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    Can someone help me find the Stationary points of

    1/(4x-x^2-3)

    I struggled through it and got a max turning point at (8,-1/35) which seems very very wrong lol

    any help me will be greatly greatly appreciated

    my working:

    y= 1/(4x-x^2-3)

    y= 1/4x - 1/x^2 - 1/3

    y= x^-1/4 - x^-2 -1/3

    dy/dx = -1/4x^2 +2/x^3

    i then tried to solve it and got x=8

    please help
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    (Original post by Michelle678)
    Can someone help me find the Stationary points of

    1/(4x-x^2-3)

    I struggled through it and got a max turning point at (8,-1/35) which seems very very wrong lol

    any help me will be greatly greatly appreciated

    my working:

    y= 1/(4x-x^2-3)

    y= 1/4x - 1/x^2 - 1/3


    y= x^-1/4 - x^-2 -1/3

    dy/dx = -1/4x^2 +2/x^3

    i then tried to solve it and got x=8

    please help
    The bit from the first bolded step to the next is wrong.
    \frac{1}{4x-x^2-3} \not= \frac{1}{4x}-\frac{1}{x^2} - \frac{1}{3}
    Personally, I think it's much easier to complete the square on the denominator. Find the smallest value this completed square form can take (as this will maximise y) and identify which value of x causes this.
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    (Original post by Michelle678)
    Can someone help me find the Stationary points of

    1/(4x-x^2-3)

    I struggled through it and got a max turning point at (8,-1/35) which seems very very wrong lol

    any help me will be greatly greatly appreciated

    my working:

    y= 1/(4x-x^2-3)

    y= 1/4x - 1/x^2 - 1/3

    y= x^-1/4 - x^-2 -1/3

    dy/dx = -1/4x^2 +2/x^3

    i then tried to solve it and got x=8

    please help
    Are you doing C3? Because you could rewrite as  (4x-x^2-3)^{-1} and differentiate using the chain rule, make it equal to zero and solve for "x".

    If your doing C1/C2, then do what Farhan said, complete the square.
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    (Original post by Farhan.Hanif93)
    The bit from the first bolded step to the next is wrong.
    \frac{1}{4x-x^2-3} \not= \frac{1}{4x}-\frac{1}{x^2} - \frac{1}{3}
    Personally, I think it's much easier to complete the square on the denominator. Find the smallest value this completed square form can take (as this will maximise y) and identify which value of x causes this.
    So there is a min tp at (2,-1)?
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    (Original post by Michelle678)
    So there is a min tp at (2,-1)?
    Almost. Your y value is not correct.

    PS: Brownie points for picking up the fact that your original answer "seems very very wrong"; many students would have been content with getting an answer without considering whether it made sense or not.
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    I did the chain rule and got a min tp at (2,1)
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    (Original post by Michelle678)
    I did the chain rule and got a min tp at (2,1)
    Sorted.
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     -(x^2 - 4x + 3)

     -((x-2)^2 - 1)

    So max value when "x = 2":

     -((2-2)^2 - 1)

     -(-1)

     1

    So it would be (2, 1)
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    Sorry, I dozed off... at least things have been sorted out.
 
 
 
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