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Thread: 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 $\displaystyle \frac{dy}{dx}$ if $\displaystyle x^3 + xy + y^3\:=\:6$

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

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

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

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



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

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

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

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

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


    Therefore:

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

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

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