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Math Help - Sketching of modulus function

  1. #1
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    Sketching of modulus function

    Hi

    Facing a problem with sketching this modulus graph.

    To get the equations of the lines I have to draw, I have labeled them as i, ii and iii.

    However, after I checked it with my graphing calculator, I realised that the graph for part ii has a positive gradient. Circled the error I faced in my attachment in red.

    Only one example was given in my text to help me solve this question. The tutorials and videos on sketching modulus graphs online only showed how to sketch one modulus function.

    Please help me to check my approach and tell me where I went wrong.

    Thanks in advance.
    Attached Thumbnails Attached Thumbnails Sketching of modulus function-problem_0001.jpg  
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    For -1<x<1 we have:

    2x+2>0,\;x-1<0

    so, for -1<x<1:

    y=|2x+2|-3|x-1|=2x+2-3(1-x)=5x-1

    Fernando Revilla
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  3. #3
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    Hello, dd86!

    Fernando found your error . . .


    (ii) Consider . -1 < x < 1

    . . . . . . . . y \:=\:|\underbrace{2x+2}_{\text{postive}}| - 3|\underbrace{x-1}_{\text{negative}}|

    becomes:. . y \:=\:2x+2 -3(1-x)

    . . . . . . . . y \:=\:2x + 2 - 3 + 3x

    . . . . . . . . y \:=\:5x-1


    The graph looks like this:

    Code:
                    |
                    | (1,4) 
                    |   *
                    |     *
                    |  *    *
                    |         *
                    | *         *
      - - - - - - - + - - - - - - * - -
                    |*              *
                    |
                    *
                    |
                   *
                    |
        *         * |
          *         |
            *    *  |
              *     |
                *   |
             (-1,-6)|
                    |
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  4. #4
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    Yep, realised my mistake. Now I've understood the whole thing clearly.

    This is what happens when you are spoon fed info without learning how to actually apply it -_-"
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  5. #5
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    2x+2<0\Rightarrow\ 2x<-2\Rightarrow\ x<-1,\;\;|2x+2|=-2x-2

    x<-1\Rightarrow\ |x-1|=1-x

    For these x, the graph is

    -2x-2-3(1-x)=x-5


    x-1<0\Rightarrow\ x<1,\;\;|x-1|=1-x

    -1<x<1\Rightarrow\ 2x+2>0\Rightarrow\ |2x+2|=2x+2

    For these x, the graph is

    2x+2-3(1-x)=5x-1

    For values of x beyond 1,

    x>1\Rightarrow\ 2x+2>0,\;\;x-1>0\Rightarrow\ 2x+2-3(x-1)=-x+5

    f(-1)=-3|-2|=-6,\;\;f(1)=4

    The graph looks like this
    Attached Thumbnails Attached Thumbnails Sketching of modulus function-modulus-graph.jpg  
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