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Difference between revisions of "2006 AMC 12A Problems/Problem 18"
 
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== Solution ==
 
== Solution ==
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Quickly verifying by plugging in values verifies that <math>-1</math> and <math>1</math> are in the domain.
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<math>f(x)+f\left(\frac{1}{x}\right)=x</math>
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Plugging in <math> \frac{1}{x} </math> into the [[function]]:
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<math>f\left(\frac{1}{x}\right)+f\left(\frac{1}{\frac{1}{x}}\right)=\frac{1}{x}</math>
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<math>f\left(\frac{1}{x}\right)+ f(x)= \frac{1}{x}</math>
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Since <math>f(x) + f\left(\frac{1}{x}\right) </math> cannot have two values:
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<math> x = \frac{1}{x} </math>
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<math> x^2 = 1 </math>
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<math> x=\pm 1</math>
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Therefore, the largest [[set]] of [[real number]]s that can be in the [[domain]] of <math>f</math> is <math>\{-1,1\} \Rightarrow E </math>
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== Solution 2==
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We know that <math>f(x) + f \left(\frac{1}{x}\right) = x.</math> Plugging in <math>x = \frac{1}{x}</math> we get <cmath>f \left(\frac{1}{x}\right) + f \left(\frac{1}{\frac{1}{x}}\right) = \frac{1}{x}</cmath> <cmath>f \left(\frac{1}{x}\right) + f(x) = \frac{1}{x}.</cmath>
 +
 +
Also notice <cmath>f \left(\frac{1}{x}\right) + f(x) = x</cmath> by the commutative property(this is the same as the equation given in the problem. We are just rearranging). So we can set <math>\frac{1}{x} = x</math> which gives us <math>x = \pm 1</math> which is answer option <math>\boxed{\mathrm{(E) \ }  \{-1,1\}}.</math>
  
 
== See also ==
 
== See also ==
 
* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
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{{AMC12 box|year=2006|ab=A|num-b=17|num-a=19}}
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{{MAA Notice}}
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[[Category: Introductory Algebra Problems]]

Latest revision as of 17:34, 18 September 2020

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 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
All AMC 12 Problems and Solutions

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