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

- anonymous

- katieb

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- anonymous

WHOA 2 LAZY 2 ANSWER BOO HOO

- anonymous

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

- anonymous

FIGURE IT OUTIT IS YOUR HW

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## More answers

- 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

MAKE ME

- 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

@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

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

|dw:1326156295808:dw|

- anonymous

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

- myininaya

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

- myininaya

|dw:1326156367729:dw|

- myininaya

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

- myininaya

|dw:1326156439379:dw|

- myininaya

now for the 2nd part we have x<1

- myininaya

|dw:1326156472771:dw|

- myininaya

something like that

- 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

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

- 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

yup

- anonymous

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

- 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

there we go

- myininaya

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

- 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

now back to our parabolas

- anonymous

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

- myininaya

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

- myininaya

so we actually have
|dw:1326157076390:dw|

- myininaya

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

- anonymous

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

- myininaya

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

- myininaya

|dw:1326157146780:dw|

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

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

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

yes most definitely

- 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

after considering the graph of the various rules of the function

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