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Math Help - Finding coordinates of stationary points???

  1. #1
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    Finding coordinates of stationary points???

    My text book proposes.....

    Find the coordinates of stationary points on a curve given by

    y=x^3 - 6x + 5x +12

    I don't think i even cover trigonometry this term but it is puzzling me.

    Also, what would be the response to determining their nature, does this mean, i.e. positive/negative?

    ?????????????????????????????????????????????????? ?????????????????????????
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bobchiba View Post
    My text book proposes.....

    Find the coordinates of stationary points on a curve given by

    y=x^3 - 6x + 5x +12

    I don't think i even cover trigonometry this term but it is puzzling me.

    Also, what would be the response to determining their nature, does this mean, i.e. positive/negative?

    ?????????????????????????????????????????????????? ?????????????????????????
    stationary points are points on a graph where the gradient is zero. the derivative gives the formula for the gradient of a curve at any given point, so you can find the stationary points by finding the derivative and setting it = 0.

    the nature of stationary points is refering to whether they are maximums or minimums. a maximum point is the point on the peak of a "hill" for example, the vertex of the -x^2 graph. a minimum point is the lowest point of a "valley" for example the vertex of the x^2 graph. they are a few ways of finding the nature of stationary points. the easiest way is to compute the second derivative and plug in the values you got from the first derivative. if the result is negative you have a maximum, if it is positive you have a minimum, if it is 0, the test is inconclusive (it could be a maximum, minimum or (usually) an inflection point.) Now let's do the problem.

    i think the question should be y=x^3 - 6x^2 + 5x +12 not y=x^3 - 6x + 5x +12

    y=x^3 - 6x^2 + 5x +12
    => y' = 3x^2 - 12x + 5
    set y' =0
    => 3x^2 - 12x + 5 = 0
    so by the quadratic formula: x = [12 +/- sqrt(144 - 4(3)(5))]/2(3)
    => x = [12 +/- sqrt(84)]/6
    so x = 2 + sqrt(84)/6 and x = 2 - sqrt(84)/6

    to find the coordinates, plug in these values for x into the ORIGINAL equation. the corresponding y-values gives the second coordinate of the stationary point.

    so the stationary points are:

    (2 + sqrt(84)/6, f(2 + sqrt(84)/6)) and (2 - sqrt(84)/6 , f(2 - sqrt(84)/6))

    again i'm in a rush, i cant work these out for you.

    now for the nature of the stationary points:
    since y' = 3x^2 - 12x + 5
    => y'' = 6x - 12
    this gives a positive for the first point and a negative for the second point.

    so (2 + sqrt(84)/6, f(2 + sqrt(84)/6)) is a minimum and (2 - sqrt(84)/6 , f(2 - sqrt(84)/6)) is a maximum
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