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Math Help - find stationary points

  1. #1
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    find stationary points

    two problems i have differentiated. need to find stationary point and state if max or min value.

    (a)y= (3x+1)^4

    dy/dx= 12(3x+1)^3

    this is where im stuck. do i make dy/dx= 0?


    (b)x=\frac{1}{4t^2+3}

    dx/dt = -8t(4t^2+3)

    many thanks
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    Quote Originally Posted by decoy808 View Post

    (a)y= (3x+1)^4

    dy/dx= 12(3x+1)^3

    this is where im stuck. do i make dy/dx= 0?

    Yep...
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  3. #3
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    Quote Originally Posted by decoy808 View Post


    (b)x=\frac{1}{4t^2+3}

    dx/dt = -8t(4t^2+3)
    This is not correct
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  4. #4
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    Quote Originally Posted by decoy808 View Post
    two problems i have differentiated. need to find stationary point and state if max or min value.

    (a)y= (3x+1)^4

    dy/dx= 12(3x+1)^3 use the chain rule for the power of a function

    this is where im stuck. do i make dy/dx= 0?


    (b)x=\frac{1}{4t^2+3}

    dx/dt = -8t(4t^2+3)

    many thanks

    Hi decoy808,

    you've erred in your derivative calculation, though it would not matter in calculating the stationary point in this case!

    \frac{d}{dx}\left(3x+1\right)^4=\frac{d}{dx}u^4=\f  rac{du}{dx}\frac{d}{du}u^4=4u^3(3)=3\left(3x+1\rig  ht)^3

    At a stationary point, this is zero, so 3x+1=0


    x=\frac{1}{4t^2+3}=\left(4t^2+3\right)^{-1}

    \frac{dx}{dt}=\frac{dx}{du}\frac{du}{dt}=-\left(4t^2+3\right)^{-2}(8t)=\frac{-8t}{\left(4t^2+3\right)^2}

    This is zero when t=0

    You would have gotten the correct answers in spite of your errors! coincidentally
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    so my value part (a) x=-1/3 ?
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    Yep..

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