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

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Isn't it just the graph of x^2 - 1 but with the part which is below the x-axis inverted so that it is above the x-axis? You will lose differentiability at x=-1 and x=1 but continuity is preserved so there is no need to define it piece-wise?
So 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.
@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.
So how would you just graph x^2-1? this is a parabola right with y-intercept -1 and x-intercept -1 and 1
-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)
now the first graph was for x>=1 so we need to play with's domain
now for the 2nd part we have x<1
something like that
Oh 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
wait something is wrong with my graph its not all above the x-axis
ok so we had \[|x^2-1|=(x^2-1) \text{ if } x^2-1 \ge 0\] \[|x^2-1|=-(x^2-1) \text{ if } x^2-1<0\] right?
don't understand what is incorrect with your graph it seems correct to me
\[(x-1)(x+1) \ge 0 \] this is 0 when x=1 and x=-1 ---|----|--- -1 1 now we need to test where (x-1)(x+1) is greater than 0 so we try the first interval (-inf,-1) we choose a number like -2 (-2-1)(-2+1)=(-3)(-1)=3>0 so on the interval (-inf,-1)=> (x-1)(x+1)>0 what about the interval (-1,1) we choose a number like 0 (0-1)(0+1)=-1(1)=-1<0 so on this interval we have that (x-1)(x+1)<0 now last interval (1,inf) we choose a number to try like 2 (2-1)(2+1)=1(3)=3>0 so for the interval (1,inf)=> (x-1)(x+1)>0 so for this |x^2-1|=x^2-1 happens on (-inf,-1) and (1,inf) now lets check out the other equality |x^2-1|=-(x^2-1)=-x^2+1 this is for x^2-1<0 (x-1)(x+1)<0 btw (-1,1)
there we go
i should have done all the work earlier i just skipped this algebra part
so do you understand |x^2-1|=x^2-1 on the intervals (-inf,-1), (1,inf) and |x^2-1|=-(x^2-1) on the interval (-1,1)
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)
|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
Thanks 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
after considering the graph of the various rules of the function

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