HelloI want some help please.

Applies -4 <x <-1 to prove that the nexts representations are independent of x.

a) |x+4| - 2 |x+1| + 3 |x|

b) |3χ+3| - |2χ+8| - 5|χ|

c) |2x-3| - |1-3x| + |-x|

d) |χ^2-1| + |χ^2-20|

sorry for the bad translation...

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- December 2nd 2012, 03:27 AMuloixHelp!!!|Absolute value of real numbers|
**Hello**I want some help please.

Applies -4 <x <-1 to prove that the nexts representations are independent of x.

a) |x+4| - 2 |x+1| + 3 |x|

b) |3χ+3| - |2χ+8| - 5|χ|

c) |2x-3| - |1-3x| + |-x|

d) |χ^2-1| + |χ^2-20|

sorry for the bad translation... - December 2nd 2012, 04:30 AMemakarovRe: Help!!!|Absolute value of real numbers|
You probably mean, "Use -4 < x < -1 to prove that the following expressions do not depend on x," i.e., are constants. The usual way of dealing with absolute values is to break the interval in question into smaller intervals where the sign of each expression under the absolute value is constant. Then on each interval absolute values can be removed depending on the sign: if e <= 0, then abs(e) = e and if e < 0, then abs(e) = -e.

For example, |2x-3| - |1-3x| + |-x| we have that

2x - 3 < 0 for x < 3/2 and 2x - 3 >= 0 for x >= 3/2

1 - 3x < 0 for x > 1/3 and 1 - 3x >= 0 for x <= 1/3

-x < 0, for x > 0 and -x >= 0 for x <= 0

Here the interval (-4, -1) from the problem statement lies left of 0, 1/3 and 3/2, so on this interval 2x - 3 < 0, 1 - 3x > 0 and -x > 0. Therefore, |2x-3| - |1-3x| + |-x| = -(2x - 3) - (1 - 3x) + (-x) = 2.