A community for students.
Here's the question you clicked on:
 0 viewing
DLS
 2 years ago
Domain:?
DLS
 2 years ago
Domain:?

This Question is Closed

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

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

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

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

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

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

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

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

satellite73
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.3if x<0 then why is the x1 thing reversed,first doubt?

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

pursultan
 2 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 : \[x \geq 0 \& x \geq 4\]

satellite73
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0So domain of the function is \[x \geq 4\]

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

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

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

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

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

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

DLS
 2 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

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

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

satellite73
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.3for kind of removing the mod right?

satellite73
 2 years 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
 2 years 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
 2 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
 2 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?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.