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The sales of a new high-tech item (in thousands) are given by the following function, where t represents time in years. Find the rate of change of sales at each time.
\[
S(t)=107-100 e^{-0.2 t}
\]
(a) After 1 year. (b) After 5 years. (c) What is happening to the rate of change of sales as time goes on? (d) Does the rate of change of sales ever equal zero?
(a) The rate of change after 1 year is thousand items per year.
(Round to three decimal places as needed.)

Step 1 ：The given function is \(S(t) = 107 - 100e^{-0.2t}\).

Step 2 ：The derivative of a constant is zero and the derivative of \(e^{-0.2t}\) is \(-0.2e^{-0.2t}\).

Step 3 ：So, the derivative of \(S(t)\) is \(S'(t) = 0 - 100(-0.2)e^{-0.2t} = 20e^{-0.2t}\).

Step 4 ：After 1 year, the rate of change is \(S'(1) = 20e^{-0.2*1} = 20e^{-0.2} = 16.597\) units per year (rounded to three decimal places).

Step 5 ：After 5 years, the rate of change is \(S'(5) = 20e^{-0.2*5} = 20e^{-1} = 7.359\) units per year (rounded to three decimal places).

Step 6 ：As time goes on, the rate of change of sales is decreasing. This is because the term \(e^{-0.2t}\) in the derivative is getting smaller as \(t\) increases.

Step 7 ：The rate of change of sales equals zero when the derivative \(S'(t)\) equals zero. However, since \(e^{-0.2t}\) is always positive for any real number \(t\), the derivative \(S'(t) = 20e^{-0.2t}\) can never be zero. Therefore, the rate of change of sales never equals zero.