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March 28, 2017

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Let F(s)=5s2+1s+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.

  • calculus - ,

    F'(1)=10+1=11

    average= (finalF(s)-initialF(s))/d
    = (5d^2+d+4-4)/d= 5d+1

    11=5d+1 solve for d.

  • calculus - ,

    I assume that your fucnction is
    F(s) = 5s^2 +s + 4.

    The instantaneous rate of change of F at s=1 is the value of the derivative at s=1. The derivative is F'(x) = 10s + 1. When s = 1, F' is 11.

    The average rate of change of F from 0 to d is [F(d) - F(0)]/d. That equals
    (5d^2 + d + 4 - 4)/d = 5d + 1

    Set 5d +1 = 11 and solve for d.

    I leave that part for you.

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