# Thread: Supremum and infimum of an equation with absolute values

1. ## Supremum and infimum of an equation with absolute values

I get sup and inf in basic functions, and even using things between absolutes | | . But when the variable you solving for cancels out like this one, I don't know how to work it out:

|x+1| + |2-x| = 3

Normally you would say then:

-3 < x+1 + 2-x < 3 right?

But the problem with this one specifically is that it cancels out. I have the answer in the back of the book (sub = 2 and inf = -1) which i'm guessing they ignored the 3s and took

the |x+1| = 0 so x = -1 and |2-x|=0 and so 2=x

But i don't know what the reasoning would be to do it like that, because whatever value you choose for x, even 10, it just cancels out so that you get 3 in the middle. I don't really get why they did this. I can copy the logic I mentioned here for future problems but I don't really get why or what is the logical way to work it out. Does the triangle inequality come into it?

Any explanations would be most appreciated

2. Originally Posted by iva
I get sup and inf in basic functions, and even using things between absolutes | | . But when the variable you solving for cancels out like this one, I don't know how to work it out:

|x+1| + |2-x| = 3

Normally you would say then:

-3 < x+1 + 2-x < 3 right?

But the problem with this one specifically is that it cancels out. I have the answer in the back of the book (sub = 2 and inf = -1) which i'm guessing they ignored the 3s and took

the |x+1| = 0 so x = -1 and |2-x|=0 and so 2=x

But i don't know what the reasoning would be to do it like that, because whatever value you choose for x, even 10, it just cancels out so that you get 3 in the middle. I don't really get why they did this. I can copy the logic I mentioned here for future problems but I don't really get why or what is the logical way to work it out. Does the triangle inequality come into it?

Any explanations would be most appreciated
So remember that the absolute value function is given by

$\displaystyle |x| =\begin{cases} -x, \text{ if } x < 0 \\ x, \text{ if } x \ge 0\end{cases}$

When $\displaystyle x < -1$ we get that $\displaystyle |x+1|=-(x+1)$ and $\displaystyle |2-x|=(2-x)$ so this gives one equation.

$\displaystyle -(x+1)+(2-x)=3 \iff -2x=2 \implies x=-1$ Now we have to check to make sure the solution works.

Next try the region $\displaystyle -1 \le x \le 2$ Here

$\displaystyle |x+1|=(x+1)$ and $\displaystyle |2-x|=(2-x)$

$\displaystyle (x+1)+(2-x)=3 \iff 3=3$ This gives an identity So every value in this region works. (check it)

Finally we have the last region.

$\displaystyle x \ge 2$ Here $\displaystyle |x+1|=(x+1)$ and $\displaystyle |2-x|=-(2-x)=x-2$

This gives the equation

$\displaystyle (x+1)+(x-2)=3 \iff 2x=4 \iff x=2$

I hope this clears things up a bit.

P.S don't forget to check your solutions as you may get solutions that fall outside the original domain of the question.

3. WOW, this was a super explanation, thanks so much!

4. One minor note:

When you use TheEmptySet's method (which is a very good method), it is not actually necessary to check if your solutions work. The solution will always work if the answer falls within the restricted domain, and it won't work if it does not.

On an exam you should of course check anyway just to make sure that you haven't made a computational error. But extraneous solutions can't occur when using this method.

So technically, in the first case x=-1 should be rejected since it doesn't fall in the interval. The fact that it is a solution shows up in the second case.

This is a minor technical point, especially since you can choose where you put the equal sign arbitrarily when defining your cases.

(If this note confuses you, you can ignore it - I only added it to make TheEmptySet's explanation more mathematically precise)