Questions: You plan to invest in an account which pays 4.5% compounded continuously. If the investment period is for 8 years, then A(P) = P e^(0.045 * 8) = P e^0.36 gives the total balance of P dollars. a. Find a formula for A'(P). b. Find and interpret A'(4000). c. Compare the approximation to the actual change. a. A'(P) = e^0.36 (Type an exact answer in terms of e.) b. A'(4000) = 1.43 dollars increased in the total amount per dollar increase in the initial investment (Round to the nearest cent as needed.) Interpret A'(4000) = 1.43 A. The future value of a 9 year investment of 4000 will be more than the future value of a 9 year investment of 4000. B. The future value of an 8 year investment of 4001 will be more than the future value of an 8 year investment of 4000. C. The future value of 8 year investment of 4000 at 5.5% will be more than the future value of an 8 year investment of 4000 at 4.5%. D. The future value of a 9 year investment of 4001 will be more than the future value of a 9 year investment of 4000.

Solution Steps

To solve the given problem, we need to follow these steps:

a. Find the derivative of the function \( A(P) = P e^{0.36} \) with respect to \( P \).

b. Evaluate the derivative at \( P = 4000 \) to find \( A'(4000) \).

c. Interpret the result of \( A'(4000) \) in the context of the problem.

1. Find the derivative \( A'(P) \):

   • The function \( A(P) = P e^{0.36} \) is a product of \( P \) and a constant \( e^{0.36} \).
   • The derivative of \( P \) with respect to \( P \) is 1, and the derivative of a constant is 0.
   • Therefore, \( A'(P) = e^{0.36} \).

2. Evaluate \( A'(4000) \):

   • Substitute \( P = 4000 \) into the derivative \( A'(P) \).

3. Interpret \( A'(4000) \):

   • The value of \( A'(4000) \) represents the rate of change of the total balance with respect to the initial investment at \( P = 4000 \).
   • This means for every additional dollar invested, the total balance increases by \( A'(4000) \) dollars.

Given the function \( A(P) = P e^{0.36} \), we need to find its derivative with respect to \( P \).

\[ A'(P) = \frac{d}{dP} \left( P e^{0.36} \right) = e^{0.36} \]

Substitute \( P = 4000 \) into the derivative \( A'(P) \):

\[ A'(4000) = e^{0.36} \approx 1.4333 \]

The value of \( A'(4000) \) represents the rate of change of the total balance with respect to the initial investment at \( P = 4000 \). This means for every additional dollar invested, the total balance increases by approximately \( 1.4333 \) dollars.

The interpretation of \( A'(4000) \) is that the future value of an 8-year investment of \( \$4001 \) will be approximately \( \$1.43 \) more than the future value of an 8-year investment of \( \$4000 \).

Final Answer