Calculus

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Let F(s)=5s2+5s+4 . Find a value of d greater than 0 such that the average rate of change of F(s) from 0 to d equals the instantaneous rate of change of F(s) at s=1. d=

  • Calculus -

    ignore the "d=". Sorry!

  • Calculus -

    the instantaneous rate of change is given by
    F'(x) which is 10s + 5
    which, when s=1, becomes F'(1) = 15

    for the "average" rate of change from 0 to d
    F(0) = 4
    F(d) = 5d^2 + 5d + 4
    average rate of change = (5d^2 + 5d + 4 - 4)/(d-0)

    so solve for d in
    (5d^2 + 5d)/d = 15
    d^2 + d = 3d
    d^2 - 2d = 0
    d(d-2) = 0
    d = 0 or d = 2 ,but you want d > 0,
    so d = 2

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