3 The function that is increasing on the interval $(-\infty, 0)$ is
A. $y=-\log _{\frac{1}{2}}(-x)$
B. $y=\frac{x}{1-x}$
C. $y=-(x+1)^{2}$
D. $y=1+x^{2}$
3 B. $y=\frac{x}{1-x}=-1+\frac{1}{1-x}$ is increasing on $(-\infty, 0)$. $y=-\log _{\frac{1}{2}}(-x)=\log _{2}(-x)$ is decreasing on $(-\infty, 0)$. $y=-(x+1)^{2}$ is decreasing on $(-1,0)$. $y=1+x^{2}$ is decreasing on $(-\infty, 0)$.