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

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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2006 AMC 12A Problems/Problem 18

Contents

Problem

Solution

Solution 2

See also