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

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

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=x1%2Bx2%2Bx36%3E0

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1apparently it is \((\infty, 0]\cup [2,\infty)\)

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1it will take a long time to do you have to solve over different intervals

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1i am sorry my answer was wrong it is \((\infty, 0]\cup [4,\infty)\)

pursultan
 one year 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 : \[x \geq 0 \& x \geq 4\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1you have to break it up in to cases if \(x>3\) then \(x1=x1,,x2=x2, x3=x3\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1so if \(x>3\) you are solving \[x1+x2+x36\geq 0\] in which case you get \[3x12\geq 0\] and so \(x\geq 4\)

pursultan
 one year ago
Best ResponseYou've already chosen the best response.0So domain of the function is \[x \geq 4\]

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1if \(x<1\) then \(x1=1x\)

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1because if \(x<1\) then \(x1<0\) and so \(x1=(x1)=1x\)

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1yes, and if \(x1\) is negative, then \(1x\) is positive

DLS
 one year 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1lets try it with numbers suppose \(x=2\) then \(21=3=3=1(2)\)

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1the definition is this \[ x = \left\{\begin{array}{rcc} x & \text{if} & x <0 \\ x & \text{if} & x\geq 0 \end{array} \right. \]

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1and so for example \[ x1 = \left\{\begin{array}{rcc} 1x & \text{if} & x <1 \\ x1& \text{if} & x \geq 1 \end{array} \right. \]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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
 one year 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
 one year 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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