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Math Help - Curve Sketching

  1. #1
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    Curve Sketching

    Did I do the question properly?

    Sketch a graph of a function f(x) that is differentiable and that satisfies the following conditions:
    • f'(x) > 0, when x < -3 and x > 1
    • f'(x) < 0, when -3 < x < 1
    • f'(-3) = 0 and f'(1) = 0
    • f'(-3) = 5 and f(1) = 1
    Legend:
    Purple is f(x)
    Green is f'(x)
    Attached Thumbnails Attached Thumbnails Curve Sketching-graph-2.png  
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  2. #2
    Moo
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    Hello,

    No, it's false... the derivative is not positive in your sketch...
    Actually, it's useless to draw f'(x). All that you have to remember is that if f'(x)>0, then f(x) will be increasing.

    If f'(x)=0, there is a horizontal tangent line (here, it's g and g1 - see attachment), at A and B, which are turning points.

    I've drawn a possible function, that satisfies the conditions

    I hope it's clear enough... (sorry for the black line... couldn't do better)
    Attached Thumbnails Attached Thumbnails Curve Sketching-derivative.png  
    Last edited by Moo; May 25th 2008 at 12:28 PM. Reason: corrections thanks to bobak :p
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  3. #3
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    Hello, Macleef!

    I believe you have a typo in the problem . . .


    Sketch a graph of a function f(x) that is differentiable
    and that satisfies the following conditions:

    {\color{blue}[1]}\;\;f'(x) > 0\text{, when }x < \text{-}3\text{ and }x > 1
    {\color{blue}[2]}\;\;f'(x) < 0\text{, when } \text{-}3 < x < 1
    {\color{blue}[3]}\;\;f'(\text{-}3) = 0\text{ and }f'(1) = 0
    {\color{blue}[4]}\;\;{\color{red}f(\text{-}3) = 5} \text{ and } f(1) = 1

    [1] says the graph is rising when x < -3\text{ and } x > 1
    [2] says the graph is falling when -3 < x < 1
    Code:
              :           :
            * :  *        :       *
          *   :     *     :     *
        *     :        *  :   *
      *       :           : *
      - - - - + - - - + - + - - - - -
             -3       0   1

    [3] says the slope is 0 at x = -3\text{ and }x = 1.
    Those are extreme points (or turning points or stationary points).
    Code:
            --o--         : 
              :           :
            * :  *        :     *
          *   :     *     :   *
        *     :        *  : *
              :         --o--
              :           :
      - - - - + - - - + - + - - - - -
             -3       0   1

    [4] tell us exactly where those extreme points are.
    We can now sketch the curve.

    Code:
           (-3,5)     |
              o       |          *
           *     *    |
         *         *  |         *
        *             |        *
                      *      *
       *              |   o
                      |    (1,1)
      - - - - + - - - + - + - - - - -
             -3       |   1
                      |
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