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Finding stationary points when derivative is in fraction form? watch

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    So i have derived the function 1/((x+5)(x-3)) using the reciprocal rule into 2x+2/((x+5)^2(x-3)^2) and am now very stuck on finding the stationary points from this.. any help would be much appreciated
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    (Original post by adam9937999)
    So i have derived the function 1/((x+5)(x-3)) using the reciprocal rule into 2x+2/((x+5)^2(x-3)^2) and am now very stuck on finding the stationary points from this.. any help would be much appreciated
    For a fixed non-zero constant or expression a, for what values of b is  \frac{a}{b} = 0?
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    Answer: no values of b.

    So it doesn't matter what b is when the numerator is a non-zero constant or expression.. so when for what values of the numerator (a) will you find that dy/dx = 0?
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    (Original post by SeanFM)
    For a fixed non-zero constant a, for what values of b is  \frac{a}{b} = 0?
    Spoiler:
    Show
    Answer: no values of b.

    So it doesn't matter what b is when the numerator is a non-zero constant.. so when for what values of the numerator (a) will you find that dy/dx = 0?
    Aaah yes! Thank you every so much fellow!
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    I've moved this into the maths forum as there are a lot more active mathematicians here, so hopefully you'll get better answers.
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    (Original post by adam9937999)
    Aaah yes! Thank you every so much fellow!
    No worries so what did you get for your value of x?
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    (Original post by SeanFM)
    No worries so what did you get for your value of x?
    -1! And that is the only stationary point as far as i am aware, so happy days.
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    (Original post by adam9937999)
    -1! And that is the only stationary point as far as i am aware, so happy days.
    Correct well done. :borat:
 
 
 
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Updated: October 27, 2016
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