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Thread: Questions in derivatives

  1. December 27th 2010, 11:40 AM #1
    r-soy
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    Questions in derivatives

    Questions in derivatives-195609.jpg
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  2. December 27th 2010, 11:43 AM #2
    dwsmith
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    Once you have the slope, identify a point on the graph and use this formula y-y_1=m(x-x_1)
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  3. December 27th 2010, 11:50 AM #3
    Archie Meade
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    Quote Originally Posted by r-soy View Post
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    You have miscalculated.

    \displaystyle\ x^{-\frac{1}{2}}=\frac{1}{1.5}\Rightarrow\frac{1}{x^{\ frac{1}{2}}}=\frac{1}{\sqrt{x}}=\frac{1}{1.5}

    \displaystyle\sqrt{x}=1.5\Rightarrow\ x=1.5^2
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  4. December 27th 2010, 08:21 PM #4
    r-soy
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    Now X = 2.25

    when I try found (y) I don't get the answer I don't now why >>

    y = (2.25) - 3 (2.25)^1/2
    =-1.125 this answer not as on my book

    Also what is the range of values of curve's slop ?
    (-in,1] is this correct ?

    plese help me
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  5. December 27th 2010, 08:32 PM #5
    pickslides
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    Quote Originally Posted by r-soy View Post

    y = (2.25) - 3 (2.25)^1/2
    =-1.125 this answer not as on my book
    check again,

    y= 2.25-3\sqrt{2.25} = -2.25
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  6. December 27th 2010, 08:39 PM #6
    r-soy
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    but when I do here different answer y = (2.25) - 3 (2.25)^1/2
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  7. December 27th 2010, 08:54 PM #7
    pickslides
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    y= 2.25-3\sqrt{2.25}

    y= 2.25-3\times 1.5

    y= 2.25-4.5

    y= -2.25
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  8. December 27th 2010, 09:43 PM #8
    r-soy
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    I know my dear it will come in that equation but when I do here y = (2.25) - 3 (2.25)^1/2 I got diffrent answer .
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  9. December 28th 2010, 02:38 AM #9
    HallsofIvy
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    What do you mean "when I do here"? Are you saying you get something other than 2.25 for y? If so you are doing the arithmetic wrong!
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