Absolute value wihin an absolute value?

[ph7]

I cant figure out how to turn |x|x-3|| into a piecewise function. Anybody know how to deal with this starting with what I should do with the inception absolute value bars.

 

[Deleted member 4993]

I cant figure out how to turn |x|x-3|| into a piecewise function. Anybody know how to deal with this starting with what I should do with the inception absolute value bars.

Try plotting the function with [-6,6] domain.

Please share your work with us .

If you are stuck at the beginning tell us and we'll start with the definitions e.g. define retained earnings.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

 

[ph7]

Try plotting the function with [-6,6] domain.

Please share your work with us .

If you are stuck at the beginning tell us and we'll start with the definitions e.g. define retained earnings.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

what will that? are you suggesting i just graph it and derive the equations for each segment visually?

 

[ph7]

but how do i use that to write a piecewise function

 

[Deleted member 4993]

but how do i use that to write a piecewise function

Did you plot it [-6,6]?

Use your graphing calculator - or wolframalfa.com

Than look at the graph - it should be pretty clear....

 

[Bob Brown MSEE]

Did you plot it [-6,6]?

Use your graphing calculator - or wolframalfa.com

Than look at the graph - it should be pretty clear....

Click here

 

[ph7]

Click here

Ok I think I understand now, thanks all

 

[HallsofIvy]

Absolute value "changes" at 0. Here you have two absolute values that involve "x" and "x- 3" so you should break this into three intervals: x< 0, \(\displaystyle 0\le x< 3\), and \(\displaystyle 3\le x\).

If x< 0 then x is also less than 3 so both |x- 3|= -(x- 3) is positive and so x|x- 3|= -x(x- 3) is negative. |x|x-3||= -(-x(x- 3)= x(x- 3).

If \(\displaystyle 0\le x< 3\) then x is positive but x-3 is negative so x|x- 3|= -x(x- 3) is positive. |x|x- 3||= -x(x- 3).

If \(\displaystyle 3\le x\) then x is also positive so x|x- 3|= x(x- 3) is positive. |x|x- 3||= x(x- 3).