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Math Help - Increasing Intervals please help!

  1. #1
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    Increasing Intervals please help!

    For the function is defined by
    On which two intervals is the function increasing

    I got -11 to 6 as one interval

    and only part of the other one which is (some # to 10)

    What is the "some #" please help this is due tonight. thank you
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  2. #2
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    what did you get for f'(x) ?
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  3. #3
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    (7x^6)(x+6)^4 + 4(x+6)^3(x^7)
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  4. #4
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    (7x^6)(x+6)^4 + 4(x+6)^3(x^7)
    fine, but did you simplify it before you set it equal to 0?

    factoring out what the two terms have in common will help alot.
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  5. #5
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    does it become [x^6(x+6)^3] 7(x+6) + 4x
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  6. #6
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    cleaned up your derivative a little bit ...
    f'(x) = 7x^6(x+6)^4 + 4x^7(x+6)^3


    now, look at the following expression ... what can you factor out of those two terms (in other words, what factors do they have in common to "pull" out?)

    7x^6y^4 + 4x^7y^3
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  7. #7
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    Hello, Lambo!

    For x \in [-11, 10], the function f(x) is defined by: . f(x) \:=\:x^7(x+6)^4
    On which two intervals is the function increasing ?

    I got -11 to 6 as one interval . . . . sorry, no
    The function is increasing when f'(x) is positive.

    The derivative is: . f'(x) \;=\;x^7\cdot4(x+6)^3 + 7x^6\cdot(x+6)^4 \;=\;x^6(x+6)^3\bigg[4x + 7(x+6)\bigg]

    So we have: . f'(x)\;=\;x^6\cdot(x+6)^3\cdot(11x+42)\;>\;0

    This product is positive when: . \begin{Bmatrix}(1) & \text{all three factors are positive} \\ (2) & \text{two factors are negative} \end{Bmatrix}


    (1) All three factors are positive.

    . . x^6 is always positive, except when x = 0

    . . (x+6)^3 \:>\:0\quad\Rightarrow\quad x \:>\:\text{-}6

    . . 11x+42 \:>\:0 \quad\Rightarrow\quad x \:>\:\text{-}\tfrac{42}{11}

    All three are satisfied if: . x \:>\:\text{-}\frac{42}{11}\:\text{ and }\:x \neq 0

    . . The interval is: . \boxed{\;\left(\text{-}\tfrac{42}{11},\:0\right) \cup (0,\:\infty)\;}


    (2) Two factors are negative.

    . . x^6 is always positive, except when x = 0

    . . (x+6)^3 \:<\:0 \quad\Rightarrow\quad x \:<\:\text{-}6

    . . 11x+42 \:<\:0 \quad\Rightarrow\quad x \:<\:\text{-}\tfrac{42}{11}

    All three are satisfied if: . x \:<\:\text{-}6

    . . The interval is: . \boxed{\;(\text{-}\infty,\:\text{-}6)\;}

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