Question

The value V of an item t years after it is purchased is given below.V = 9000exp(-0.6301t), 0 ≤ t ≤ 10Part b only-456.21 wrong-129.40 wrong

Answer 1

• V'(4)=-456.10

,

• V'(6)=-129.35

Explanation:

Given the function V which is defined as follows:

`V=9000e^(-0.6301t),0\leq t\leq10`

First, find the derivative of V with respect to time.

`\begin{gathered} (dV)/(dt)=9000(d)/(dt)(e^(-0.6301t)) \\ \text{ The derivative of }e^(kt)=ke^(kt) \\ V^(\prime)(t)=9000(-0.6301)e^(-0.6301t) \end{gathered}`

I. When t=4

`\begin{gathered} V^(\prime)(t)=9000(-0.6301)e^(-0.6301t) \\ V^(\prime)(4)=9000(-0.6301)e^(-0.6301*4) \\ =-456.10 \end{gathered}`

The rate of change when t=4 is -456.10 (correct to 2 decimal places).

II. When t=6

`\begin{gathered} V^(\prime)(t)=9000(-0.6301)e^(-0.6301t) \\ V^(\prime)(4)=9000(-0.6301)e^(-0.6301*6) \\ =-129.35 \end{gathered}`

The rate of change when t=6 is -129.35 (correct to 2 decimal places).