2006 AMC 12A Problems/Problem 18

Problem

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 $f$?

$\mathrm{(A) \ } \{x|x\ne 0\}\qquad \mathrm{(B) \ } \{x|x<0\}$

$\mathrm{(C) \ } \{x|x>0\}$$\mathrm{(D) \ } \{x|x\ne -1\;\rm{and}\; x\ne 0\;\rm{and}\; x\ne 1\}$

$\mathrm{(E) \ } \{-1,1\}$

Solution

Quickly verifying by plugging in values verifies that $-1$ and $1$ are in the domain.

$f(x)+f\left(\frac{1}{x}\right)=x$

Plugging in $\frac{1}{x}$ into the function:

$f\left(\frac{1}{x}\right)+f\left(\frac{1}{\frac{1}{x}}\right)=\frac{1}{x}$

$f\left(\frac{1}{x}\right)+ f(x)= \frac{1}{x}$

Since $f(x) + f\left(\frac{1}{x}\right)$ cannot have two values:

$x = \frac{1}{x}$

$x^2 = 1$

$x=\pm 1$

Therefore, the largest set of real numbers that can be in the domain of $f$ is $\{-1,1\} \Rightarrow E$

Solution 2

We know that $f(x) + f \left(\frac{1}{x}\right) = x.$ Plugging in $x = \frac{1}{x}$ we get $$f \left(\frac{1}{x}\right) + f \left(\frac{1}{\frac{1}{x}}\right) = \frac{1}{x}$$ $$f \left(\frac{1}{x}\right) + f(x) = \frac{1}{x}.$$

Also notice $$f \left(\frac{1}{x}\right) + f(x) = x$$ by the commutative property(this is the same as the equation given in the problem. We are just rearranging). So we can set $\frac{1}{x} = x$ which gives us $x = \pm 1$ which is answer option $\boxed{\mathrm{(E) \ } \{-1,1\}}.$

See also

 2006 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.