absolute value question
how can we demonstrate that |x − 5| + |x +1| ≠ 4 without removing the absolute value symbol?
your help is really appreciated!
RowanH: In general, |x|+|y| ≠ |x+y|. Consider x = 1, y = -1.
|x|+|y| = 2
|x+y| = 0.
So. |x|+|y| ≠ |x+y| because 2 ≠ 0.
zineb.aj: I'm not sure why you wouldn't want to remove the absolute value symbol. As far as I know, there is no easy way to do so. When dealing with absolute value problems like these, we define the absolute value as a piecewise function. Let's look at the simplest absolute value function:
|x| = { x , x >= 0
{ -x, x < 0.
Because the text does not show math symbols well, assume that >= is greater than or equal to, and <= is less than or equal to.
Why is it defined this way? For x < 0, the function is negative. The negative on the -x is just multiplying an already negative number by -1 to get its corresponding positive number. That way, |x| >= 0, which is necessary by definition.
First, let's define each absolute value as a piecewise function and instead solve for |x − 5| + |x +1| = 4 to then solve for the original equation.
|x-5| = { x - 5, x - 5 >= 0 ====> { x - 5, x >= 5
{ -(x-5), x - 5 < 0. ====> { -x+5, x < 5.
|x+1| = { x + 1, x >= -1
{ -x - 1, x < -1
We have three intervals to solve the equation on: x < -1; -1<= x< 5; x >=5. There is no interval for x < 5 by itself or x >= -1 by itself because they are already included by the three intervals we have.
For x < -1 (so x < 5 is also true), our equation becomes
(-x+5) + (-x-1) = 4. ==> -2x +4 = 4. ==> -2x = 0 ==> x = 0.
Plug x = 0 back into the equation |x-5| + |x+1| = 4 to see if it is true.
| -5 | + |1| = 5+1 = 6.
6 = 4? False. Why was this false? x = 0 is outside of our interval of definition, because it is not less than -1. Therefore, there is no x satisfying the equation |x − 5| + |x +1| = 4 for x < -1. As a result, all x < -1 does satisfy the equation |x − 5| + |x +1| ≠ 4.
For -1<= x< 5, our equation becomes
(-x+5) + (x+1) = 4 ==> 6 = 4 which is false for all x on the interval -1<= x< 5. This tells us that |x − 5| + |x +1| ≠ 4 is true for all x on the interval -1<= x< 5.
For x >=5 (so x >= -1 is also true), our equation becomes:
(x-5) + (x+1) = 4 ==> 2x - 4 = 4 ==> 2x = 8 ==> x = 4.
Plug x = 4 into |x-5| + |x+1| = 4 to see if it is true.
|4 -5 | + |4+1| = | -1| + |5| = 6
6 = 4? False. Why? x = 4 is also outside of our interval of definition, because 4 is not greater than or equal to 5. Therefore, there is no x satisfying |x-5| + |x+1| = 4 for x >= 5. As a result, all x on x>=5 does satisfy |x − 5| + |x +1| ≠ 4.
Therefore, in interval notation ( I recommend searching this up if you are not familiar with it) all x in (-∞,∞) satisfies |x − 5| + |x +1| ≠ 4. This means for all real values of x, |x − 5| + |x +1| ≠ 4.
Graphing y = |x − 5| + |x +1| on Desmos, we see that the graph never equals 4.
I would have thought it depends on the value of x.
Since bothe terms on the left are magnitudes, my instinct tells me I can add them together:
|x - 5 + x + 1| =/= 4
simplifying:
|2x-4| =/= 4
But if x = 0, or x = 4, |2x-4| is equal to 4
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