Help : How to prove the following simple function is decreasing function? I am stuck on figuring out why the following function is a decreasing function when I read a paper. The function is following
$f(x) = \frac{c^x-c\cdot x+x-1}{2^x-x-1}$,
where $1<c<2$, and the independent variable $x$ is on interval $(2,+\infty)$, i.e. $2<x<+\infty$.
I have plotted the function curves of f(x) with different c, and found that it seems to be monotonic function. However, it is too hard to me to prove it. 

The function curves of f(x) with different c is here:  function curves of f(x)
The first derivative of $f(x)$ is by the following:
\begin{equation}
f'(x)={\frac {{c}^{x}\ln  \left( c \right) -c+1}{{2}^{x}-x-1}}-{\frac {
 \left( {c}^{x}-cx+x-1 \right)  \left( {2}^{x}\ln  \left( 2 \right) -1
 \right) }{ \left( {2}^{x}-x-1 \right) ^{2}}}
\end{equation}
And the second derivative of $f(x)$ is as follows:
\begin{equation}
\begin{split}
f''(x)&= {\frac {{c}^{x} \left( \ln  \left( c \right)  \right) ^{2}}{{2}^{x}-x-
1}}-2\,{\frac { \left( {c}^{x}\ln  \left( c \right) -c+1 \right) 
 \left( {2}^{x}\ln  \left( 2 \right) -1 \right) }{ \left( {2}^{x}-x-1
 \right) ^{2}}}\\&+2\,{\frac { \left( {c}^{x}-cx+x-1 \right)  \left( {2}^
{x}\ln  \left( 2 \right) -1 \right) ^{2}}{ \left( {2}^{x}-x-1 \right) 
^{3}}}\\&-{\frac { \left( {c}^{x}-cx+x-1 \right) {2}^{x} \left( \ln 
 \left( 2 \right)  \right) ^{2}}{ \left( {2}^{x}-x-1 \right) ^{2}}}
\end{split}
\end{equation}
 A: The derivative function results from WolframAlpha makes it a bit difficult to interpret the behavior of this set of functions, which does have a subtle issue that needs to be handled carefully.
It is helpful to start with looking at the asymptotic behavior, where things are simpler.  It is clear, as $ \ x \ $ tends to positive infinity, that this function tends to $ \ \left(  \frac{c}{2} \right)^x \ $ , which is an exponentially-decaying function for $ \ 1 \ < \ c \ < \ 2 \ $ .
If we look at the derivative of this, we have $$ \ \frac{[\ln c] \ (2c)^x \ - \ [\ln 2] \ (2c)^x}{2^{2x}} \ \ = \ \ \left(\ln  \frac{c}{2} \right) \ \frac{(2c)^x}{2^{2x}} \ \ , $$
for which the logarithmic factor is negative, making the derivative function here negative.
When we work out the derivative of the complete expression of the function, we have
$$ \ \frac{(2^x - x - 1) \ ([\ln c] \ c^x - [c-1])\ - \ (c^x - [c-1] x - 1 ) \ ([\ln 2]  2^x - 1)}{(2^x - x - 1)^2} $$
$$ = \ \ \frac{\ln  \frac{c}{2}\ (2c)^x \ - \ ( [1 + x] \ [\ln c] \ - \ 1) \ c^x \ - \  ( [c - 1 + \ln 2]  \ + \ [ c - 1 ] \ [\ln 2] \ x \ ) 2^x \ +  \ ( c - 2)}{(2^x - x - 1)^2} \ \ , $$
with the first numerator term as before; the linear terms in the numerator have canceled.  (And I will say that this calculation is an open invitation to commit sign errors...)
The denominator is always positive for $ \ x \ > \ 2 \ $ , so we only need focus on the numerator.  The first term is negative, as we already saw in our simplified version of the function; the last term is also negative for   $ \ 1 \ < \ c \ < \ 2 \ $  . The other "assembled" terms are linear functions times exponential "growth" functions. For the third term, we find
$$ [c - 1 + \ln 2]  \ + \ [ c - 1 ] \ [\ln 2] \ x \ > \ 0 \ \ \Rightarrow \ \ x \ > \ -\frac{c \ - \ 1 \ + \ln 2}{(c - 1) \ \ln 2} \ \ ; $$
so the third term in the numerator of the derivative is negative for the values of $ \ x \ $ and $ \ c \ $ under discussion. 
In the second term, the factor  $ \ [1 + x] \ [\ln c] \ - \ 1 \ $ complicates matters a bit: it is positive for $ \ x \ > \ \frac{1}{\ln c} \ -  \ 1 \ = \ X $ , so the second term in the numerator of the derivative is positive for $ \ 0 \ < \ x \ < X \ $ .  This is not an issue for $ \ e^{1/3} \ \approx \ 1.396 \ < \ c \ < \ 2 \ $ (for which $ \ X \ < \ 2 \ $ ), but otherwise we must examine the behavior of the term for smaller values of $ \ c \ $ .  What we find is that the second term is always less than $ \ +1 \ $ for $ \ 0 \ < \ x \ < X \ $ , so it is always less positive than the third term in the numerator is negative for $ \ x \ > \ 2 \ $ .
We can conclude that the entire numerator is negative for $ \ x \ > \ 2 \ $ and $ \ 1 \ < \ c \ < \ 2 \ $ .  It proves to be important to include all of the terms in order to see the behavior accurately (answering a diffculty raised earlier in the development of this post), but it also requires some graphical assistance to aid in understanding how certain of the terms in the derivative sum together.
That issue seems to be connected with an interesting feature of these curves.  If we consider the family of functions $$ \ f(c) \ = \frac{c^x-(c\ - 1) x-1}{2^x-x-1}$$
for $ \ 1 \ \le \ c \ \le \ 2 \ $ on $ \ x \ > \ 2 \ $ ,  we see that $ \ f(1) \ $ is identically the zero function and that $ \ f ( 2 ) \ $ is identically $ \ y \ = 1 \ $ . For $ \ c \ < \ e^{1/2} \ $ , the curves are concave upward only; they simply have a "decaying" behavior as seen in Jinliang Xu's graphs.  Above this critical value for $ \ c \ $ , an inflection point moves into the region and an increasingly "long" portion of the curve is concave downward as the curves "flatten out" toward $ \ y \ = \ 1 $ for small values of $ \ x \ $ , while remaining monotonically decreasing.  These curves then require more careful analysis (which I missed early on in exploring this family).
