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x^(2)-1 if x^(2) -1 >=0
and
-x^(2) + 1 x^(2) -1 <0

FIGURE IT OUTIT IS YOUR HW

MAKE ME

|dw:1326156295808:dw|

-x^(2) + 1 if x < 1 violates -x^(2) + 1 if x^(2) -1 <0. Take x = 1/2

now what does -x^2+1 look like
(note we will worry about the domains in a sec)

|dw:1326156367729:dw|

now the first graph was for x>=1 so we need to play with's domain

|dw:1326156439379:dw|

now for the 2nd part we have x<1

|dw:1326156472771:dw|

something like that

wait something is wrong with my graph its not all above the x-axis

yup

don't understand what is incorrect with your graph it seems correct to me

there we go

i should have done all the work earlier
i just skipped this algebra part

now back to our parabolas

so (1,-1) does not exist in the domain?

so x^2-1 we had |dw:1326157039943:dw|
but we didn't have it for (-1,1)

so we actually have
|dw:1326157076390:dw|

that is for the |x^2-1|=x^2-1 for the intervals (-inf,-1),(1,inf)

why is (-1, 1) excluded from the domain I don't understand

now for the middle part (-1,1) we have |x^2-1|=-(x^2-1)

|dw:1326157146780:dw|

|x^2-1|=x^2-1 for x^2-1>0 right?

now did you solve x^2-1>0?

you will get that |x^2-1|=x^2-1 is only true for the intervals (-inf,-1),(1,inf)

now for |x^2-1|=-(x^2-1) is only true for the intervals (-1,1)

oh, because it is the only area of the graph that will make the graph less than zero

yes x^2-1<0 on the interval (-1,1)

so when solving these it is important to make a chart and plug in numbers between the intervals

yes most definitely

after considering the graph of the various rules of the function