<s>[INST] Find the value of $x$ for which $ff=gf$. Functions $f$ and $g$ are defined by $f:x \mapsto \frac{1}{2x+1}$, $x \neq \frac{-1}{2}$ and $g:x \mapsto x+1$. Find the value of $x$ for which $ff=gf$.
So I started in this way:
$f[f(x)]=g[f(x)]$
$f(\frac{1}{2x+1}) = g(\frac{1}{2x+1})$
$\frac{1}{2(\frac{1}{2x+1})+1}=\frac{1}{2x+1}+1$
Then I solve it in simple algebra. Is this right?
I get the wrong answer, my book says $\frac{-5}{6}$ but I get $\frac{-11}{2}$
Helppp
 [/INST] $\frac{1}{2(\frac{1}{2x+1})+1}=\frac{1}{2x+1}+1 \Leftrightarrow$
$\frac{2x+1}{2x+3}=\frac{2x+2}{2x+1}\Leftrightarrow$
$(2x+1)^2=(2x+2)(2x+3) \Leftrightarrow$
$4x^2+4x+1=4x^2+10x+6 \Leftrightarrow$
$6x+5=0$. So $x=-\frac{5}{6}$. Your book is right.
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