<s>[INST] Prove only one solution exists for $f(\zeta)$ and $g(\zeta)$ using a convex function argument I want to prove that the following only have one solution, for $\zeta\in[0,1]$, at $\zeta =1$.
$$f(\zeta)=\frac{1}{1+\zeta}$$
$$g(\zeta) =
\frac{
(1-\zeta)(2-\zeta)\zeta
-
(2-\zeta)^2\log\left(2 - \zeta \right)
}
{\zeta(\zeta-4)(\zeta-1)^2}$$
These are plotted below. Note that $f(0)=1$, $\lim_{\zeta\rightarrow 0}g(\zeta)=\infty$ and $f(1)=1/2=g(1)$.
Both functions are convex and monotonically decreasing over this region, therefore if they intersect at $\zeta=1$ this is the only place they can? Can I shown they are both convex over $\zeta\in[0,1]$ by showing their second derivatives wrt $\zeta$ are +ve over this region? Or is this thinking too simplistic?

Note, the limit as $g(\zeta)$ as $\zeta\rightarrow 1$ is proven below:
\begin{align}
\lim_{\zeta\rightarrow 1} g(\zeta) 
%%$
&=
%%%
\lim_{\zeta\rightarrow 1}
\left(
\frac
{4\log(2 - \zeta) - 7\zeta - 2\zeta \log(2 - \zeta) + 3\zeta^2 + 4}
{4\zeta^3 - 18\zeta^2 + 18\zeta - 4}
\right)\\
%%%
&=
%%%
\lim_{\zeta\rightarrow 1}
\left(
\frac
{6\zeta - 2\log(2 - \zeta) - 9}
{12\zeta^2 - 36\zeta + 18}
\right)\\
%%%
&=
%%%
\frac
{6 - 9}
{12 - 36 + 18}\\
%%%
&=
%%%
\frac
{1}
{2}.
\end{align}
EDIT: So the second derivative of $g(\zeta)$ is pretty horrific. It does  have a simplified form, but it is also pretty terrible.
http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))
$$g'' = 
\frac
{(x - 1) x (x (x (x (x (2 x - 7) - 13) + 124) - 224) + 64) 
- 2 (x (x (x (x (3 (x - 12) x + 184) - 472) + 588) - 304) + 64) \log(2 - x))}
{(x - 4)^3 (x - 1)^4 x^3}$$
It is clear from the numerics that it is greater than $0$ in the region $\zeta\in[0,1]$.
http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))+%3D+0
 [/INST] we want to solve g(x)=f(x) for x in [0,1] 
by doing the calculations we end up in this equation
$$ \ log(2-x)= \frac{4x^3-10x^2+6x}{(x+1)(2-x)^2}$$
$$h(x)=log(2-x)$$ and
$$ k(x)=\frac{4x^3-10x^2+6x}{(x+1)(2-x)^2}$$
h(x) is convex and k(x) is concave in [0,1] (1)
$$ k'(x)=\frac{-2(x^3+8x^2-17x+6)}{(x-2)^3(x+1)^2} $$
so k(1)=h(1)=0 and k'(1)=h'(1)=-1 (2)
from (1) and (2) we have that 
$$ h(x)>=k(x)$$ for every x in [0,1] and k(x)=h(x) only for x=1
</s>