Difference between revisions of "2006 AMC 12A Problems/Problem 18"

Revision as of 22:08, 11 July 2006

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 $f$ ?

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

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

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

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

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 $\{-1,1\} \Rightarrow E$

@@ Line 14: / Line 14: @@
 == Solution ==
+<math>f(x)+f\left(\frac{1}{x}\right)=x</math>
+Plugging in <math> \frac{1}{x} </math> into the function:
+<math>f\left(\frac{1}{x}\right)+f\left(\frac{1}{\frac{1}{x}}\right)=\frac{1}{x}</math>
+<math>f\left(\frac{1}{x}\right)+ f(x)= \frac{1}{x}</math>
+Since <math>f(x) + f\left(\frac{1}{x}\right) </math> cannot have two values:
+<math> x = \frac{1}{x} </math>
+<math> x^2 = 1 </math>
+<math> x=\pm 1</math>
+Therefore, the largest set of real numbers that can be in the domain of <math>f</math> is <math>\{-1,1\} \Rightarrow E </math>
 == See also ==
 * [[2006 AMC 12A Problems]]