Problem: Let f(t)=t1−t,t≠1f(t)=\dfrac{t}{1-t}, t \neq 1f(t)=1−tt​,tî€ =1. If y=f(x)y=f(x)y=f(x), then xxx can be expressed as:
Answer Choices:
A. f(1y)f\left(\dfrac{1}{y}\right)f(y1​)
B. −f(y)-\mathrm{f}(\mathrm{y})−f(y)
C. −f(−y)-\mathrm{f}(-\mathrm{y})−f(−y)
D. f(−y)f(-y)f(−y)
E. f(y)f(y)f(y) Solution:
Since y=x1−x,y−yx=xy=\dfrac{x}{1-x}, y-y x=xy=1−xx​,y−yx=x and x=y1+y∴x=−−y1−(−y)=−f(−y)x=\dfrac{y}{1+y} \quad \therefore x=-\dfrac{-y}{1-(-y)}=-f(-y)x=1+yy​∴x=−1−(−y)−y​=−f(−y).