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Math Help - Differentiation Problem

  1. #1
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    Differentiation Problem

    Hi

    I have a question on a paper that I have to solve though I have become unsure as to what to do.

    The problem is as follows

    Given that y=16x+x^-1, find the two values of x for which dy/dx = 0.

    When I differentiate I get 16 + x^-2 or 16 + 1/x^2

    to remove the fraction I multiply both sides by x^2 giving

    16x^2 + 1 = 0

    so dy/dx = 0 -> 16x^2 + 1 = 0.

    I thought to obtain the two values of x you have to factorise the quadratic, and I am not sure how to do this given the form of the derived equation.

    Any help with this would be gratefully appreciated.

    Thank You

    David
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  2. #2
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    Quote Originally Posted by DavidRUK View Post
    Hi

    I have a question on a paper that I have to solve though I have become unsure as to what to do.

    The problem is as follows

    Given that y=16x+x^-1, find the two values of x for which dy/dx = 0.

    When I differentiate I get 16 + x^-2 or 16 + 1/x^2 Mr F says: This is wrong. It should be 16 - x^-2 or 16 - 1/x^2.

    to remove the fraction I multiply both sides by x^2 giving

    16x^2 + 1 = 0

    so dy/dx = 0 -> 16x^2 + 1 = 0.

    I thought to obtain the two values of x you have to factorise the quadratic, and I am not sure how to do this given the form of the derived equation.

    Any help with this would be gratefully appreciated.

    Thank You

    David
    ..
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  3. #3
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    Quote Originally Posted by DavidRUK View Post

    Given that y=16x+x^-1, find the two values of x for which dy/dx = 0.

    When I differentiate I get 16 + x^-2 or 16 + 1/x^2

    This is not completly correct

     <br />
y=16x+x^{-1} \implies \frac{dy}{dx}= 16-x^{-2}<br />

    Now finish it off as you were.
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  4. #4
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    Other than that error in differentiating it looks like your main doubt was:
    I thought to obtain the two values of x you have to factorise the quadratic, and I am not sure how to do this given the form of the derived equation.
    Since there is no x^1 term in the quadratic, you don't need to factorise anything.

    You have \frac{dy}{dx}= 16- \frac{1}{x^2} = 0

    I wont tell you the full solution unless you want me too; I'm sure you can see what to do. You need to do some simple algebra and then remember to think about whether you need +.
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  5. #5
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    Sorry about that. I got so caught up in the problem i forgot to bring the -1 down to change the sign.

    I am still not sure how to find two values for x though.
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  6. #6
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    Quote Originally Posted by DavidRUK View Post
    Sorry about that. I got so caught up in the problem i forgot to bring the -1 down to change the sign.

    I am still not sure how to find two values for x though.
    I don't see how you can be studying calculus and not know how to solve an equation like 16- \frac{1}{x^2} = 0. That is very worrying and suggests life is going to be very tough for you unless you get remedial help in pre-calculus ....

    The most obvious approach is

    \frac{1}{x^2} = 16

    \Rightarrow x^2 = \frac{1}{16}

    etc.
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  7. #7
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    You have no idea...
    Try doing anything with a screaming baby and a sick wife..
    Also I have been up since 4:30AM.

    Not good.

    Thanks though.. I will go and practice.
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  8. #8
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    Quote Originally Posted by DavidRUK View Post
    You have no idea...
    Try doing anything with a screaming baby and a sick wife..
    Also I have been up since 4:30AM.

    Not good.

    Thanks though.. I will go and practice.
    I'm sorry to hear that.

    (Note that 99 times out of 100 there is no "screaming baby and a sick wife" ....)
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