Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 3"

(Standardized answer choices; minor edits)
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==Problem==
 
==Problem==
The domain of the function <math>f(x)=\sqrt{4+2x}</math> is
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The [[domain]] of the [[function]] <math>f(x)=\sqrt{4+2x}</math> is
  
A. <math>(-2,+\infty)</math>
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<math>\mathrm{(A)}\ (-2,+\infty)\qquad\mathrm{(B)}\ [0,+\infty)\qquad\mathrm{(C)}\ [-2,+\infty)\qquad\mathrm{(D)}\ [-2,0]\qquad\mathrm{(E)}\ \mathbb{R}</math>
  
B. <math>[0,+\infty)</math>
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==Solution==
 
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<math>2x+4</math> must be non-negative, so <math>x+2</math> must be non-negative.
C.  <math>[-2,+\infty)</math>
 
 
 
D. <math>[-2,0]</math>
 
 
 
E. <math>R</math>
 
  
 
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Therefore, all <math>x\ge-2</math> are in the domain <math>\mathrm{(C)}</math>.
==Solution==
 
2x+4 must be non-negative. Therefore, x+2 must be non-negative. Therefore, all x greater than or equal to -2 are in the domain. <math>\mathrm {(A)}</math>
 
  
 
==See also==
 
==See also==
 
{{CYMO box|year=2006|l=Lyceum|num-b=2|num-a=4}}
 
{{CYMO box|year=2006|l=Lyceum|num-b=2|num-a=4}}

Revision as of 10:49, 27 April 2008

Problem

The domain of the function $f(x)=\sqrt{4+2x}$ is

$\mathrm{(A)}\ (-2,+\infty)\qquad\mathrm{(B)}\ [0,+\infty)\qquad\mathrm{(C)}\ [-2,+\infty)\qquad\mathrm{(D)}\ [-2,0]\qquad\mathrm{(E)}\ \mathbb{R}$

Solution

$2x+4$ must be non-negative, so $x+2$ must be non-negative.

Therefore, all $x\ge-2$ are in the domain $\mathrm{(C)}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 2
Followed by
Problem 4
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