Difference between revisions of "1983 AHSME Problems/Problem 8"

Revision as of 17:37, 26 January 2019

Let $f(x) = \frac{x+1}{x-1}$ . Then for $x^2 \neq 1, f(-x)$ is

$\textbf{(A)}\ \frac{1}{f(x)}\qquad \textbf{(B)}\ -f(x)\qquad \textbf{(C)}\ \frac{1}{f(-x)}\qquad \textbf{(D)}\ -f(-x)\qquad \textbf{(E)}\ f(x)$

We find $f(-x) = \frac{-x+1}{-x-1} = \frac{x-1}{x+1} = \frac{1}{f(x)}$ , so the answer is $\fbox{A}$ .

Revision as of 12:02, 7 June 2016 (view source) Katzrockso (talk \| contribs) (Created page with "== Problem 8 == Let <math>f(x) = \frac{x+1}{x-1}</math>. Then for <math>x^2 \neq 1, f(-x)</math> is <math>\textbf{(A)}\ \frac{1}{f(x)}\qquad \textbf{(B)}\ -f(x)\qquad \textb...")		Revision as of 17:37, 26 January 2019 (view source) Sevenoptimus (talk \| contribs) m (Slightly cleaned up and added more explanation to the solution) Newer edit →
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	== Solution ==		== Solution ==
−	<math>\frac{-x+1}{-x-1}~~\implies~~\frac{x-1}{x+1}=\frac{1}{f(x)}</math>, <math>\fbox{A}</math>	+	We find <math>f(-x) = \frac{-x+1}{-x-1} = \frac{x-1}{x+1} = \frac{1}{f(x)}</math>, so the answer is <math>\fbox{A}</math>.