Modulus inequality some different concepts

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Consider, |x + 1| - |x| < 2

 means, we have to find those values for x for which the difference of distance from -1 and 0 is less than 2

 By distance method discussed in http://iitmathseasy.blogspot.in/2013/03/modulus-equality-solving-concept-ii.html
"d" denotes distance

 from figure, it is clear that if we take any value less than -1 , consider a < - 1
d( - 1,a) < d(0,a) ( because nearer to -1 than 0)

 therefore , d( - 1,a) - d(0,a) < 0
means, for any point less than -1
|x + 1| - |x| < 0 < 2

 e.g. take x = - 3
d( - 1, - 3) = 2  ( put x = -3, |x + 1| )
 and, d(0, - 3) = 3 ( put x = -3 , |x| )

 therefore, d( - 1, - 3) - d(0, - 3) = 2 - 3 = - 1 < 2 ( difference will always be -1, see figure )
i.e. |x + 1| - |x| = - 1 < 2

 therefore , for all values less than -1 inequality satisfies.

 consider , any point between -1 and 0 including -1 and 0

 since, d( - 1,0) = 1 , 
therefore for no points the difference between difference from -1 and 0 will exceed 2.

 implies that for all values between and including -1 and 0 inequality satisfies.

 i.e. d( - 1,a) - d(0,a) \le 1 < 2

 |x + 1| - |x| \le 1 < 2 , ( -1 gives -1 and 0 gives 1, )

 consider, any point greater than 0,

 clearly from figure difference will always be 1
|x + 1| - |x| = 1 < 2

 therefore, for all values greater than 0 inequality satisfies.

 considering all the statements, inequality satisfies for all Real Numbers.
x \in ( - \infty ,\infty )
or x \in R

 Second method,

 |x + 1| - |x| < 2
from graph,

 there are two curves, 

 y = |x + 1| - |x| and y = 2

 and we have to find those values of x for which y = |x + 1| - |x| is less than y = 2 i.e. y = |x + 1| - |x| will below the line y = 2

 consider, y = |x + 1| - |x|
two critical points -1 and 0
for, x \le - 1
y = - x - 1 - ( - x)
y = - 1

 for,  - 1 \le x \le 0
y = x + 1 - ( - x)
y = 2x + 1 ,

 we need two points to draw a line.
put x = -1
y = -1
and for x = 0
y = 1

 for, x \ge 0
y = x + 1 - (x)

 y = 1

 plot the graph as mentioned above, 

 as cleared from the above graph, y = |x + 1| - |x| is always less than or below to y = 2,

  therefore for x \in ( - \infty ,\infty ) , |x + 1| - |x| < 2










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