Q:

Solve $f'(x)<0$ when $f(x)=x-\log_e {(2x + 1)}$

given :$$f(x)=x-\log_e {(2x + 1)}$$
  question : solve $f'(x)<0$

Attempt :
$$f'(x) < 0 = 1 - \frac{1}{2x+1} <0$$
$$\frac{1}{2x+1} > 1$$
$$2x+1 < 1$$
$$x < 0$$
For the logarithm function to be defineable, $x>-0.5$
But i am wrong, why? 

A:

(revised) question. Given the real function $f(x)=x-\log_e {(2x + 1)}$, for which 
$x$ does the derivative satisfy $f'(x)<0$?
For $2x+1>0 \Leftrightarrow x>-\frac{1}{2}$, the function is well defined. Computing the derivative we obtain 
$$f'(x)=1-\frac{2}{2x+1}$$
for all $x>-\frac{1}{2}$
so the question for which $x$ we have $f'(x)<0$ means asking for which $x$ does $f'(x)$ exist and for which of those $x$ do we have $f'(x)<0$. Solving $$f(x)=1-\frac{2}{2x+1}=0,$$
we obtain $$1=\frac{2}{2x+1}$$
and thus, taking the inverse on both sides, $x+\frac{1}{2}=1$ such that $x=\frac{1}{2}$.
Because $f'(0)=-1<0$ and there is only one $x$ where $f'(x)=0$, the function $f'(x)$ has negative sign for all $x\in (-\frac{1}{2},\frac{1}{2})$.

