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

Latest revision as of 09:50, 27 April 2008

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

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

$2x+4$ must be non-negative, so $x+2$ must be non-negative. Therefore, all $x\ge-2$ are in the domain $\mathrm{(C)}$ .

@@ Line 1: / Line 1: @@
 ==Problem==
-The domain of the function <math>f(x)=\sqrt{4+2x}</math> is
+The [[domain]] of the [[function]] <math>f(x)=\sqrt{4+2x}</math> is
-A. <math>(-2,+\infty)</math>
-B. <math>[0,+\infty)</math>
-C.  <math>[-2,+\infty)</math>
-D. <math>[-2,0]</math>
-E. <math>R</math>
+<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>
 ==Solution==
-x+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>
+<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>.
 ==See also==
 {{CYMO box|year=2006|l=Lyceum|num-b=2|num-a=4}}