Problem: Let f(x)=x+1xβ1f(x)=\dfrac{x+1}{x-1}f(x)=xβ1x+1β. Then for x2β 1,f(βx)x^{2} \neq 1, f(-x)x2ξ =1,f(βx) is
Answer Choices:
A. 1f(x)\dfrac{1}{f(x)}f(x)1β
B. βf(x)-f(x)βf(x)
C. 1f(βx)\dfrac{1}{f(-x)}f(βx)1β
D. βf(βx)-f(-x)βf(βx)
E. f(x)f(x)f(x)
Solution:
f(βx)=βx+1βxβ1=xβ1x+1=1(x+1xβ1)=1f(x).f(-x)=\dfrac{-x+1}{-x-1}=\dfrac{x-1}{x+1}=\dfrac{1}{\left(\dfrac{x+1}{x-1}\right)}=\dfrac{1}{f(x)}. f(βx)=βxβ1βx+1β=x+1xβ1β=(xβ1x+1β)1β=f(x)1β.