How to deal with 3 absolute values in an equation?

[nigahiga]

|x-3| + |x+2| - |x-4| = 3

The solution is -6 and 2. What is the method of finding the x? Split into piecewise? If so, how would one split it into piecewise function?

 

[JeffM]

|x-3| + |x+2| - |x-4| = 3

The solution is -6 and 2. What is the method of finding the x? Split into piecewise? If so, how would one split it into piecewise function?

$\displaystyle u \ge 0 \implies |u| = u;\ u < 0 \implies |u| = - u.$ Correct?

$\displaystyle f(x) = |x - 3| + |x + 2| + |x - 4|.$ Translate into a piecewise function by eliminating absolute values.

$\displaystyle x \ge 4 \implies x - 4 \ge 0\ and\ x - 3 \ge 1\ and\ x + 2 \ge 6 \implies f(x) = (x - 3) + (x + 2) + (x - 4) \ge 1 + 6 + 0 = 7.$

No solution if x $\displaystyle \ge$ 4.

Now what?

 

[stapel]

|x-3| + |x+2| - |x-4| = 3

The solution is -6 and 2. What is the method of finding the x? Split into piecewise? If so, how would one split it into piecewise function?

Consider cases. The sign of |x - 3| changes when you go from x < 3 to x > 3, so that can be an interval endpoint. The same can be said for x = -2 and x = 4. So you've got intervals (-infinity, -2), [-2, 3), [3, 4), and [4, +infinity). For each interval, you know the sign of the three absolute values.

So, for each interval, take the bars off, apply the correct sign to each absolute-value expression, and solve the resulting equation. Don't forget, though, to check that the solution for that interval is actually in that interval! For instance, as the previous poster pointed out, there's a problem for the fourth interval:

When x > 4, then all four absolute-value expressions are positive (or at least non-negative), so we get:

(x - 3) + (x + 2) - (x - 4) = 3

x - 3 + x + 2 - x + 4 = 3

x + x - x - 3 + 2 + 4 = 3

x + 3 = 3

x = 0

But... since 0 is not in fact greater than 4, then "x = 0" cannot possibly be a solution for the interval "x > 4".

 

[HallsofIvy]

|x-3| + |x+2| - |x-4| = 3

The solution is -6 and 2. What is the method of finding the x? Split into piecewise? If so, how would one split it into piecewise function?

The simplest thing to do is separate the number line into sub-intervals: x< -2, -2< x< 3, 3< x< 4, and x> 4. If x< -2, all three of x-3= x- (3), x+ 2=x- (-2), and x- 4=x- (4) are negative: |x- 3|+ |x+ 2|- |x- 4|= -(x- 3)- (x+ 2)+ x- 4= -x- 3= 3 so -x= 6, x= -6. That IS "less than -2" x=-6 is solution. If -2< x< 3 then x+ 2 is positive but x- 3 and x- 4 are still negative: |x- 3|+ |x+ 2|- |x- 4|= -(x- 3)+ (x+ 2)+ x- 4= x+ 1= 3 so x= 2. That IS "between -3 and 3" so it is also a solution. If 3< x< 4, the x+ 2 and x- 3 are positive but x- 4 is still negative: |x- 3|+ |x+ 2|- |x- 4|= (x- 3)+ (x+ 2)+ x- 4= 3x- 5= 3. 3x= 8 so x= 8/3. That is NOT between 3 and 4 so is NOT a solution. Finally, if x> 4, all three of x- 3, x+ 2, and x- 4 are positive so |x- 3|+ |x+ 2|- |x- 4|= x- 3+ x+ 2- x+ 4= x+ 3= 3 so x= 0. That is NOT between 3 and 4 so is NOT a solution.

 

[nigahiga]

So when dealing with multiple absolute value equations, extraneous solutions are bound to come out? That is, if the solutions are out of the "domain" of the specific boundaries you are testing?

 

[JeffM]

So when dealing with multiple absolute value equations, extraneous solutions are bound to come out? That is, if the solutions are out of the "domain" of the specific boundaries you are testing?

There was ONE equation.

What extraneous solutions? if the assumption that x lies in a specific interval leads to a solution where x is outside the interval that was the basis for the solution, the solution is wrong, not extraneous.

On the assumption that x is in the interval (a, b) then x is not in the interval (a, b) is just nonsense.

 

[nigahiga]

There was ONE equation.

What extraneous solutions? if the assumption that x lies in a specific interval leads to a solution where x is outside the interval that was the basis for the solution, the solution is wrong, not extraneous.

On the assumption that x is in the interval (a, b) then x is not in the interval (a, b) is just nonsense.

Okay thank you! Last question, let's say the absolute value equation is |x-8|-|x-2|+|x|=6. Then what would the intervals you are testing be?

Would it be x<0 , 0<x<2 , 2<x<8 , 8<x ??? or
would it be x<0 , 0≤x≤2 , 2<x<8 , 8≤x or
would it be x≤0 , 0<x<2 , 2≤x≤8 , 8<x or
would it be x<0 , 0≤x<2 , 2≤x<8 , 8≤x or........

