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Math Help - How to find Maximum and Minimum Points?

  1. #1
    Member
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    How to find Maximum and Minimum Points?

    1. y=(1-x^2)(4-x^2)

    2. y=3x^4-16x^3+18x^2

    Answer:
    1. (-\sqrt{\frac{5}{2}}, -2) (\sqrt{\frac{5}{2}}, -2)

    2. (1, 5) (3, -27)
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  2. #2
    Senior Member
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    The condition is : dy/dx = 0

    but i get three values for each question, not two
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  3. #3
    Senior Member I-Think's Avatar
    Joined
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    Differentiation

    The only way I know to solve this question is to employ some calculus and differentiate. Differentiate to find \frac{dy}{dx}, then set \frac{dy}{dx}=0, and solve the resulting equation.

    Eg. No.2

    \frac{dy}{dx}=12x^3-48x^2+36x
    =12x(x^2-4x+3)
    \frac{dy}{dx}=12x(x^2-4x+3)=0
    =(12x)(x-1)(x-3)=0
     x=0,1,3

    Place these results into the original equation

    When x=1
    y=3(1)^4-16(1)^3+18(1)^2=5
    When  x=3
    y=3(3)^4-16(3)^3+18(3)^2=-27
    When  x=0,y=0

    So your turning points are
    (0,0)

    Do the same for no.1
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