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Given that f(x) = 2x^2 - 8x + 9, 0 ≤ x ≤ 3.
(a) Find the average value of f.
(b) Find c such that f(average) = f(c).

  • Math -

    average value
    = 1/(3-0) ∫(2x^2 - 8x + 9) dx from x = 0 to x = 3
    = (1/3) [(2/3)x^3 - 4x^2 + 9x[ from 0 to 3
    = (1/3) (18 - 36 + 27 - 0 )
    = 3

    b)
    f(c) = f(3)
    2c^2 - 8c + 9 = 18 - 16 + 9
    2c^2 - 8c -2 = 0
    c^2 - 4c - 1 = 0

    c = (4 ± √20)/2
    = 2 ± √5

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