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Math Help - critical points

  1. #1
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    critical points

    please try to solve these questions.
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  2. #2
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    Hello, m777!

    2) Find \frac{dy}{dx} if x^3 + xy + y^3\:=\:6

    Differentiate implicitly: . 3x^2 + x\left(\frac{dy}{dx}\right) + y + 3y^2\left(\frac{dy}{dx}\right) \;= \;0

    Then: . x\left(\frac{dy}{dx}\right) + 3y^2\left(\frac{dy}{dx}\right) \;= \;-3x^2 - y

    Factor: . \left(x^2 + 3y^2\right)\!\cdot\!\frac{dy}{dx} \;= \;-(3x^2 + y)

    Therefore: . \frac{dy}{dx} \;= \;-\frac{3x^2+y}{x+3y^2}



    3) a) What are the critical points of f(x) ?
    b) On what interval(s) is f(x) increasing or decreasing?
    . . . . f'(x) \:=\:(x-1)^2(x+2)

    (a) Critical points occur where f'(x) = 0
    . . .This happens when: . x\:=\:\text{-}2,\,1

    To the left of x = \text{-}2, we have: . f'(\text{-}3) \:=\:(\text{-}4)^2(-1) \:=\:\text{-}16 .Decreasing: \searrow

    Between x = \text{-}2 and x = 1, we have: . f'(0) \:=\:(-1)^2(2^2) \:=\:+4 .Increasing: \nearrow

    To the right of x = 1, we have: . f'(2)\:=\:(1^2)(4^2) \:=\:+16 .Increasing: \nearrow


    Therefore:

    . . f(x) is decreasing on (\text{-}\infty,\,\text{-}2)

    . . f(x) is increasing on (\text{-}2,\,1) \cup (1,\,\infty)

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