Thread: Absolute Value problem

First of all, does anyone know where I can find an explanation of these types of questions on the internet? I've never solved absolute value questions before so I don't know exactly what to do. And my book just solves the question but doesn't explain it at all...

I understand this much:





And from here I don't understand - the book starts answering the question by the boundaries:

(i) for

the inequality and this inequality produces

So I won't continue the rest of the answer. If I understand this part I assume I'll understand the rest. How did they get (i) as the boundary? Where did that inequality come from??

Well, that's how I do it:





Expanding and simplifying,



Now, since x^2 - 2x + 4 < 0 has no real solutions,

x(x-4) < 0

The critical values are x = 0 and x = 4.

I do my table, and get as answer: 0 < x < 4.

Hello, jayshizwiz!

How about a graphic solution?

The graph of
. . is an up-opening parabola with -intercepts 1 and 2.

Its graph looks like this:

Code:

        |
       *|                      *
        |
        |
        *                     *
        | 
        |*                   *
        | *                 *
    - - + - * - - - - - - * - - - -
        |   1  *       *  2
        |          *
        |

The graph of looks like this:

Code:

        |
       o|                      o
        |

        |
        o                     o
        | 
        |o         o         o
        | o   o         o   o
  - - - + - o - - - - - - o - - - -
        |   1             2
        |

Everything below the -axis is reflected upward.



The graph of looks like this:

Code:

            |
            |         *
            |       *
            |     *
            |   *
           2| *
            *
          * |
        *   |
      *     |
- - * - - - + - - - - - -
  * -2      |
*           |
            |

The graph of looks like this:

Code:

            |
            |         ∆
            |       ∆
            |     ∆
            |   ∆
            | ∆
            ∆
          ∆ |
∆       ∆   |
  ∆   ∆     |
- - ∆ - - - + - - - - - -
    -2      |

Sketch the two graphs on a set of coordinate axes.

Code:

            |
         o  |                         ♥(4,6)
            |                       ∆
            |                     ∆
            |                   ∆
            |                 ∆
            |               ∆
          o |             ∆         o
            |           ∆
            |         ∆
            |       ∆
           o|     ∆                o
            |   ∆
       (0,2)| ∆
            ♥                     o
          ∆ | 
∆       ∆   |o         o         o
  ∆   ∆     | o   o         o   o
- - ∆ . . - + - o - - - - - - o - - - -
   -2       |   1             2
            |

When is the "parabola" below the "line"?

Answer: .

Thanks guys!

wow, Soroban your graphs are awesome. And I like how you put little hearts at (4, 6) and (0,2)!

Just to make sure, is 0<x<4 the only answer you guys got??

Originally Posted by jayshizwiz

Thanks guys!

wow, Soroban your graphs are awesome. And I like how you put little hearts at (4, 6) and (0,2)!

Just to make sure, is 0<x<4 the only answer you guys got??

Well, is there anywhere else where the "parabola" is less than the "line"?

Well, you can see the graph now. The graph of |x+2| is above that of |x^2 - 3x + 2| only within that range.

When you go towards the left, the parabolic part of |x^2 - 3x + 2| goes up 'faster' than the graph of |x+2|, hence, |x+2| will never be above the parabola again.

The same thing occurs on the right.