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Math Help - Find stationery point where x to -ve power. How?

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    Find stationery point where x to -ve power. How?

    Hi

    To find stationery points I always differentiate the function which in my previouas work has always resulted in a quadratic equation so set to zero and factorise or find roots.

    But what to do if gradient is eg x^-2 ???

    the function is 4x^2 + 1/x

    dy/dx = 8x - x^-2

    8x - x^-2 = 0 ??? How do I solve this?

    Angus
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  2. #2
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    Quote Originally Posted by angypangy View Post
    To find stationery points I always differentiate the function which in my previouas work has always resulted in a quadratic equation so set to zero and factorise or find roots.
    But what to do if gradient is eg x^-2 ???
    the function is 4x^2 + 1/x
    dy/dx = 8x - x^-2
    8x - x^-2 = 0 ??? How do I solve this?
    8x-x^{-2}=\frac{8x^3-1}{x^2}.
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    Sorry I am probably being a bit thick but any more hints? I am not understanding your point?
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  4. #4
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    Or: 8x- x^{-2}= 0 is the same as 8x= x^{-2} so, multiplying both sides by x^2, 8x^3= 1. That is the same as x^3= \frac{1}{8}
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  5. #5
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    Are you saying that you cannot solve \frac{8x^3-1}{x^2}=0~?.
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  6. #6
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    Sorry I was confused with something I read about the quotient rule which I have not yet covered and I thought I needed to use somehow. Yes understand now.
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