Problem: Let f(t)=t1βt,tβ 1f(t)=\dfrac{t}{1-t}, t \neq 1f(t)=1βttβ,tξ =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(y1β)
B. βf(y)-\mathrm{f}(\mathrm{y})βf(y)
C. βf(βy)-\mathrm{f}(-\mathrm{y})βf(βy)
D. f(βy)f(-y)f(βy)
E. f(y)f(y)f(y) Solution:
Since y=x1βx,yβyx=xy=\dfrac{x}{1-x}, y-y x=xy=1βxxβ,yβyx=x and x=y1+yβ΄x=ββy1β(βy)=βf(βy)x=\dfrac{y}{1+y} \quad \therefore x=-\dfrac{-y}{1-(-y)}=-f(-y)x=1+yyββ΄x=β1β(βy)βyβ=βf(βy).