Question

An investment of $900 is invested monthly at a rate of 4%. The function that models this 0.04 situation is f(x)=900 1+ where x represents time in years, and f(x) represents the worth of the investment. What is the rate of change for this function over the interval 3x10?

Answer 1

The rate of change for this function over the interval 3x10 is 900(0.04)(1.04)^9.

Step-by-step explanation:

The rate of change for an investment can be found by taking the derivative of the function that models the situation. In this case, the function that models the investment is f(x) = 900(1+0.04)^x, where x represents time in years and f(x) represents the worth of the investment. Taking the derivative of this function, we get f'(x) = 900(0.04)(1+0.04)^(x-1). So, the rate of change for this function at any given time is 900(0.04)(1+0.04)^(x-1).

Therefore, to find the rate of change over the interval 3x10, we can substitute x = 10 into the equation. The rate of change for this function over the interval 3x10 is 900(0.04)(1+0.04)^(10-1) = 900(0.04)(1.04)^9.