<s>[INST] $(f(1 - f(x)) = 1 - x^9$, $f(1) = 0$ and $f'(1) < 0$, then where is the real number $r$ such that $f(r) = r^{99}$? If $f(1 - f(x)) = 1 - x^9$, $f$: R $\to$ R is differentiable, $f(1) = 0$ and $f'(1) < 0$, how to show there is a real number $r$ such that $$f(r) = r^{99}?$$
Edit: Taylor Theorem makes no use. I try to take $a = 1 - {1 \over n}$, where n is also a real no. Then as $f'(1) < 0$, $f$ is decreasing at $x = 1$ and $f(a) > f(1) = 0$. By mean-value theorem, $$f(a) - f(1) = f'(a_0)(a - 1), a < a_0 < 1$$
Let $g(x) = f(x) - x^{99}$, then clearly $g(1) < 0$ and $g(r) = 0$.
How about showing $g(a) > 0$? Or to take $a$ as something else?
 [/INST] After almost 4 years I finally think of these. Let me know anywhere wrong.
Firstly if $f(1 - f(x)) = 1 - x^9$, then let $a \in \mathbb{R}$ so that $f(a) = 1$, hence \begin{align} f(0) = f(1 - f(a)) = 1 - a^9. (1) \end{align}
Also we rearrange the terms in expressing $f(1 - f(x))$ to be $f(x) = 1 - f^{-1}(1 - x^9)$, then \begin{align} f(0) = 1 - f^{-1}(1 - 0) = 1 - a. (2) \end{align}
Combining $(1)$ & $(2)$, we have $a^9 - a = 0$, which $a = 1$ (rejected, or $f(1)$ has $2$ values), $0$ or $-1$ (rejected).
If $a = -1$, i.e. $f(-1) = 1$, then $f(1 - f(-1)) = f(0) = 1 - (-1)^9 = 2$. If we let $g(x)$ as I tried above, $g(0) = 2 > 0$. However, by differentiating $f(1 - f(x))$ against $x$, we have $f'(1 - f(x))\cdot -f'(x) = -9x^8$.
\begin{align} \Rightarrow f'(1 - f(-1)) \cdot -f'(-1) = f'(0)f'(-1) &= -9; \\ f'(1 - f(0)) \cdot -f'(0) = f'(-1)f'(0) &= 0 \end{align}
produces contradiction, so we have to reject this possibility.
If $a = 0$, i.e. $f(0) = 1$, then $g(0) = 1 > 0$ as well without the confusion on (the product of) the derivatives. Now the mean-value theorem becomes useful that


*

*differentiabilty of $f$ (so does of $g$) guarantees continuity on at least $[0, 1]$.

*$g(0) = 1 > 0 > -1 = g(1)$.


So $\exists \text{ } r \in [0, 1] \Rightarrow g(r) = 0$, i.e. $f(r) = r^{99}$. $\Box$
</s>