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

Revision as of 20:48, 31 January 2007

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

@@ Line 34: / Line 34: @@
 == See also ==
 * [[2006 AMC 12A Problems]]
-*[[2006 AMC 12A Problems/Problem 17|Previous Problem]]
-*[[2006 AMC 12A Problems/Problem 19|Next Problem]]
+{{AMC box|year=2006|n=12A|num-b=17|num-a=19}}
 [[Category:Introductory Algebra Problems]]

{{{header}}}
Preceded by Problem 17	AMC 12A 2006	Followed by Problem 19

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

Revision as of 20:48, 31 January 2007

Problem

Solution

See also