Absolute Value Function

Let's say that we have two numbers, $A$ and $B$ , that are drawn on a number line. What is the distance between them? Or in other words, how far are they from each other?

Let's take a look at an example, where $A = 3$ and $B = 7$ .

We could count. By doing this, we get the answer $4$ . But when we have very large numbers, such as $A = 123893292$ and $B = 5$ , then drawing them on a number line and counting the distance between them becomes easily tiresome. We might observe the following: Number $B$ has traveled $7$ numbers, where number $A$ has traveled $3$ numbers. How many numbers more must number $A$ travel in order to reach number $B$ ? That's Basic Math! Number $A$ must travel $7 - 3 = 4$ numbers more until it reaches number $B$ . But, the amount of numbers that number $A$ had to travel until it reached number $B$ is also the distance between the two numbers.

What can we conclude from this? The distance $d$ between two numbers, $A$ and $B$ , is the same as the difference between our two numbers: $d = B - A$ .

Right? No, not exactly. We knew in our calculation that number $B$ was further away than number $A$ . So we calculated $d = B - A$ and not $d = A - B$ which would imply that number $A$ is further away than number $B$ . As a general formula we could say then:

If number $B$ is further away than number $A$ , then the distance $d$ between these two numbers is $d = B - A$ .

But what if we actually calculated $A - B$ instead of $B - A$ ? Then our answer would be $A - B = 3 - 7 = (-4)$ . This means number $B$ has to travel $-4$ numbers ( $4$ numbers backwards) until it reached number $A$ .

This also somehow shows the distance between the two numbers. What we don't like about it, is the minus sign. Wouldn't it be dreamy if we had something that gives us the positive version of every number we gave it? We indeed have something like that, the so-called absolute value (function) of a number $x$ , written as $abs(x)$ or $|x|$ .

So, we can say that the distance $d$ between two numbers $A$ and $B$ is $d = |A - B|,$ without worrying about which of these numbers is further away.

You can actually prove that $A - B = -(B - A)$ .

So what does an equation like the following mean? $3 = | x - 2 |$ $| x - 2 |$ is the distance between $x$ and $2$ . So, $3 = | x - 2 |$ means that we want to know all values of $x,$ such that the distance between $x$ and $2$ is exactly $3$ . And only the numbers $x = 5$ and $x = (-1)$ have the distance $3$ from the number $2$ .

In a similar way, we get for $2 = |x| = | x - 0 |$ that only the numbers $x = 2$ and $x = (-2)$ have the distance $2$ from the number $0$ .

So, the actually problem was to solve the following equation

$2 = |x - 2| + |x - 4|$

$|x - 2|$ is the distance between $2$ and $x$ .

$|x - 4|$ is the distance between $4$ and $x$ .

Their sum has to be $2$ .

Firstly, let's think about what the distance between $2$ and $4$ is. It's obviously $d = |4 - 2| = 2$ . And now, let's draw a number $C$ on the the number line between $2$ and $4$ .

The black line you see in the picture, is the distance between number $C$ and $2$ .

The red line you see in the picture, is the distance between number $C$ and $4$ .

Wait a second. As we just just saw, the distance between $2$ and $4$ is $2$ , so the "red line + black line" is always $2$ as well! It doesn't matter how big the red line and black line are each, but their sum is always $2$ .

With this we have found numbers that satisfy the condition

$2 = |x - 2| + |x - 4|,$

i.e. numbers whose distance to $2$ and distance to $4$ together is exactly $2$ . All numbers that are between $2$ and $4$ (inclusive)!

One final thought may be, if there are more numbers with this property. If we put our Number $C$ before $2$ , then distance between all numbers $C$ that are before $2$ and the number $4$ will be greater than $2$ , but never exactly $2$ . And if $|x - 4|$ is already greater than $2$ , then the expression $|x - 2| + |x - 4|$ will also always be greater than $2$ , because $|x - 2|$ is always $0$ or positive (non-negative).

Similar, if we put our number $C$ after $4$ , then the distance between $C$ and $2$ will always be greater than $2$ .

Think about the geometrically interpretation about the following inequalities. Which values of $x$ satisfy them and what is created when they are drawn on a number line?

$x < |1|$ $x < |0|$ $x < |a|$ $|x| < 5$ $|x| > 5$ $|x| < a$ $|x - 3| < 4$ $|x - 3| > 4$ $|x - 2| < a$ $|x - a| < \epsilon$ $|x - a| > \epsilon$

The problem gave us this equation:

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

Remember the triangle inequality theorem applied to absolute values:

|x + y| ≤ |x| + |y|

If we use this theorem to the RHS of the equation we get this:

|(x - 2) + (x - 4)| ≤ |x - 2| + |x - 4|

And by replacing the RHS by the LHS from the first equation we get:

|(x - 2) + (x - 4)| ≤ 2

|2x - 6| ≤ 2

Then if we use the following property of inequalities that involve absolute value:

|x| ≤ c ⇔ c ≤ x ≤ c

We have:

-2 ≤ 2x - 6 ≤ 2

4 ≤ 2x ≤ 8

2 ≤ x ≤ 4

Which gives us the interval [2,4] for x that satisfy the problem equation.

by drawing graph of function... easily ans is [2,4] . it takes 2-3 seconds to solve it if you know to draw this function's graph..

If $x \le 4$ then $2=|x-2|+|x-4|=x-2+x-4=2x-6 \implies 2=2x-6 \implies 2x-8=0$

$2x-8=0 \implies x=-\frac{-8}{2}=4$

If $2 \le x \le 4$ then $2=|x-2|+|x-4|=x-2+4-x=2$ so we canceled out.

If $x \le 2$ then $2=|x-2|+|x-4|=2-x+4-x=6-2x \implies 2=6-2x \implies 2x-4=0$

$2x-4=0 \implies x=-\frac{-4}{2}=2$

So the answer is $\boxed{\large{2,4}}$

Put value x=0, all but two solutions fall away.

That leaves with either "no solution" or [2,4].

Attempt [2,4], success. :P

Put x = 7 To get first point .Put x = 14 to get 2nd point then use sqrt ( ( Y2-Y1) ^2 + ( x2-x1)^2 )