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Thread: Derivative and graphing!!!

  1. #1
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    Red face Derivative and graphing!!!



    The graphs of the function (given in blue) and (given in red) are plotted above. Find each of the following:
    = ?
    = ?

    2. If

    find the values of and that make differentiable everywhere. m= ? and b=?
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  2. #2
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    Hello, Kayla_N!



    The graphs of the function f (in blue) and g (in red) are plotted above.

    Find the following: . (1)\;(fg)'(1)\qquad (2)\;\left(\frac{f}{g}\right)\!'(1)
    We have piecewise functions . . .

    f(x) \;=\;\bigg\{ \begin{array}{cccc}\frac{3}{2}x & & 0 < x < 3 \\ \frac{9}{2}-x & & x > 3 \end{array}

    g(x) \;=\;\bigg\{\begin{array}{cccc}4-x & & x < 2 \\ \frac{1}{2}x + 2 & & x > 2 \end{array}


    If x = 1, we have: . \begin{array}{ccc}f(x) \;=\; \frac{3}{2}x \\ g(x) \;=\; 4 - x \end{array}



    (1)\;\;(fg) \;=\;\tfrac{3}{2}x(4 - x) \;=\;6x - \tfrac{3}{2}x^2

    Then:. . (fg)' \;=\;6 - 3x

    Therefore: . (fg)'(1) \;=\;6-3(1) \;=\;\boxed{3}



    (2)\;\;\left(\frac{f}{g}\right) \;=\;\frac{\frac{3}{2}x}{4-x} \;=\;\frac{3}{2}\cdot\frac{x}{4-x}

    Then: . \left(\frac{f}{g}\right)\!' \;=\;\frac{3}{2}\cdot\frac{(4-x)\!\cdot\!1 - x(-1)}{(4-x)^2} \;=\;\frac{6}{(4-x)^2}

    Therefore: . \left(\frac{f}{g}\right)\!'(1) \;=\;\frac{6}{(4-1)^2} \;=\;\frac{6}{9} \;=\;\boxed{\frac{2}{3}}

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  3. #3
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    Thanks for explaining steps by steps. Thats really helped and i understand it now.

    PS> can someone please help me on number 2?
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