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Math Help - Find the stationary points on the curve y=(3x-1)(x-2)^4

  1. #1
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    Find the stationary points on the curve y=(3x-1)(x-2)^4

    Find the stationary points on the curve y=(3x-1)(x-2)^4

    I differentiated it using the udv + vdu, then i tried to work it out and i got no were close to the answer i cant figure how to do my working out properly.

    back of book says it (2,0) and (2/3, 3/13/81)

    can sum1 plz show me how they would do working out on this question.

    thx
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  2. #2
    Senior Member DivideBy0's Avatar
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    f(x) = (3x-1)(x-2)^4

    f'(x) = (3x-1)\cdot \frac{d}{\,dx}((x-2)^4) + (x-2)^4 \cdot \frac{d}{\,dx}(3x-1)

    The derivatives are:

    \frac{d}{\,dx} ((x-2)^4) = 4(x-2)^3

    And:

    \frac{d}{\,dx} (3x-1) = 3

    So:

    \Rightarrow f'(x) = 4(3x-1)(x-2)^3 + 3(x-2)^4

    Setting f'(x) = 0, we get:

    4(3x-1)(x-2)^3 + 3(x-2)^4 =0

    Now we can't simply divide by (x-2), or we will be missing out on a solution, so instead we factor out:

    (x-2)^3(4(3x-1)+3(x-2))=0

    (x-2)^3(12x-4+3x-6)=0

    (x-2)^3(15x-10)=0

    \mbox{$\Rightarrow x = 2$ or $x=\frac{2}{3}$}\\

    Then, after substituting these values back into the original equation, we get the stationary points as:

    2,0),\left(\frac{2}{3},\frac{256}{81}\right)\rbrac e}" alt="\lbrace{(x,y)2,0),\left(\frac{2}{3},\frac{256}{81}\right)\rbrac e}" />
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  3. #3
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    thanx a lot
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  4. #4
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    Hello, KavX!

    Find the stationary points on the curve: .  y\:=\:(3x-1)(x-2)^4
    Advice: After doing a Product Rule, think of factoring.

    We have: . y' \;=\;(3x-1)\cdot4(x-2)^3\cdot 1 \:+\: 3\cdot(x-2)^4

    Simplify: . y' \;=\;4(x-2)^3(3x-1) + 3(x-2)^4

    Factor: . . y' \;=\;(x-2)^3\left[4(3x-1) + 3(x-2)\right]

    Simplify: . y' \;=\;(x-2)^3(15x-10) \;=\;5(x-2)^3(3x-2)


    To solve y' = 0, we have: . 5(x-2)^3(3x-2) \;=\;0

    Got it?

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  5. #5
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    aahh yer thanx for that
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