3) Martin has his money invested in a stock portfolio. The value, \( v(x) \), of his portfolio can be modeled with the function \( v(x) \) - \( 30,000(0,78)^{x} \), where \( x \) is the number of years since he made his investment. Find the rate of change of the value of his portfolio Kate of change \( = \) I

To find the **rate of change** of the value of Martin's portfolio, we need to compute the derivative of the portfolio value function \( v(x) \) with respect to \( x \). Given: \[ v(x) = 30,000 \times (0.78)^x \] **Step 1: Differentiate \( v(x) \) with respect to \( x \)** The function \( v(x) = 30,000 \times (0.78)^x \) is an exponential function. The derivative of \( a^x \) with respect to \( x \) is \( a^x \ln(a) \). So, \[ v'(x) = \frac{d}{dx} \left[ 30,000 \times (0.78)^x \right] = 30,000 \times (0.78)^x \times \ln(0.78) \] **Step 2: Simplify the Expression** \(\ln(0.78)\) is a negative number since \( 0.78 < 1 \), which indicates that the portfolio value is decreasing over time. \[ v'(x) = 30,000 \times (0.78)^x \times \ln(0.78) \] **Final Answer:** \[ v'(x) = 30,\!000 \cdot (0.78)^x \times \ln(0.78) \] **Interpretation:** This expression represents the **rate of change** of Martin's portfolio value at any year \( x \). Since \( \ln(0.78) \) is negative, the portfolio is decreasing at a rate proportional to its current value. **Example Calculation:** If you want to find the rate of change after, say, 2 years: \[ v'(2) = 30,000 \times (0.78)^2 \times \ln(0.78) \approx 30,000 \times 0.6084 \times (-0.2485) \approx -4,545.6 \] This means that after 2 years, the portfolio is decreasing at a rate of approximately \$4,545.60 per year. **Conclusion:** The rate of change of Martin's portfolio value with respect to time is given by: \[ v'(x) = 30,\!000 \cdot (0.78)^x \times \ln(0.78) \] **Answer:** After simplifying, the rate of change is 30 000 · 0.78ˣ multiplied by ln 0.78. Thus, v ′(x) = 30,000 · (0.78)^x · ln(0.78)