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DLS
 3 years ago
Domain:?
DLS
 3 years ago
Domain:?

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DLS
 3 years ago
Best ResponseYou've already chosen the best response.3\[\LARGE \sqrt{x1+x2+x36}\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3I want to do this with graphical method,would be appreciated

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3no where near,absolutely wrong.

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3tell me what do you get for x=3?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i= x1%2Bx2%2Bx36%3E0

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3yes that is the correct graph,i need some good explanation..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0apparently it is \((\infty, 0]\cup [2,\infty)\)

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3tell me what do u get for x=3 like posted above

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it will take a long time to do you have to solve over different intervals

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3if x<0 then why is the x1 thing reversed,first doubt?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i am sorry my answer was wrong it is \((\infty, 0]\cup [4,\infty)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0taking positive and negative value of piecewise function and solving the square root part greater than equal to 0 you get the following : \[x \geq 0 \& x \geq 4\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you have to break it up in to cases if \(x>3\) then \(x1=x1,,x2=x2, x3=x3\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so if \(x>3\) you are solving \[x1+x2+x36\geq 0\] in which case you get \[3x12\geq 0\] and so \(x\geq 4\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So domain of the function is \[x \geq 4\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3If i have x1 then for x<0 it becomes x1?why?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if \(x<1\) then \(x1=1x\)

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3i wrote that too but why :

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because if \(x<1\) then \(x1<0\) and so \(x1=(x1)=1x\)

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3but modulus function makes it positive

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, and if \(x1\) is negative, then \(1x\) is positive

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3why is the sign getting reversed? i thought   will make it +ve no matter if x>0 or x<0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lets try it with numbers suppose \(x=2\) then \(21=3=3=1(2)\)

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3seems legit but very non intuitive for me first time :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the definition is this \[ x = \left\{\begin{array}{rcc} x & \text{if} & x <0 \\ x & \text{if} & x\geq 0 \end{array} \right. \]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3for kind of removing the mod right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and so for example \[ x1 = \left\{\begin{array}{rcc} 1x & \text{if} & x <1 \\ x1& \text{if} & x \geq 1 \end{array} \right. \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, it seems like absolute value should be easy right? just "make it positive" but in fact it is a very annoying piecewise function and hard to work with because of that fact

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3hmm..okay..clear enough..thanks for your time :") why do i have 3 medals anyway :P

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3and the last thing,how do i get the points on the graph? i am supposed to find dy/dx at all these intervals right? If they turn out to be +ve then increasing slope..and vice versa..but what about the points?
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