$(f(1 - f(x)) = 1 - x^9$, $f(1) = 0$ 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 $f'(1) &lt; 0$, 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 $f'(1) &lt; 0$, $f$ is decreasing at $x = 1$ and $f(a) &gt; f(1) = 0$. By mean-value theorem, $$f(a) - f(1) = f'(a_0)(a - 1), a &lt; a_0 &lt; 1$$

Let $g(x) = f(x) - x^{99}$, then clearly $g(1) &lt; 0$ and $g(r) = 0$.

How about showing $g(a) &gt; 0$? Or to take $a$ as something else?
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 $(1)$ &amp; $(2)$, 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, $g(0) = 2 &gt; 0$. 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) &amp;= -9; \\ f'(1 - f(0)) \cdot -f'(0) = f'(-1)f'(0) &amp;= 0 \end{align}

produces contradiction, so we have to reject this possibility.

If $a = 0$, i.e. $f(0) = 1$, then $g(0) = 1 &gt; 0$ 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 $[0, 1]$.
$g(0) = 1 &gt; 0 &gt; -1 = g(1)$.


So $\exists \text{ } r \in [0, 1] \Rightarrow g(r) = 0$, i.e. $f(r) = r^{99}$. $\Box$
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 $(1)$ &amp; $(2)$, 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, $g(0) = 2 &gt; 0$. 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) &amp;= -9; \\ f'(1 - f(0)) \cdot -f'(0) = f'(-1)f'(0) &amp;= 0 \end{align}

produces contradiction, so we have to reject this possibility.

If $a = 0$, i.e. $f(0) = 1$, then $g(0) = 1 &gt; 0$ 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 $[0, 1]$.
$g(0) = 1 &gt; 0 &gt; -1 = g(1)$.


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