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2006 AMC 12A Problems/Problem 18

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Problem

The function \displaystyle f has the property that for each real number \displaystyle x in its domain, \displaystyle 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

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

See also

2006 AMC 12A (Problems • Resources)
Preceded by
Problem 17 		Followed by
Problem 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
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