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Math Help - Use of the Signum Function

  1. #1
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    Smile Use of the Signum Function

    This is a tough one... And I really need someones help


    Graph the following equation:

    x(y-1/2 sgn(x)- 1/2 sgn (x-1)) (x-1) (y-2(x-3)^2) √(2y-y^2 )= 0
    Last edited by FutureMD2be; April 15th 2009 at 04:29 AM.
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by FutureMD2be View Post
    This is a tough one... And I really need someones help
    Graph the following equation:
    x(y-\frac{1}{2} sgn(x)- \frac{1}{2} sgn (x-1)) (x-1) (y-2(x-3)^2) \sqrt{(2y-y^2 )}= 0
    Have you considered the cases:

    x = 0 , x = 1, and the intervals (-infinity, 0), (0, 1), and (1, infinity)?

    at x = 0, for example, you get:
    -1 (y- 18) \sqrt{2y-y^2} = 0

    I tried for the first open interval but get a pretty messy implicit relationship. Are you asked to sketch the graph by hand?

    Good luck!
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  3. #3
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    Hello, FutureMD2be!

    Graph the following equation:

    . . \underbrace{x}_{[1]}\cdot\underbrace{\left[y-\tfrac{1}{2}\text{sgn}(x)- \tfrac{1}{2}\text{sgn}(x-1)\right]}_{[2]}  \cdot \underbrace{(x-1)}_{[3]} \cdot \underbrace{\left[y-2(x-3)^2\right]}_{[4]}\cdot \underbrace{\sqrt{2y-y^2}}_{[5]} \;=\; 0
    The function is a product of five factors set equal to zero.

    We can consider each factor separately.


    [1]\;\,x = 0 . . . vertical line (the y-axis)


    [3]\;\,x-1\:=\:0 \quad\Rightarrow\quad x \:=\:1 . . . vertical line


    [4]\;\,y -2(x-3)^2\:=\:0 \quad\Rightarrow\quad y \:=\:2(x-3)^2 . . . parabola, opens upward, vertex (3,0)


    [5]\;\;\sqrt{2y-y^2} \:=\:0 \quad\Rightarrow\quad y(2-y)\:=\:0\quad\Rightarrow\quad y\:=\;0,\:y\:=\:2 . . . horizontal lines



    [2]\;\;y \:=\:\tfrac{1}{2}\text{sgn}(x) + \tfrac{1}{2}\text{sgn}(x-1)
    Recall the definition: . \text{sgn}(x) \;=\;\begin{Bmatrix}1 & x > 0 \\ 0 & x = 0 \\ \text{-}1 & x < 0 \end{Bmatrix}

    We must consider five cases . . .

    x > 1\!:\;\;y \:=\:\tfrac{1}{2}(1) + \tfrac{1}{2}(1) \:=\:1

    x = 1\!:\;\;y \:=\:\tfrac{1}{2}(1) + \tfrac{1}{2}(0) \:=\:\tfrac{1}{2}

    0 < x < 1\!:\;\;y \:=\:\tfrac{1}{2}(1) + \tfrac{1}{2}(\text{-}1) \:=\: 0

    x = 0\!:\;\;y \:=\:\tfrac{1}{2}(0) + \tfrac{1}{2}(\text{-}1) \:=\:-\tfrac{1}{2}

    x < 0\!:\;\;y \:=\:\tfrac{1}{2}(\text{-}1) + \tfrac{1}{2}(\text{-}1) \:=\:\text{-}1

    The graph looks like this:


    Code:
                    |
                   1+       o=======
                    |
                    +       *
                    |
          - - - - - o=======o - - - -
                    |       1
                    *
                    |
             =======o-1
                    |

    Now plot all five graphs on the same set of coordinate axes.

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