Question

The function $f$ has the property that for each real number $x$ in its domain, $1/x$ is also in its domain and $f(x) + f\!\left(\frac{1}{x}\right) = x.$ What is the largest set of real numbers that can be in the domain of $f?$

$\{x \mid x \neq 0\}$

$\{x \mid x \lt 0\}$

$\{x \mid x \gt 0\}$

$\{x \mid x \neq -1 \text{ and } x \neq 0 \text{ and } x \neq 1\}$

$\{-1, 1\}$

Solution:

Replacing $x$ by $1/x$ gives $f\!\left(\tfrac{1}{x}\right) + f(x) = \tfrac{1}{x}.$ Together with $f(x) + f\!\left(\tfrac{1}{x}\right) = x,$ this requires $x = \tfrac{1}{x},$ so $x = \pm 1.$

Both values are consistent, with $f(1) = \tfrac{1}{2}$ and $f(-1) = -\tfrac{1}{2}.$ So the largest possible domain is $\{-1, 1\}.$

Thus, the correct answer is E.

2006 AMC 12A Problem 18

Solution:

Problem 18 in Other Years