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27 June, 11:06

Consider the equation below. f (x) = 2x3 + 3x2 - 336x (a) Find the interval on which f is increasing. (Enter your answer in interval notation.) (-[infinity],-8],[ (7,[infinity]) ] Find the interval on which f is decreasing. (Enter your answer in interval notation.)

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  1. 27 June, 11:13
    0
    The function f (x) = 2x³ + 3x² - 336x

    is

    Increasing on: (-∞, - 8) U (7, ∞)

    and

    Decreasing on: (-8, 7)

    Step-by-step explanation:

    Given the function

    f (x) = 2x³ + 3x² - 336x

    We need to find the interval on which f is increasing, and the interval on which it is decreasing.

    First, we take the derivative of f with respect to x to obtain

    f' (x) = 6x² + 6x - 336

    Next, set f' (x) = 0 and solve for x.

    f' (x) = 6x² + 6x - 336 = 0

    Dividing by 6, we have

    x² + x - 56 = 0

    (x - 7) (x + 8) = 0

    x = 7 or x = - 8

    Now we have the interval

    (-∞, - 8) ∪ (-8, 7) ∪ (7, ∞)

    Let us substitute a value from the interval (-∞, - 8) into the derivative to determine if the function is increasing or decreasing.

    Let us choose x = - 9, then

    f' (-9) = 6 (-9) ² + 6 (-9) - 336

    = 96 > 0

    Since 96 is positive, the function is increasing on (-∞, - 8).

    Again, substitute a value from the interval (-8, 7) into the derivative to determine if the function is increasing or decreasing.

    Choose - 2

    f' (-2) = 6 (-2) ² + 6 (-2) - 336

    = - 324 < 0

    Since - 324 is negative, then the function is decreasing on (-8, 7)

    Lastly, substitute a value from the interval (7, ∞) into the derivative

    Choose x = 9

    6 (9) ² + 6 (9) - 336

    = 204

    The function is increasing on (8, ∞)

    f' (x) > 0

    Collating these, we have the function to be

    Increasing on:

    (-∞, - 8) U (7, ∞)

    Decreasing on:

    (-8, 7)
  2. 27 June, 11:29
    0
    f is increasing = Solution Interval = x ∈ (-∞, - 8] ∪ [7, ∞)

    f is decreasing = Solution Interval = x ∈ [-8, 7]

    Step-by-step explanation:

    The given function is

    f (x) = 2x³ + 3x² - 336x

    We are required to find the interval on which f (x) is increasing and decreasing.

    f (x) is increasing:

    f (x) = 2x³ + 3x² - 336x

    Taking the derivative of f (x) with respect to x yields,

    f' (x) = 3 * (2x²) + 2 * (3x) - 336

    f' (x) = 6x² + 6x) - 336

    f' (x) ≥ 0

    6x² + 6x - 336 ≥ 0

    You can use either quadratic formula or use your calculator to solve the quadratic equation,

    (x + 8) + (x - 7) ≥ 0

    (x + 8) ≥ 0 or (x - 7) ≥ 0

    x ≤ - 8 or x ≥ 7

    so

    x ∈ (-∞, - 8]

    x ∈ [7, ∞)

    Solution Interval = x ∈ (-∞, - 8] ∪ [7, ∞)

    f (x) is decreasing:

    f (x) = 2x³ + 3x² - 336x

    We have already found the derivative f' (x)

    f' (x) = 6x² + 6x - 336

    f' (x) ≤ 0

    6x² + 6x - 336 ≤ 0

    (x - 7) + (x + 8) ≤ 0

    (x - 7) ≤ 0 or (x + 8) ≤ 0

    x ≤ 7 or x ≥ - 8

    -8 ≤ x ≤ 7

    Solution Interval = x ∈ [-8, 7]
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