اگر $f(x)=\left\{ \begin{array}{*{۳۵}{l}}|x|sign(-x) & [x]\ge ۰ \\۲-sign(-x) & [x]<۰ \\\end{array} \right.$ باشد، حاصل $f(\frac{۱}{۲})+f(-\frac{۱}{۳})$ کدام است؟
$x=\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\Rightarrow \left[ \frac{1}{2} \right]=0\,\,\,\,\,\,\,\,\,\Rightarrow f(\frac{1}{2})=|\frac{1}{2}|\text{sign}(-\frac{1}{2})=\frac{1}{2}\times (-1)=-\frac{1}{2}$ $x=-\frac{1}{3}\,\,\,\,\Rightarrow \left[ -\frac{1}{3} \right]=-1\,\,\,\Rightarrow f(-\frac{1}{3})=2-\text{sign}(\frac{1}{3})=2-1=1$ $f(\frac{1}{2})+f(-\frac{1}{3})=-\frac{1}{2}+1=\frac{1}{2}$