Laplace transforms for f(t)

[R.K.4.7]

Hi guys, her I have attached a pic where I can't understand , how did they get the answers 6, 2t+4, and 4t.
I get that when we solve the equation, we get 6. But why is the eqn changed, I mean why did they add a -ve sign?
I notice, there are 2 -ve signs in the 1st eqn,
1 -ve in the 2nd & no change in the 3rd eqn.
This gets me to the main doubt, when are we supposed to add that -ve sign ?
Thanks guys

 

[stapel]

Hi guys, her I have attached a pic where I can't understand , how did they get the answers 6, 2t+4, and 4t.
I get that when we solve the equation, we get 6. But why is the eqn changed, I mean why did they add a -ve sign?
I notice, there are 2 -ve signs in the 1st eqn,
1 -ve in the 2nd & no change in the 3rd eqn.
This gets me to the main doubt, when are we supposed to add that -ve sign ?

You've posted almost a full page of text (which is curved and sideways, making it quite difficult to read). You've made reference to... what I'm guessing is some portion of the page of text, but I'm not sure what. And you've typed out things that are probably typoes (like "her" for what I think is meant to be "here"), abbreviations ("eqn" for what is almost certainly "equation"), and high-school crypto-cisms, such as "-ve".

Please reply with a legible post of the text in question, a clarification of the line or lines which you are finding confusing, and a clear statement of your query. Thank you!

 

[HallsofIvy]

I am also not certain what it is you are asking about. Is it just the absolute value?

You are given that f(x)= |x- 1|+ |x+ 1|+ |x- 2|+ |x+ 2|

|a| is defined as "a if \(\displaystyle a\ge 0\), -a if \(\displaystyle a< 0\)".

So |x- 1| is x- 1 if \(\displaystyle x- 1\ge 0\), which is the same as saying \(\displaystyle x\ge 1\), -(x- 1)= 1- x if \(\displaystyle x< 1\).

|x+ 1| is x+ 1 if \(\displaystyle x+ 1\ge 0\), which is the same as saying \(\displaystyle x\ge -1\), -(x+ 1)= -x- 1 if \(\displaystyle x< -1\).

|x- 2| is x- 1 if \(\displaystyle x- 2\ge 0\), which is the same as saying \(\displaystyle x\ge 2\), -(x- 2)= 2- x if \(\displaystyle x< 2\).

|x+ 2|= x+ 2 if \(\displaystyle x+ 2\ge 0\), which is the same as saying \(\displaystyle x\ge -2\), -(x+ 2)= -x- 2 if \(\displaystyle x< -2\).

In order of size, those "break points", where the absolute value changes, are -2, -1, 1, and 2.

If x< -2, all four of x- 1, x+ 1, x- 2, and x+ 2 are negative so that sum of absolute values is (1- x)+ (-x- 1)+ (2- x)+ (-x- 2)= -4x.

If \(\displaystyle -2\le x< -1\), x+ 2 is positive but x- 1, x+ 1, and x- 2 are still negative so the sum is (1- x)+ (-x-1)+ (2- x)+ (x+2)= -2x+ 4.

If \(\displaystyle -1\le x< 1\), x+ 2 and x+ 1 are positive but x- 1 and x- 2 are still negative so the sum is (1- x)+ (x+ 1)+ (2- x)+ (x+ 2)= 6.

If \(\displaystyle 1\le x< 2\), x+ 2, x+ 1, and x- 1 are positive but x- 2 is still negative so the sum is (x- 1)+ (x+ 1)+ (2- x)+ x+ 2= 2x+ 4.

If \(\displaystyle 2\le x\) all four of x+ 2, x+ 1, x- 1, and x- 2 are positive so the sum is (x- 1)+ (x+ 1)+ (x- 2)+ (x+ 2)= 4x.