attempt 1: 10 attempts remaining. the value of a laptop $l(t)$, in dollars, after $t$ years is modeled by…

attempt 1: 10 attempts remaining. the value of a laptop $l(t)$, in dollars, after $t$ years is modeled by: $l(t)=2200 - 2e^{0.3t}$ a) find the value of the laptop after 3 years. round to two decimal places as needed. $l(3)=$ b) how fast is the value of the laptop decreasing at that time? round to two decimal places as needed. $l(3)=$ dollars

attempt 1: 10 attempts remaining. the value of a laptop $l(t)$, in dollars, after $t$ years is modeled by: $l(t)=2200 - 2e^{0.3t}$ a) find the value of the laptop after 3 years. round to two decimal places as needed. $l(3)=$ b) how fast is the value of the laptop decreasing at that time? round to two decimal places as needed. $l(3)=$ dollars

Answer

Explanation:

Step1: Substitute t = 3 into L(t)

Substitute $t = 3$ into $L(t)=2200 - 2e^{0.3t}$. [ \begin{align*} L(3)&=2200-2e^{0.3\times3}\ &=2200 - 2e^{0.9}\ &\approx2200-2\times2.4596\ &=2200 - 4.9192\ &\approx2195.08 \end{align*} ]

Step2: Differentiate L(t)

Differentiate $L(t)=2200 - 2e^{0.3t}$ with respect to $t$. Using the rule that the derivative of a constant is 0 and the derivative of $e^{ax}$ is $ae^{ax}$, we get $L'(t)=- 2\times0.3e^{0.3t}=-0.6e^{0.3t}$.

Step3: Substitute t = 3 into L'(t)

Substitute $t = 3$ into $L'(t)=-0.6e^{0.3t}$. [ \begin{align*} L'(3)&=-0.6e^{0.3\times3}\ &=-0.6e^{0.9}\ &\approx-0.6\times2.4596\ &\approx - 1.48 \end{align*} ]

Answer:

a. $2195.08$ b. $1.48$ (the negative sign just indicates decrease, we take the absolute - value for the rate of decrease)