Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

DLSBest ResponseYou've already chosen the best response.3
\[\LARGE \sqrt{x1+x2+x36}\]
 11 months ago

DLSBest ResponseYou've already chosen the best response.3
I want to do this with graphical method,would be appreciated
 11 months ago

DLSBest ResponseYou've already chosen the best response.3
no where near,absolutely wrong.
 11 months ago

DLSBest ResponseYou've already chosen the best response.3
tell me what do you get for x=3?
 11 months ago

satellite73Best ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=x1%2Bx2%2Bx36%3E0
 11 months ago

DLSBest ResponseYou've already chosen the best response.3
yes that is the correct graph,i need some good explanation..
 11 months ago

satellite73Best ResponseYou've already chosen the best response.1
apparently it is \((\infty, 0]\cup [2,\infty)\)
 11 months ago

DLSBest ResponseYou've already chosen the best response.3
tell me what do u get for x=3 like posted above
 11 months ago

satellite73Best ResponseYou've already chosen the best response.1
it will take a long time to do you have to solve over different intervals
 11 months ago

DLSBest ResponseYou've already chosen the best response.3
tell me over one interval
 11 months ago

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

satellite73Best ResponseYou've already chosen the best response.1
i am sorry my answer was wrong it is \((\infty, 0]\cup [4,\infty)\)
 11 months ago

pursultanBest 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 : \[x \geq 0 \& x \geq 4\]
 11 months ago

satellite73Best ResponseYou've already chosen the best response.1
you have to break it up in to cases if \(x>3\) then \(x1=x1,,x2=x2, x3=x3\)
 11 months ago

satellite73Best 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 \(x\geq 4\)
 11 months ago

pursultanBest ResponseYou've already chosen the best response.0
So domain of the function is \[x \geq 4\]
 11 months ago

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

satellite73Best ResponseYou've already chosen the best response.1
if \(x<1\) then \(x1=1x\)
 11 months ago

DLSBest ResponseYou've already chosen the best response.3
i wrote that too but why :
 11 months ago

satellite73Best ResponseYou've already chosen the best response.1
because if \(x<1\) then \(x1<0\) and so \(x1=(x1)=1x\)
 11 months ago

DLSBest ResponseYou've already chosen the best response.3
but modulus function makes it positive
 11 months ago

satellite73Best ResponseYou've already chosen the best response.1
yes, and if \(x1\) is negative, then \(1x\) is positive
 11 months ago

DLSBest 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
 11 months ago

satellite73Best ResponseYou've already chosen the best response.1
lets try it with numbers suppose \(x=2\) then \(21=3=3=1(2)\)
 11 months ago

DLSBest ResponseYou've already chosen the best response.3
seems legit but very non intuitive for me first time :P
 11 months ago

satellite73Best ResponseYou've already chosen the best response.1
the definition is this \[ x = \left\{\begin{array}{rcc} x & \text{if} & x <0 \\ x & \text{if} & x\geq 0 \end{array} \right. \]
 11 months ago

DLSBest ResponseYou've already chosen the best response.3
for kind of removing the mod right?
 11 months ago

satellite73Best ResponseYou've already chosen the best response.1
and so for example \[ x1 = \left\{\begin{array}{rcc} 1x & \text{if} & x <1 \\ x1& \text{if} & x \geq 1 \end{array} \right. \]
 11 months ago

satellite73Best 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
 11 months ago

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

DLSBest 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?
 11 months ago
See more questions >>>
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.