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Mathematics 8 Online
OpenStudy (anonymous):

Domain and Range f(x)= √3x+2

OpenStudy (anonymous):

\[ {x \in \mathbb{R} : x \ge 0 }\]

OpenStudy (anonymous):

Domain^

OpenStudy (anonymous):

\[f \in \mathbb{R} : f \ge 2\]

OpenStudy (anonymous):

range ^

OpenStudy (anonymous):

then again maybe it is \[f(x)=\sqrt{3x+2}\]

OpenStudy (anonymous):

i should of put parenthesis around it huh? XD

OpenStudy (anonymous):

@satellite73 your right

OpenStudy (anonymous):

@zepdrix :D

OpenStudy (anonymous):

yeah true I did the domain & range based on: \[f(x) = \sqrt(3 x)+2\]

OpenStudy (anonymous):

sorry about that :X

OpenStudy (anonymous):

@zepdrix help lol wth...

OpenStudy (anonymous):

i know how to get Domain by setting the equation >_ to zero then solve. But how do you get the Range?

OpenStudy (anonymous):

\[3x+2\geq =\iff 3x\geq -2\iff x\geq-\frac{2}{3}\]

OpenStudy (anonymous):

for the domain

OpenStudy (anonymous):

range is \([0,\infty)\) since the square root means the positive square root

OpenStudy (anonymous):

always ?

OpenStudy (anonymous):

and 0 is included?

OpenStudy (anonymous):

yes, \(\sqrt{x}\geq 0\) always

OpenStudy (anonymous):

zeros is included because the square root of zero is zero

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