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.