Question

Find the range of $f\left(x\right)=\frac{\left\{x\right\}}{\left\{x\right\}+1}$

Answer 1

Let $y=\frac{\left\{x\right\}}{\left\{x\right\}+1}$

$\Rightarrow y\left\{x\right\}+y=\left\{x\right\}$ as $0\le \left\{x\right\}<1$

Thus,$0\le \frac{y}{1-y}<1$

$\Rightarrow \frac{y}{1-y}\ge 0\Rightarrow \frac{y}{y-1}\le 0$

$\Rightarrow y\in [0,1)$

$\frac{y}{y-1}-1<0\Rightarrow \frac{2y-1}{y-1}>0$

$\Rightarrow t\in (-\infty ,\frac{1}{2})\cup (1,\infty )$

Hence range is $[0,\frac{1}{2})$