Andrea invested $\$3,000$ in an online currency. If the value of the currency increases at a rate of 14\% per year, the situation can be modeled using the equation A(t) = 3,000(1.14)^t, where A is the amount in the investment after t years. What speed does a increase at the beginning

To find the initial rate of increase of the investment at the beginning (t = 0), we need to calculate the derivative of the function \( A(t) \) with respect to \( t \) and then evaluate that derivative at \( t = 0 \).

The function modeling the investment is given by: \[ A(t) = 3000(1.14)^t \]

Now, we differentiate \( A(t) \) with respect to \( t \): \[ A'(t) = \frac{d}{dt}[3000(1.14)^t] \] Using the chain rule, we have: \[ A'(t) = 3000 \cdot \ln(1.14) \cdot (1.14)^t \]

Now, let's evaluate \( A'(t) \) at \( t = 0 \): \[ A'(0) = 3000 \cdot \ln(1.14) \cdot (1.14)^0 = 3000 \cdot \ln(1.14) \cdot 1 = 3000 \cdot \ln(1.14) \]

Now we compute \( \ln(1.14) \): Using a calculator, we find: \[ \ln(1.14) \approx 0.130 \]

Thus, we can compute \( A'(0) \): \[ A'(0) \approx 3000 \cdot 0.130 \approx 390 \]

So, the speed of increase at the beginning (at \( t = 0 \)) is approximately \( $390 \) per year.