what is the difference between absolute value and 'regular ones '

example: y= abs x-4

Y=x-4....if i give 1 to x the result is the same...but what is the difference?

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- Jun 9th 2010, 11:43 PMalessandromangionedifference of abs value
what is the difference between absolute value and 'regular ones '

example: y= abs x-4

Y=x-4....if i give 1 to x the result is the same...but what is the difference? - Jun 10th 2010, 12:19 AMProve It
The absolute value represents the "size" of the function.

E.g. $\displaystyle |3| = 3$ since it is 3 units from 0.

$\displaystyle |-5| = 5$ since it is 5 units from 0.

In other words

$\displaystyle |X| = \begin{cases}\phantom{-}X\textrm{ if }X\geq 0\\-X\textrm{ if }x<0\end{cases}$

So in your case

$\displaystyle |x - 4| = \begin{cases}\phantom{-(}x - 4\phantom{)}\textrm{ if }x-4\geq 0\\-(x-4)\textrm{ if }x-4<0\end{cases}$

$\displaystyle |x - 4| = \begin{cases}x-4\textrm{ if }x \geq 4\\4-x\textrm{ if }x<4\end{cases}$ - Jun 20th 2010, 05:28 PMmfetch22
Another useful alternative definition gives the same return as the absolute value expression:

$\displaystyle |x| = \sqrt{x^2}$

This is the consquence of the fact that a number squared always returns a positive number. So in your case:

$\displaystyle |x-4| = \sqrt{(x-4)^2}$

But as for evaluating abosolute values you cannot simply write:

$\displaystyle |x-4| = \sqrt{(x-4)^2} = (x-4)$

Because this will not give the correct values, you must evaluate each value to get the proper number in this alternative definition. But otherwise, this definition only really comes in handy in some situations, otherwise simply use the definition that the post above mine gave. That is the proper definition, mine is simply a "band-aid" definition that comes in handy in certain situations.