2006 AMC 12A Problem 18

Question

The function

f

has the property that for each real number

x

in its domain,

\frac{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,\ x \neq 0,\ x \neq 1\}

\{-1, 1\}

Substitute

\frac{1}{x}

in place of

x

to get a second equation

Comparing the two equations forces

x = \frac{1}{x}

Replacing

x

by

\frac{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.

Problem 18 in other years