Problem: Let f(x)=x+1x−1f(x)=\dfrac{x+1}{x-1}f(x)=x−1x+1​. Then for x2≠1,f(−x)x^{2} \neq 1, f(-x)x2î€ =1,f(−x) is
Answer Choices:
A. 1f(x)\dfrac{1}{f(x)}f(x)1​
B. −f(x)-f(x)−f(x)
C. 1f(−x)\dfrac{1}{f(-x)}f(−x)1​
D. −f(−x)-f(-x)−f(−x)
E. f(x)f(x)f(x)
Solution:
f(−x)=−x+1−x−1=x−1x+1=1(x+1x−1)=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(−x)=−x−1−x+1​=x+1x−1​=(x−1x+1​)1​=f(x)1​.