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Math Help - interval for f'(x)>0

  1. #1
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    interval for f'(x)>0

    the question is
    on what interval is f'(x)>0 for f(x)=2x^3 - 3x^2 - 36x +2

    i found f'(x) = 6x^2 - 6x -36

    then what do i have to do next to find interval?
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  2. #2
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    Now you have to address inequality 6x^2-6x-36 > 0

    That is equivalent to x^2-x-6 > 0

    factoring (x-3)(x+2) > 0

    you conclude it by separating cases
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  3. #3
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    Quote Originally Posted by haebinpark View Post
    the question is
    on what interval is f'(x)>0 for f(x)=2x^3 - 3x^2 - 36x +2

    i found f'(x) = 6x^2 - 6x -36

    then what do i have to do next to find interval?
    Factor 6(x-3)(x+2). You'll find that this takes a zero at x = -2 an x = 3. Between these zero 6(x-3)(x+2) >0 , otherwise it's negative.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    f'(x) = 6x^2 - 6x -36

    You should notice f ' is a parabola opening up. Locate the zeroes then f ' is negative between the zeroes and positive otherwise

    Set f ' = 0

    6x^2 - 6x -36 = 0

    x^2 -x -6 = 0

    Now factor to find the zeroes
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  5. #5
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    so

    f'(x) = 0 when x = 3 and x= -2

    f'(x) > 0 when x > 3 and x < -2


    am i able to write answer like -2<x<3 or should i leave it as f'(x) > 0 when x > 3 and x < -2 ?
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  6. #6
    MHF Contributor Calculus26's Avatar
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    Actually you would write x > 3 OR x < -2

    as you are taking the union of (-inf, -2) U (3,inf)


    -2 < x < 3 is where f ' < 0

    at least you didn't write 3 < x < -2 like a lot of students would which is
    nonsense
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  7. #7
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    As the product is greater than zero we have two possibilities

    1) x-3 >0 \wedge x+2 > 0

    We conclude that x \in (3,\infty)

    2) x-3 <0 \wedge x+2 < 0

    We conclude that x \in (-\infty, -2)

    Then the interval is x \in (-\infty,-2)\cup (3,\infty)
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