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Math Help - Calculus Graphing

  1. #1
    Member
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    Calculus Graphing

    I am having trouble with this problem and asking if someone can show me a this step by step so i can do good on my test on Friday. THANKS


    Sketch the graph of a continuous function f that satisfies all of the stated conditions.

    f(0)=4; f(-3)=f(3)=0; f^l(-3)=0; f^l(0)is undefined;
    f^l(x) >0 if -3 <x<0; f^l(x) <0 if x <-3 or x >0;
    f^ll(x) > 0 if x <0 or <x <2; f^ll(x)<0 if x >2
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  2. #2
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    Hello, uniquereason81!

    Sketch the graph of a continuous function f(x) such that:

    . . f(0)=4,\;\; f(\text{-}3)=f(3)=0\;\;{\color{blue}[1]}

    . . f'(\text{-}3)=0,\;\;f'(0)\text{ is un{d}efined }\;{\color{blue}[2]}

    . . f'(x) \:=\:\bigg\{\begin{array}{cc} > 0 & \text{if }-3 <x<0 \\ <0 & \text{if }x <-3\text{ or }x >0 \end{array}\;\;{\color{blue}[3]}

    . . f''(x) \:=\:\bigg\{\begin{array}{cc}> 0 & \text{if }x <0\text{ or }0<x <2 \\ <0 & \text{if }x >2 \end{array}\;\;{\color{blue}[4]}

    From [1], we have three points: . 0,4),\3,0)" alt="(\text{-}3,0),\0,4),\3,0)" />

    From [2], there is a horizontal tangent at (-3,0).
    . . And the graph "goes vertical" at x = 0.
    Code:
                    |
                    o
                    |
                    |
                    |
       --=o=--------+---------o--
         -3         |         3
                    |

    From [3], f(x) is "uphill" for -3 < x < 0
    . . and "downhill" everywhere else.
    Code:
                    |
                    *
    \             / | \
      \        /    |    \
        \   /       |       \
    ------*---------+---------*--
         -3         |         3 \
                    |             \

    From [4], f(x) is concave down for x > 2
    . . and concave up everywhere else.
    Code:
                    |
     *              *
                   *|*
      *          *  |    *
       *      *     |    :  *
    ------*---------+----+----*--
         -3         |    2    3
                    |          *

    There!


    There is a "cusp" at (0,4).
    .
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