With an initial deposit of $100, and and interest rate of 18%, the balance in a
ID: 2855046 • Letter: W
Question
With an initial deposit of $100, and and interest rate of 18%, the balance in a bank account after t years is
f(t) = 100(1.18)t dollars.
(c) Estimate the instantaneous rate of change at t = 0.5 by computing the average rate of change over intervals to the left and right of t = 0.5. (Round your answers to four decimal places.)
Average rate of change
*The answer to each of them is not 118*
Interval [0.5, 0.51] [0.5, 0.501] [0.5, 0.5001] Average rate of change The answer is not 118 to all already been missed several timesExplanation / Answer
f(t) = 100(1.18)t
average rate of change in [0.5, 0.51]
=[f(0.51) -f(0.5)]/(0.51-0.5)
=[100(1.18)(0.51) -100(1.18)(0.5)]/(0.51-0.5)
=17.9944
average rate of change in [0.5, 0.501]
=[f(0.501) -f(0.5)]/(0.501-0.5)
=[100(1.18)(0.501) -100(1.18)(0.5)]/(0.501-0.5)
=17.9810
average rate of change in [0.5, 0.5001]
=[f(0.5001) -f(0.5)]/(0.5001-0.5)
=[100(1.18)(0.5001) -100(1.18)(0.5)]/(0.5001-0.5)
=17.9796
=============================
average rate of change in [0.49, 0.5]
=[f(0.5) -f(0.49)]/(0.5-0.49)
=[100(1.18)(0.5) -100(1.18)(0.49)]/(0.5-0.49)
=17.9646
average rate of change in [0.499, 0.5]
=[f(0.5) -f(0.499)]/(0.5-0.499)
=[100(1.18)(0.5) -100(1.18)(0.499)]/(0.5-0.499)
=17.9780
average rate of change in [0.4999, 0.5]
=[f(0.5) -f(0.4999)]/(0.5-0.4999)
=[100(1.18)(0.5) -100(1.18)(0.4999)]/(0.5-0.4999)
=17.9793
instantaneous rate of change at t = 0.5 is 17.9795 approximately
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