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    For each, Find the co-ordinates of any stationary points and state weather it is a maximum, minimum or point of inflexion.

    a) y=(2x-3)^6 For this one i've got the right co-ordinates but it's meant to be a mnimun point but i keep getting points of inflexion

    b)y=x+(1/x)

    c)y=((x+3)^4)-4x

    d)y=(0.5x-1)^3

    Help with any of these would really really be appreciated.
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    Have you tried differentiating it again and substituting in your x value into f"(x)?

    If you have and it comes out as equal to zero, it doesn't necessarily make it a point of inflection I don't think :/
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    Yea did that and f"x = 0. i'll investigate a bit more and see what happens Thanks. Any idea about the others?
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    General method is:

    1) Find derivative and set to 0 to find stationary point/max/min
    So for a) it would be

     \frac{d}{dx}[(2x-3)^6] = 0

    implies

     12(2x-3)^5 = 0

    so  x = \frac{3}{2}  is min/max/stationary point

    Now need to find second order derivative:

     \frac{d^2}{dx^2}[(2x-3)^6] = 120(2x-3)^4  and put in  x = \frac{3}{2} In this case that gives 0, so you gotta carry on with the differentiating. We see that we're gonna get zero until the 6th order derivative, i.e.

    EDIT - made a mistake with the differentiating sorry!

     \frac{d^6}{dx^6}[(2x-3)^6] = 46080

    This is a positive value, meaning the point must be a minimum.

    For the others, is it the method or the differentiating you're having trouble with?
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    Okay thanks but i have no idea what this 6th order derivative thing is

    For the others i'm not really sure.. i can't get started with b) cause i'm pretty sure if you defferentiate it d/dx=0 ?
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    Ok no worries about that 6th order derivative thing!

    For b)

    \frac{d}{dx}[x + \frac{1}{x}] = 1 - \frac{1}{x^2}

    Writing it like this might make that more obvious

    \frac{d}{dx}[x + \frac{1}{x}] = \frac{d}{dx}[x + x^{-1}] =  \frac{d}{dx}[x] +  \frac{d}{dx}[x^{-1}] = 1 -x^{-2}

    Is that a bit clearer?
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    Yea that helps a lot, thanks.
    Whenever you equal that to zero then do you get x=1 and x=-1?
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    Whenever you equal that to zero then do you get x=1 and x=-1?
    Yeah thats right, so you got two points to determine the nature of in this question!
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    Whoo Can you help with c) ? I keep getting the wrong co-ordinates?
    I'm definately going to fail in life lol...
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    Man these are just all about practice - when I did my A levels I did like a million of them, then you screw up enough times to learn from the mistakes!

    So c) y=((x+3)^4)-4x

    Always remember that expressions can be split up when differentiating like this

     \frac{d}{dx}[(x+3)^4 - 4x] = \frac{d}{dx}[(x+3)^4] +  \frac{d}{dx}[-4x]

    So we get the result as

      4(x+3)^3 - 4 = 0

    Giving

      (x+3)^3 = 1

    So   (x+3) = 1

    So   x = -2

    Putting this back into the original expression gives

     y = 1^4 - (-2 \times 4) = 1 - (-8) = 9
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    Ahh okay. i understand. Thank you very much, you're a great help. Still not sure about a but i'll ask my teacher to explain 2moro
 
 
 
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Updated: January 19, 2010
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