anonymous
  • anonymous
How would I graph an expression like |x^(2)-1| I know I would first convert it to a piece wise function where:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
WHOA 2 LAZY 2 ANSWER BOO HOO
anonymous
  • anonymous
x^(2)-1 if x^(2) -1 >=0 and -x^(2) + 1 x^(2) -1 <0
anonymous
  • anonymous
FIGURE IT OUTIT IS YOUR HW

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anonymous
  • anonymous
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?
anonymous
  • anonymous
MAKE ME
anonymous
  • anonymous
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.
myininaya
  • myininaya
@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
  • myininaya
So how would you just graph x^2-1? this is a parabola right with y-intercept -1 and x-intercept -1 and 1
myininaya
  • myininaya
|dw:1326156295808:dw|
anonymous
  • anonymous
-x^(2) + 1 if x < 1 violates -x^(2) + 1 if x^(2) -1 <0. Take x = 1/2
myininaya
  • myininaya
now what does -x^2+1 look like (note we will worry about the domains in a sec)
myininaya
  • myininaya
|dw:1326156367729:dw|
myininaya
  • myininaya
now the first graph was for x>=1 so we need to play with's domain
myininaya
  • myininaya
|dw:1326156439379:dw|
myininaya
  • myininaya
now for the 2nd part we have x<1
myininaya
  • myininaya
|dw:1326156472771:dw|
myininaya
  • myininaya
something like that
anonymous
  • anonymous
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
myininaya
  • myininaya
wait something is wrong with my graph its not all above the x-axis
myininaya
  • myininaya
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?
anonymous
  • anonymous
yup
anonymous
  • anonymous
don't understand what is incorrect with your graph it seems correct to me
myininaya
  • myininaya
\[(x-1)(x+1) \ge 0 \] 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)
myininaya
  • myininaya
there we go
myininaya
  • myininaya
i should have done all the work earlier i just skipped this algebra part
myininaya
  • myininaya
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)
myininaya
  • myininaya
now back to our parabolas
anonymous
  • anonymous
so (1,-1) does not exist in the domain?
myininaya
  • myininaya
so x^2-1 we had |dw:1326157039943:dw| but we didn't have it for (-1,1)
myininaya
  • myininaya
so we actually have |dw:1326157076390:dw|
myininaya
  • myininaya
that is for the |x^2-1|=x^2-1 for the intervals (-inf,-1),(1,inf)
anonymous
  • anonymous
why is (-1, 1) excluded from the domain I don't understand
myininaya
  • myininaya
now for the middle part (-1,1) we have |x^2-1|=-(x^2-1)
myininaya
  • myininaya
|dw:1326157146780:dw|
myininaya
  • myininaya
|x^2-1|=x^2-1 for x^2-1>0 right?
myininaya
  • myininaya
now did you solve x^2-1>0?
myininaya
  • myininaya
you will get that |x^2-1|=x^2-1 is only true for the intervals (-inf,-1),(1,inf)
myininaya
  • myininaya
now for |x^2-1|=-(x^2-1) is only true for the intervals (-1,1)
anonymous
  • anonymous
oh, because it is the only area of the graph that will make the graph less than zero
myininaya
  • myininaya
yes x^2-1<0 on the interval (-1,1)
anonymous
  • anonymous
so when solving these it is important to make a chart and plug in numbers between the intervals
myininaya
  • myininaya
yes most definitely
anonymous
  • anonymous
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
anonymous
  • anonymous
after considering the graph of the various rules of the function

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