An investment worth $1 million in 2005 has been growing at a rate of
f(t) = 0.141(1.17t)
 million dollars per year
where t is the number of years since 2005.
(a) Calculate how much the investment will have grown between 2005 and 2016. (Round your answer to three decimal places.)
$1.112 

  seenKey 4.153  million

How much is it projected to grow between 2016 and 2021? (Round your answer to three decimal places.)
$   

 million

(b) Recover the function for the model that gives future value of an investment in million dollars t years since 2005. Hint: First, find the general antiderivative of 
f(t)
. Then, use the initial investment amount to find the specific antiderivative. Your final answer should not have a 
C
 in it.
F(t) =    


Let's break this down step by step, focusing first on part (a) and then part (b).

### Part (a)
We are given the rate of growth function for the investment:
\[
f(t) = 0.141(1.17^t)
\]
where $ t $ is the number of years since 2005.

#### 1. Calculate how much the investment will have grown between 2005 and 2016
To find the total growth over a period of time, we need to integrate the rate of growth function from 2005 to 2016. This means we are integrating $ f(t) $ from $ t = 0 $ (corresponding to 2005) to $ t = 11 $ (corresponding to 2016).

\[
\text{Growth} = \int_0^{11} 0.141(1.17^t) \, dt
\]

#### 2. How much will it grow between 2016 and 2021?
For this part, we'll integrate the same function from $ t = 11 $ (2016) to $ t = 16 $ (2021):

\[
\text{Growth} = \int_{11}^{16} 0.141(1.17^t) \, dt
\]

Let's calculate both parts now.### Part (a) - Results:
1. **The investment growth from 2005 to 2016** is approximately \$4.150 million.
2. **The projected growth from 2016 to 2021** is approximately \$6.020 million.

### Part (b)
We are asked to recover the function $ F(t) $ that gives the future value of the investment based on the growth rate function $ f(t) $.

#### 1. General Antiderivative of $ f(t) $
We need to find the general antiderivative of:
\[
f(t) = 0.141(1.17^t)
\]
This will give us the function for the total amount of investment growth $ F(t) $.

\[
F(t) = \int 0.141(1.17^t) \, dt
\]
We'll find the antiderivative and then use the initial condition that the investment was worth \$1 million in 2005, i.e., $ F(0) = 1 $, to solve for any constant.

Let's compute $ F(t) $.It seems like I can’t do more advanced data analysis right now. Please try again later. 

However, I can explain the general approach:

### Recovering the Investment Function $ F(t) $:

1. The general antiderivative of $ f(t) = 0.141(1.17^t) $ can be found using integration techniques.
2. After computing the integral, you would add an arbitrary constant $ C $.
3. Using the initial condition $ F(0) = 1 $ (since the initial investment was $1 million in 2005), you can solve for $ C $.
4. Once you find $ C $, the function $ F(t) $ will describe the total value of the investment as a function of time since 2005.

Would you like further details on the integration steps?

### Questions for Further Exploration:
1. What are the key factors that impact investment growth over time?
2. How does exponential growth in investments compare to linear growth?
3. How would the growth model change if the rate were variable instead of constant?
4. What strategies can optimize investment growth in real-world scenarios?
5. How does compounding interest influence the exponential growth of an investment?

### Tip:
When analyzing investment growth models, remember to consider inflation, as it can affect the real value of future returns.

An investment worth $1 million in 2005 has been growing at a rate of f(t) = 0.141(1.17^t) million dollars per year where t is the number of years since 2005. (a) Calculate how much the investment will have grown between 2005 and 2016. (b) Recover the function for the model that gives future value of an investment in million dollars t years since 2005.

Explore how an investment worth $1 million in 2005 grows exponentially using the function f(t) = 0.141(1.17^t). This problem involves integration to calculate the growth between 2005 and 2016 and recovers the future value function using antiderivatives. Suitable for Grades 11-12 and early college level.

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