the sales of a new high - tech item (in thousands) are given by s(t)=100 - 90e^(-0.2t) where t represents…

the sales of a new high - tech item (in thousands) are given by s(t)=100 - 90e^(-0.2t) where t represents time in years. find the rate of change of sales at each time. 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.) b. the rate of change after 5 years is thousand items per year. (round to three decimal places as needed.) c. what is happening to the rate of change of sales as time goes on? a. it always decreases. b. it always increases. c. first it decreases, then it increases. d. first it increases, then it decreases. d. does the rate of change of sales ever equal zero? yes

the sales of a new high - tech item (in thousands) are given by s(t)=100 - 90e^(-0.2t) where t represents time in years. find the rate of change of sales at each time. 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.) b. the rate of change after 5 years is thousand items per year. (round to three decimal places as needed.) c. what is happening to the rate of change of sales as time goes on? a. it always decreases. b. it always increases. c. first it decreases, then it increases. d. first it increases, then it decreases. d. does the rate of change of sales ever equal zero? yes

Answer

Explanation:

Step1: Find the derivative of $S(t)$

The derivative of $S(t)=100 - 90e^{-0.2t}$ using the chain - rule. The derivative of a constant is 0, and the derivative of $e^{-0.2t}$ is $- 0.2e^{-0.2t}$. So, $S^\prime(t)=-90\times(-0.2)e^{-0.2t}=18e^{-0.2t}$.

Step2: Calculate the rate of change after 1 year

Substitute $t = 1$ into $S^\prime(t)$. $S^\prime(1)=18e^{-0.2\times1}=18e^{-0.2}\approx18\times0.81873 = 14.737$.

Step3: Calculate the rate of change after 5 years

Substitute $t = 5$ into $S^\prime(t)$. $S^\prime(5)=18e^{-0.2\times5}=18e^{-1}\approx18\times0.36788 = 6.622$.

Step4: Analyze the behavior of $S^\prime(t)$ as $t$ increases

As $t$ increases, the exponent $-0.2t$ becomes more negative. Since $y = e^x$ is a positive - valued function and $S^\prime(t)=18e^{-0.2t}$, as $t$ increases, $S^\prime(t)$ always decreases because the value of $e^{-0.2t}$ decreases as $t$ increases.

Step5: Check if $S^\prime(t)$ can be zero

Set $S^\prime(t)=0$, so $18e^{-0.2t}=0$. Since $e^{-0.2t}>0$ for all real - valued $t$ (the exponential function $y = e^x$ is always positive for any real $x$), and $18>0$, the equation $18e^{-0.2t}=0$ has no solutions.

Answer:

a. $14.737$ b. $6.622$ c. A. It always decreases. d. No