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anonymous
 5 years ago
How would I graph an expression like
x^(2)1
I know I would first convert it to a piece wise function where:
anonymous
 5 years ago
How would I graph an expression like x^(2)1 I know I would first convert it to a piece wise function where:

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0WHOA 2 LAZY 2 ANSWER BOO HOO

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^(2)1 if x^(2) 1 >=0 and x^(2) + 1 x^(2) 1 <0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0FIGURE IT OUTIT IS YOUR HW

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Isn't it just the graph of x^2  1 but with the part which is below the xaxis inverted so that it is above the xaxis? You will lose differentiability at x=1 and x=1 but continuity is preserved so there is no need to define it piecewise?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So x^(2)1 if x^(2) 1 >=0 and x^(2) + 1 x^(2) 1 <0 x^(2)  1 if x >= 1 and x^(2) + 1 if x < 1 Thus I can just input values into both sides, anyone able to provide me with a more difficult piece wise function. I know it is elementary I just want to refresh my memory.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2@Hermella please don't yell at people to just figure it out on their own. They come on this site to get help. If you don't want to help, then just ignore their question.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2So how would you just graph x^21? this is a parabola right with yintercept 1 and xintercept 1 and 1

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2dw:1326156295808:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^(2) + 1 if x < 1 violates x^(2) + 1 if x^(2) 1 <0. Take x = 1/2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2now what does x^2+1 look like (note we will worry about the domains in a sec)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2dw:1326156367729:dw

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2now the first graph was for x>=1 so we need to play with's domain

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2dw:1326156439379:dw

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2now for the 2nd part we have x<1

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2dw:1326156472771:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh thanks for going through this step by step, it is clear to me now. Could you give me an example of a piece wise function that is not continuous if not I understand

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2wait something is wrong with my graph its not all above the xaxis

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2ok so we had \[x^21=(x^21) \text{ if } x^21 \ge 0\] \[x^21=(x^21) \text{ if } x^21<0\] right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't understand what is incorrect with your graph it seems correct to me

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2\[(x1)(x+1) \ge 0 \] this is 0 when x=1 and x=1  1 1 now we need to test where (x1)(x+1) is greater than 0 so we try the first interval (inf,1) we choose a number like 2 (21)(2+1)=(3)(1)=3>0 so on the interval (inf,1)=> (x1)(x+1)>0 what about the interval (1,1) we choose a number like 0 (01)(0+1)=1(1)=1<0 so on this interval we have that (x1)(x+1)<0 now last interval (1,inf) we choose a number to try like 2 (21)(2+1)=1(3)=3>0 so for the interval (1,inf)=> (x1)(x+1)>0 so for this x^21=x^21 happens on (inf,1) and (1,inf) now lets check out the other equality x^21=(x^21)=x^2+1 this is for x^21<0 (x1)(x+1)<0 btw (1,1)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2i should have done all the work earlier i just skipped this algebra part

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2so do you understand x^21=x^21 on the intervals (inf,1), (1,inf) and x^21=(x^21) on the interval (1,1)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2now back to our parabolas

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so (1,1) does not exist in the domain?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2so x^21 we had dw:1326157039943:dw but we didn't have it for (1,1)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2so we actually have dw:1326157076390:dw

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2that is for the x^21=x^21 for the intervals (inf,1),(1,inf)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why is (1, 1) excluded from the domain I don't understand

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2now for the middle part (1,1) we have x^21=(x^21)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2dw:1326157146780:dw

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2x^21=x^21 for x^21>0 right?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2now did you solve x^21>0?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2you will get that x^21=x^21 is only true for the intervals (inf,1),(1,inf)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2now for x^21=(x^21) is only true for the intervals (1,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, because it is the only area of the graph that will make the graph less than zero

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2yes x^21<0 on the interval (1,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so when solving these it is important to make a chart and plug in numbers between the intervals

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks I think it is a bit more clear in my head how to solve these now I need to focus on the output of the piece wise function and use that to guide me on how to graph the function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0after considering the graph of the various rules of the function
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