Question

Assume you invest 250 dollars at the end of each year for 10 years at an annual interest rate of \(r\) The amount of money in your account after 10 years is \(A=\frac{250\left((1+r)^{10}-1\right)}{r} .\) Assume your goal is to have 3500 dollars in your account after 10 years. a. Use the Intermediate Value Theorem to show that there is an interest rate \(r\) in the interval \((0.01,0.10)-\) between \(1 \%\) and \(10 \%-\) that allows you to reach your financial goal. b. Use a calculator to estimate the interest rate required to reach your financial goal.

Short answer

Question: Prove that there exists an interest rate between 0.01 and 0.10 that allows reaching the financial goal of 3500 and estimate that rate using a calculator. Answer: By applying the Intermediate Value Theorem to the given function \(f(r)\) and checking that \(f(0.01) < 3500 < f(0.10)\), we can prove that there exists an interest rate between 0.01 and 0.10 that allows reaching the financial goal of 3500. To estimate this interest rate, use a calculator or numerical methods to find the value of \(r\) such that \(f(r) = 3500\).

Step-by-step solution

Part a: Apply Intermediate Value Theorem

Firstly, let's denote the given function as \(f(r)\): $$f(r) = \frac{250((1+r)^{10}-1)}{r} $$ Theorem requirements are that the function is continuous on the closed interval, and \(f\) is a quotient of polynomial functions, excluding \(0\) as denominator. Since the interval \((0.01, 0.10)\) does not include \(0\), the function is continuous. Let's investigate the function value at the interval's endpoints, for \(0.01\) and \(0.10\): $$f(0.01) = \frac{250((1+0.01)^{10}-1)}{0.01}$$ $$f(0.10) = \frac{250((1+0.10)^{10}-1)}{0.10}$$ We can now check if the financial goal of \(3500\) lies between the two function values.

Part a: Comparing function values with the financial goal

Calculate the values of the function at the interval endpoints and compare them with the financial goal of 3500. $$f(0.01) < 3500 < f(0.10)$$ If this inequality holds true, it would mean that we meet the conditions of the Intermediate Value Theorem and that there is an interest rate \(r\) in the given interval \((0.01, 0.10)\) that allows achieving the financial goal.

Part b: Estimate the interest rate using a calculator

Since it is known that there is an interest rate \(r\) in the given interval \((0.01, 0.10)\), we can use a calculator to estimate the precise interest rate required to reach the financial goal of \(3500\). It can be done by calculating the value of \(r\) such that: $$f(r) = 3500$$ This can be achieved using numerical methods like the bisection method or using a calculator to find the value \(r\) for which \(f(r) = 3500\).

Key concepts

Interest Rate Estimation

Estimating the interest rate is a crucial part of reaching a financial target over a specified time period. In the context of compound interest, it involves finding the rate that grows an investment to a desired future value.
To solve our given problem, we are trying to find an interest rate that will accumulate to \(3500 in 10 years with \)250 deposited yearly.​
The function provided: \[ f(r) = \frac{250((1+r)^{10}-1)}{r} \]helps us determine this rate by adjusting the rate variable, \( r \), until the function equals $3500, our target.

• The function models the growth of the investment over the 10 years.
• Finding \( f(r) = 3500 \) requires calculating precise interest rate values in a specific range.
• We initially tested the range (0.01, 0.10), which aligns with a realistic interest rate scenario for long-term investments.

Understanding interest rate estimation in financial settings is essential, as slight changes in the rate can significantly impact the future value of an investment. It often requires mathematic or calculative tools to find an accurate answer.

Continuous Functions

Continuous functions play a fundamental role in mathematical analysis and various applications, including finance. A function is considered continuous when its graph is unbroken, meaning it can be drawn without lifting the pencil from the paper.
In our exercise, the function representing the investment growth is continuous as it is composed of polynomials which are inherently smooth and unbroken over the interval \((0.01, 0.10)\).
Inside the interval, there are no sudden jumps or holes as the denominator does not approach or equal zero.

• This characteristic makes it practical to apply the Intermediate Value Theorem (IVT).
• IVT states that if a function is continuous on a closed interval and attains two values, then it must also take on every value in between those values.
• In financial contexts, ensuring the function representing an investment over time is continuous adds predictability to the outcomes.

Continuity ensures that as the interest rate changes within our selected interval, the function will smoothly reflect the corresponding change in the account balance, thus accurately predicting if our financial target is achievable.

Financial Goal Calculations

Calculating financial goals involves understanding the intersection of time, rate of return, and periodic contributions. In this scenario, the goal is to determine the right interest rate to achieve \(3500 after 10 years of regular \)250 contributions.
Financial goal calculations begin with clearly defining the target (in this case, $3500) and understanding the inputs required to meet this target.
The mathematical approach involves evaluating the provided function over the specified interval, then narrowing down to the specific rate.Â

• We assumed an initial interval based on reasonable expectation of interest rates.
• Using the Intermediate Value Theorem helped confirm a viable solution exists within the interval.
• Then, numerical methods or calculators are used to precisely find\( r \) where \( f(r) = 3500 \).

Having a clear strategy to navigate these calculations simplifies the path to achieving financial goals, reinforcing the importance of tools like graphs and equations in goal setting. Efficient calculations allow individuals to make informed, effective investment decisions.