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

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

TareKKowshiK Group TitleBest ResponseYou've already chosen the best response.0
Domain={R}
 one year ago

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.3
and x=1 and x=2 too
 one year ago

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

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

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

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.3
tell me over one interval
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.3
and 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?
 one year ago
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