To rephrase the question, I am confused as to where to put the "or equal to" sign (≤ and ≥). Tacking the equal sign onto all the intervals is not possible right? Does the placement of the equals sign not matter as long as it does not occupy the same number? E.g. (a,b) U [b,c) as opposed to (a,b] U [b,c) <---- since that is just (a,c) ?

Ahhh! My logic is not very good. Please help me out!

 

[nigahiga]

Or do you leave out the equals sign altogether when testing intervals?

 

[JeffM]

Okay thank you! Last question, let's say the absolute value equation is |x-8|-|x-2|+|x|=6. Then what would the intervals you are testing be?

Would it be x<0 , 0<x<2 , 2<x<8 , 8<x ??? or
would it be x<0 , 0≤x≤2 , 2<x<8 , 8≤x or
would it be x≤0 , 0<x<2 , 2≤x≤8 , 8<x or
would it be x<0 , 0≤x<2 , 2≤x<8 , 8≤x or........

To rephrase the question, I am confused as to where to put the "or equal to" sign (≤ and ≥). Tacking the equal sign onto all the intervals is not possible right? Does the placement of the equals sign not matter as long as it does not occupy the same number? E.g. (a,b) U [b,c) as opposed to (a,b] U [b,c) <---- since that is just (a,c) ?

Ahhh! My logic is not very good. Please help me out!

You need to get what we are doing. When we are dealing with a function that includes absolute values, we can restate it in terms of a piecewise function that has no absolute values.

The fundamental idea is this: u < 0 \implies |u| = - u,\ but\ u \ge 0 \implies |u| = u.[/tex]

If you remember this fundamental idea, the proper relationship symbol is obvious.

$\displaystyle x - 8 \ge 0 \implies x \ge 8 \implies x - 2 \ge 6 \ge 0\ and\ x \ge 8 \ge 0.$

So $\displaystyle f(x) = |x - 8| + |x - 2| + |x|\ and\ x \ge 8 \implies f(x) = x - 8 + x - 2 + x = 3x - 10.$

This function applies only in the interval $\displaystyle [8, \infty).$

OK Now let's deal with the equation appropriate for that interval $\displaystyle 3x - 10 = 6 \implies 3x = 16 \implies x < 8.$

But that's a contradiction. x cannot be greater than or equal to 8 and also less than 8. Nonsense.

$\displaystyle Obviously\ x - 2 \ge 0 \implies x \ge 2 \ge 0.$ But we already know that a valid answer will require x < 8.

$\displaystyle 2 \le x < 8 \implies x \ge 0\ and\ x - 2 \ge 0\ but\ x - 8 < 0.$

So $\displaystyle f(x) = |x - 8| + |x - 2| + |x|\ and\ 2 \le x < 8 \implies f(x) = -(x - 8) + x - 2 + x = - x + 8 + x - 2 + x = x + 6.$

This function applies only in the interval $\displaystyle [2, 8).$

OK Now let's deal with the equation appropriate for that interval $\displaystyle x + 6 = 6 \implies x = 0 \implies x < 2.$

But that's a contradiction. x cannot be greater than or equal to 2 and also less than 2. Nonsense.

$\displaystyle Obviously\ x \ge 0 \implies x \ge 0.$ But we already know know that a valid answer will require x < 2.

So $\displaystyle f(x) = |x - 8| + |x - 2| + |x|\ and\ 0 \le x < 2 \implies f(x) = -(x - 8) - (x - 2) + x = - x + 8 - x + 2 + x = -x + 10.$

This function applies only in the interval $\displaystyle [0, 2).$

OK Now let's deal with the equation appropriate for that interval $\displaystyle -x + 10 = 6 \implies x = 4 \implies x > 2.$

But that's a contradiction. x cannot be less than 2 and also greater than 2. Nonsense.

$\displaystyle Obviously\ x < 0 \implies x < 0\ and\ x - 2 < 0\ and\ x - 8 < 0.$

So $\displaystyle f(x) = |x - 8| + |x - 2| + |x|\ and\ x < 0 \implies f(x) = -(x - 8) - (x - 2) - x = - x + 8 - x + 2 - x = -3x + 10.$

This function applies only in the interval $\displaystyle (- \infty, 0).$

OK Now let's deal with the equation appropriate for that interval $\displaystyle -3x + 10 = 6 \implies -3x = -4 \implies x =\dfrac{4}{3} > 0.$

But that's a contradiction. x cannot be less than 0 and also greater than 0. Nonsense.

So this is not a valid equation. But that is really beside the point. To find the intervals, you are looking for where a term expressed in absolute value terms is equal to itself, not its additive inverse. Because this non-equation had three terms in absolute value terms, we ended up with four intervals: one where every absolute value term equaled its additive inverse, an interval where two absolute value terms equaled their respective additive inverses, an interval where one absolute value term equaled its additive inverse, and an interval where no absolute value term equaled its additive inverse.