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Math Help - Of the function

  1. #1
    ADY
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    Of the function

    Please could some one check this and help where required:

    The Function:

    f(x) = x^3 + 6x^2 -15x -24

    a) Find its stationary points

    So, f ' (x) = 3x^2 + 6(2x) - 15(1) = 3x^2 + 12x - 15 = 3(x^2 + 4x - 5) = 0

    The equation f ' (x) = 0 is equivilent to: 3(x^2 + 4x - 5) that is 3(x + 5)(x - 1)

    So the Stationary points are X = -5 and X = 1

    b) Classify the LH Stationary point with 1st Derivative Test

    To classify we take X = -5 and choose the test point XL = -6 then we have:

    f ' (XL) = f ' (-6) = 3(-6 + 5)(-6 - 1) = 21 > 0

    c) Classify the RH Stationary point with 2nd Derivative Test

    So,  f '' (x) = 3(2x) + 12 = 6x + 12 = 6(x + 2)

    The stationary point XR = 1, so the test point will be XR = 2
    <br />
f '' (2) = 6(2 + 2)= 24 > 0

    So f has a local minimum at X = 1

    d) Find the coordinates of each stationary point

    so,

    XL - f (-5) = -125 + 150 - (-75) - 24 = 76
    XR - f (1) = 1 + 6 - 15 - 24 = -32

    Hence the graph passes through the point (-5,76) & (1,-32)

    Thanks for your help

    A
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  2. #2
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    Quote Originally Posted by ADY View Post
    Please could some one check this and help where required:

    The Function:

    f(x) = x^3 + 6x^2 -15x -24

    a) Find its stationary points

    So, f ' (x) = 3x^2 + 6(2x) - 15(1) = 3x^2 + 12x - 15 = 3(x^2 + 4x - 5) = 0

    The equation f ' (x) = 0 is equivilent to: 3(x^2 + 4x - 5) that is 3(x + 5)(x - 1)
    You mean 3(x^2+4x- 5)= 0 and 3(x+5)(x-1)= 0, of course.

    So the Stationary points are X = -5 and X = 1

    b) Classify the LH Stationary point with 1st Derivative Test

    To classify we take X = -5 and choose the test point XL = -6 then we have:

    f ' (XL) = f ' (-6) = 3(-6 + 5)(-6 - 1) = 21 > 0
    Just checking at one point is not in general sufficient. Here it is- if you knew that, you should have said how you knew. Also, you haven't answered the question! You haven't said what kind of point that is.

    f'= 3(x-1)(x+5). If x is close to -5, then x-1 is close to -5-1= -4 and so is negative. For x< -5, x+ 5< 0 so f'= 3(x-1)(x+5) is the product of one positive and two negative numbers- f' is positive so f is increasing coming up to -5. For x> -5, x+5> 0 so f'= 3(x-1)(x+5) is the product of two positive and one negative number. f'< 0 so f is decreasing past -5. Since f was increasing and then started decreasing, x= -5 is a maximum.

    c) Classify the RH Stationary point with 2nd Derivative Test

    So,  f '' (x) = 3(2x) + 12 = 6x + 12 = 6(x + 2)

    The stationary point XR = 1, so the test point will be XR = 2
    <br />
f '' (2) = 6(2 + 2)= 24 > 0
    No, no, no! For the second derivative test, you evaluate f" at the stationary point! f"(1)= 6(1+2)= 3> 0.

    So f has a local minimum at X = 1
    Yes, but you were lucky!

    d) Find the coordinates of each stationary point

    so,

    XL - f (-5) = -125 + 150 - (-75) - 24 = 76
    XR - f (1) = 1 + 6 - 15 - 24 = -32

    Hence the graph passes through the point (-5,76) & (1,-32)

    Thanks for your help

    A
    Why are you taking points exactly "1" away from the stationary point? There is nothing special about integers!
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  3. #3
    ADY
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    Quote Originally Posted by HallsofIvy View Post


    No, no, no! For the second derivative test, you evaluate f" at the stationary point! f"(1)= 6(1+2)= 3> 0.


    Yes, but you were lucky!


    Why are you taking points exactly "1" away from the stationary point? There is nothing special about integers!
    So for the Second Der. test, how do you make it 3? I make it 18?
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  4. #4
    ADY
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    Quote Originally Posted by HallsofIvy View Post
    Why are you taking points exactly "1" away from the stationary point? There is nothing special about integers!
    Well to work out the y coordinates of each the stationary point dont we use -5 and 1 ?
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