# Prove that there is a differentiable function $f$ such that $[f(x)]^5 + f(x) + x = 0$. Spivak Ch. 12, Q. 14.

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.

• 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. – Moya Apr 24 at 20:22
• @Moya But the inverse function theorem as I learned it (and it was rigorous proof) requires that $f$ be previously known to be differentiable. – user_hello1 Apr 24 at 20:25
• 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 – Moya Apr 24 at 20:26

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.
• Since $\forall y~g'(y) \gt 0$, it follows that $g(y)$ is monotonically increasing so the global result follows. – Robert Shore Apr 24 at 21:57
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.