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

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==Solution==
 
==Solution==
<math>2x+4</math> must be non-negative, so <math>x+2</math> must be non-negative.
+
<math>2x+4</math> must be non-negative, so <math>x+2</math> must be non-negative. Therefore, all <math>x\ge-2</math> are in the domain <math>\mathrm{(C)}</math>.
 
 
Therefore, all <math>x\ge-2</math> are in the domain <math>\mathrm{(C)}</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}}

Latest revision as of 10:50, 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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