Find the intervals in which the function f given by f (x) = 2x² – 3x is (a) strictly increasing (b) strictly decreasing

Question

Find the intervals in which the function f given by f (x) = 2x² – 3x is
(a) strictly increasing (b) strictly decreasing

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Ximena 2 weeks 2021-10-04T16:29:42+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-04T16:30:52+00:00

    AnswEr :

    Given Expression,

     \sf \: f(x) = 2 {x}^{2}  - 3x

    We have to find the intervals for which the given function is

    • Strictly Increasing
    • Strictly Decreasing

    Firstly,let us find the derivative of the above function w.r.t. x,

     \longrightarrow \:  \sf \:  \dfrac{f(x) }{dx}  =  \dfrac{d(2 {x}^{2} - 3x) }{dx}  \\  \\  \longrightarrow \:  \sf \: f'(x) = 4x - 3

    Equating f'(x) = 0 because f'(x) becomes a “constant function” at 0. This would help us find the intervals at which the given function is increasing and decreasing.

     \longrightarrow \sf \: 4x - 3 = 0 \\  \\  \longrightarrow \sf \: 4x = 3 \\  \\  \longrightarrow  \underline{ \boxed{\sf \: x =  \dfrac{3}{4} }}

    • The point (3/4,0) divides the x-axis into two intervals (-∞,3/4) and (3/4,∞).

    Strictly Increasing

    • A function is said to be strictly increasing if all values of f'(x) > 0.

    Therefore,

     \sf \: f'(x) > 0 \\  \\  \implies \:  \sf \: 4x - 3 > 0 \\  \\  \implies \sf \: x >  \dfrac{3}{4}

    f(x) is strictly increasing in the interval (3/4,∞)

    Strictly Decreasing

    • A function is said to be strictly decreasing if f'(x) < 0

    Therefore,

     \sf \: f'(x) < 0 \\  \\  \implies \:  \sf \: x <  \dfrac{3}{4}

    f(x) is strictly decreasing in the interval (-∞,3/4)

    To sum it up,

    \begin{array}{|c|c|c|} \cline{1-3}\sf Given \ Function & \sf f'(x) & \sf Intervals \\ \cline{1-3} \sf 2x^2 - 3x & \sf 4x-3 &  \sf Increasing (\dfrac{3}{4},\infty) \\   \cline{3-3} & & \sf Decreasing (-\infty, \dfrac{3}{4}) \\  \cline{1-3} \end{array}

    0
    2021-10-04T16:31:07+00:00

    given, f(x) = 2x² – 3x

    differentiate f(x) with respect to x,

    f'(x) = 4x – 3 ——(1)

    (a) when f(x) is strictly increasing function :

    f'(x) > 0

    from equation (1),

    4x – 3 > 0 => x > 3/4

    e.g., x ∈ (3/4, ∞ )

    Therefore, the given function (f) is strictly increasing in interval x ∈ (3/4, ∞ ) .

    (b) when f(x) is strictly decreasing function :

    f'(x) < 0

    from equation (1),

    4x -3 < 0 => x < 3/4

    e.g., x ∈ (-∞ , 3/4)

    Therefore, the given function (f) is strictly decreasing in interval x ∈ (-∞ , 3/4)

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    Step-by-step explanation:

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