Let f(x)=(x+1)/(x-1). Then for x^2≠1, f(-x) is

1983 AMC 12 Problem 8

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Let

f(x)=\frac{x+1}{x-1}.

Then for

x^2\ne1,

f(-x)

\frac1{f(x)}

-f(x)

\frac1{f(-x)}

-f(-x)

f(x)

Substitute

-x

directly into the formula

Multiply numerator and denominator of

f(-x)

-1

We have

\begin{aligned} f(-x)&=\frac{-x+1}{-x-1}\\ &=\frac{x-1}{x+1}=\frac1{f(x)}. \end{aligned}

The restriction

x^2\ne1

makes all displayed quantities defined.

Therefore, the correct answer is A.