+y+x=0$, or in other words $x=-y^5-y$.

This set clearly determines $x$ as a function of $y$, so if $g(y)=-y^5-y$ has an inverse, the inverse function will work as the $f$ you're looking for.

Answer 2

Consider the relation $y^5+y+x=0$; it defines $x$ as a function $g$ of $y$: $x=-(y^5+y)$. This function is a decreasing bijection from $\mathbf R$ to $\mathbf R$ a since its derivative is $-(5y^4+1)\le -1<0$ and its limits at $\pm\infty$ are $\mp\infty$.

Let $f$ be its reciprocal; the relation $x=-(y^5+y)$ can be expressed as $$x=-\bigl(f(x)^5+f(x)\bigr).$$