Given a positive integer $m\ge 3$, consider the function $f:[0,1]\longrightarrow \mathbb{R}$ given by \begin{equation*} f(x)=(1-x)^{m-2}-\frac{3x}{2-x} \end{equation*} It is easy to see that $f(x)=0$ has unique solution in $[0,1]$, $\phi_m$.
What is the order of $\phi_m$ as $m\rightarrow \infty$?
On
Too long for comments.
LutzL provided an elegant answer to the question and a very good estimate of the solution for large values of $m$.
If we use $$x_0=\frac{W_0(\frac23(m-2.5))}{(m-2.5)}$$ as the starting value of Newton method, the second iterate will be given by $$x_1=\frac{(x_0-2) (1-x_0)^{m-2}+3 x_0}{(1-x_0)^{m-3} (x_0 (m-1)-2 m+3)-3}+x_0$$
Just a few numbers (I cannot resist when I see equations !) $$\left( \begin{array}{cccc} m & x_0 & x_1 & \text{solution} \\ 3 & 0.515255 & 0.349032 & 0.354249 \\ 4 & 0.378096 & 0.284280 & 0.288654 \\ 5 & 0.308381 & 0.244906 & 0.248165 \\ 6 & 0.264321 & 0.217346 & 0.219832 \\ 7 & 0.233313 & 0.196597 & 0.198553 \\ 8 & 0.210018 & 0.180237 & 0.181814 \\ 9 & 0.191726 & 0.166909 & 0.168210 \\ 10 & 0.176897 & 0.155786 & 0.156877 \\ 20 & 0.105401 & 0.098020 & 0.098344 \\ 30 & 0.078014 & 0.074025 & 0.074178 \\ 40 & 0.062937 & 0.060368 & 0.060456 \\ 50 & 0.053219 & 0.051396 & 0.051453 \\ 60 & 0.046362 & 0.044988 & 0.045028 \\ 70 & 0.041231 & 0.040151 & 0.040180 \\ 80 & 0.037229 & 0.036352 & 0.036374 \\ 90 & 0.034008 & 0.033280 & 0.033297 \\ 100 & 0.031354 & 0.030737 & 0.030751 \\ 200 & 0.018225 & 0.018022 & 0.018025 \\ 300 & 0.013187 & 0.013082 & 0.013083 \\ 400 & 0.010454 & 0.010388 & 0.010389 \\ 500 & 0.008718 & 0.008672 & 0.008673 \\ 600 & 0.007508 & 0.007475 & 0.007475 \\ 700 & 0.006614 & 0.006588 & 0.006588 \\ 800 & 0.005923 & 0.005902 & 0.005902 \\ 900 & 0.005372 & 0.005355 & 0.005355 \\ 1000 & 0.004921 & 0.004907 & 0.004907 \end{array} \right)$$
For any $a>0$ you get $(1-x)^{m-2}\approx 0$ on $x\in[a,1]$ for $m$ large enough. The value of the second term will be decidedly positive on $[a,1]$, so that there can be no root there. The root will have to be close to zero.
Introduce exponentials for the non-zero terms $1-x$ and $2-x$ to get $$ e^{(m-2)\ln(1-x)}e^{\ln(1-x/2)}=\frac32x $$ In a first approximation, assuming that $mx^2$ is small, this gives $$ (m-2.5)x\cdot e^{(m-2.5)x}=\frac23(m-2.5)\implies x=\frac{W_0(\frac23(m-2.5))}{(m-2.5)}$$ for the approximate value, and then also $$ x=\frac23e^{-(m-2.5)x}=\frac23e^{W_0(\frac23(m-2.5))} $$ Asymptotically this means $x_m=O(m^{-1}\log(m))$