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23 October, 14:51

Consider the function as representing the value of an ounce of palladium in U. S. dollars as a function of the time t in days.

R (t) = 30t - 3t^2; t = 3

a. Find the average rate of change of R (t) over the time intervals [t, t + h], where t is as indicated and h = 1, 0.1, and 0.01 days. (Use smaller values of h to check your estimates.)

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  1. 23 October, 15:30
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    the average rate of change rc is 13, 12.1 and 12.01 for h=1, 0.1 and 0.01 respectively

    Step-by-step explanation:

    for

    R (t) = 30*t - 3*t²; t = 3

    the average rate of change of R (t) over the time interval [t, t + h] is

    rc = [R (t+h) - R (t) ] / [ (t+h) - t) = [R (t+h) - R (t) ] / h = 1/h * [ 30 * (t+h) - 3 * (t+h) ² - (30*t - 3*t²) ] = (1/h) * (30*h - (3*t² + 6*t*h + t²) + 3*t²) = 30 - 6*t + h

    then

    rc = 30 - 6*t + h

    for t=3 and h=1

    rc = 30 - 6*3 + 1 = 13

    for t=3 and h=0.1

    rc = 30 - 6*3 + 0.1 = 12.1

    for t=3 and h=0.01

    rc = 30 - 6*3 + 0.01 = 12.01

    for t=3 and h=0.001

    rc = 30 - 6*3 + 0.01 = 12.001

    when h goes smaller, the average rate of change gets closer to the instantaneous rate of change of R (t) in t=3 (the derivative of R in t=3), that is

    R' (t) = 30 - 6*t
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