Answer:964.60$ 6 years after (b) the car does not have a specific value of (t) at which it depreciates most rapidly.
Step-by-step explanation:
a) Let’s find the value of the car 6 years after it was purchased using the given formula:
The value of the car after (t) years is given by:
[ V(t) = 11000 \left(\frac{2}{3}\right)^t ]
Substitute (t = 6):
[ V(6) = 11000 \left(\frac{2}{3}\right)^6 ]
Calculating:
[ V(6) = 11000 \left(\frac{64}{729}\right) ]
[ V(6) = 11000 \cdot 0.0877 ]
[ V(6) = 964.7 ]
Therefore, the value of the car 6 years after it was purchased is approximately $964.70.
b) To find the value of (t) when the car depreciates most rapidly, we need to determine the critical points of the function (V(t)). The rate of depreciation is given by the derivative of (V(t)):
[ \frac{dV}{dt} = 11000 \cdot \frac{2}{3} \cdot \ln\left(\frac{2}{3}\right) \cdot \left(\frac{2}{3}\right)^t ]
Setting the derivative equal to zero to find critical points:
[ \frac{dV}{dt} = 0 ]
[ 11000 \cdot \frac{2}{3} \cdot \ln\left(\frac{2}{3}\right) \cdot \left(\frac{2}{3}\right)^t = 0 ]
Solving for (t):
[ \ln\left(\frac{2}{3}\right) \cdot \left(\frac{2}{3}\right)^t = 0 ]
Since (\ln\left(\frac{2}{3}\right)) is nonzero, we have:
[ \left(\frac{2}{3}\right)^t = 0 ]
This equation has no real solutions, which means the car does not have a maximum rate of depreciation. Instead, it continuously decreases over time.
Therefore, the car does not have a specific value of (t) at which it depreciates most rapidly.
