Question

Suppose \(\$ 5000\) is invested in a savings account for 10 years ( 120 months), with an annual interest rate of \(r,\) compounded monthly. The amount of money in the account after 10 years is \(A(r)=5000(1+r / 12)^{120}.\) a. Use the Intermediate Value Theorem to show there is a value of \(r\) in \((0,0.08)-\) an interest rate between \(0 \%\) and \(8 \%-\) that allows you to reach your savings goal of \(\$ 7000\) in 10 years. b. Use a graph to illustrate your explanation in part (a); then, approximate the interest rate required to reach your goal.

Short answer

If so, what is the approximate interest rate? Answer: Yes, there exists an interest rate (r) in the interval (0, 0.08) that allows the investment to grow from $5000 to $7000 in 10 years. The approximate interest rate is 4.9%.

Step-by-step solution

Understand the given function and goal

The given function is \(A(r)=5000(1 + \frac{r}{12})^{120}\). The goal is to reach a savings of \(7000 (; A(r) = 7000)\). We want to use the Intermediate Value Theorem to prove that there exists an interest rate \(r\) in \((0,0.08)\) that achieves this goal.

Find values of the function at the endpoints of the interval \((0, 0.08)\)

First, let's find the value of the function \(A(r)\) at the endpoints of the interval \((0, 0.08)\): For \(r=0\), \(A(0)=5000(1 + \frac{0}{12})^{120}=5000\)\ For \(r=0.08\), \(A(0.08)=5000(1 + \frac{0.08}{12})^{120} \approx 7508.63\)

Apply the Intermediate Value Theorem

The Intermediate Value Theorem states that if a continuous function \(f(x)\) is defined on an interval \([a, b]\) and \(k\) is a value between \(f(a)\) and \(f(b)\), then there exists a value \(c\) in the interval \((a, b)\) such that \(f(c) = k\). In our case,\ \(A(r)\) is continuous,\ \(0<r<0.08\), and\ \(7000\) lies between \(A(0)=5000\) and \(A(0.08) \approx 7508.63\).\ According to the Intermediate Value Theorem, there exists an interest rate \(r\) in \((0,0.08)\) such that \(A(r) = 7000\).

Graph the function and approximate the interest rate

Next, plot the function \(A(r)=5000(1 + \frac{r}{12})^{120}\) and find the point where it crosses the horizontal line at \(y=7000\) to obtain an approximate interest rate that reaches the goal. Doing this, we find that the graph intersects at approximately \(r \approx 0.049\) or \(4.9 \%\). This means that an interest rate of about \(4.9 \%\) compounded monthly is needed to reach the savings goal of $7000 in 10 years.

Key concepts

Compounded Interest

When we talk about investments or savings, compounded interest is a popular term. It refers to earning interest not only on the initial amount of money put into an account but also on the interest that accumulates over time. For example, if you start with $100 in a savings account with compounded interest, you'll earn interest in the first month on the $100. Next month, interest is calculated on your new total, which includes the interest from the prior month. This creates a chain reaction of growing your savings, often more quickly than with simple interest.

• Interest is accumulated on both the initial sum and the interest added each time.
• The more frequently interest is compounded, the more you earn. Monthly compounding, like in our problem, yields more than annual compounding.

Using the compound interest formula, helps us see how much our savings can grow over time, especially in accounts such as a savings account.

Savings Account

A savings account is a type of bank account that earns interest, allowing your money to grow at a small but steady rate. It is generally considered a low-risk way to set aside funds for future use. In the context of our problem, a savings account with a fixed deposit of $5000 and compounded interest over 10 years demonstrates how your savings can increase significantly.

• These accounts are accessible and may have a fixed or variable interest rate.
• They are an ideal option for short-term goals due to easy access and dependable returns.
• Generally, there are no lock-in periods making it flexible for withdrawal.

The savings account in our problem showcases how consistent saving and interest compounding can significantly boost total savings.

Interest Rate

The interest rate is an essential factor in determining how much you earn on your savings. It refers to the proportion of a loan or deposit balance that is paid as interest to the lender annually. In simple terms, it is what the bank pays you for keeping money in your account. In our example, an annual interest rate is compounded monthly, which affects how much total interest you earn over time.

• A higher interest rate means more earnings on your savings, while a lower one means less.
• When evaluating savings options, understanding how frequency of compounding affects total return is important.

Our problem involves finding an interest rate between 0% and 8% that will meet the savings goal using compounded interest.

Continuous Function

A continuous function is a mathematical concept where the value of the function changes smoothly without any jumps or breaks. In the context of the Intermediate Value Theorem, continuity ensures that any value between the function's outputs can be achieved with at least one input value.

• The function used to calculate the account balance, being continuous, allows us to use the Intermediate Value Theorem.
• This theorem helps us show there's an interest rate within a specified range that will ensure the goal is met.

Using a continuous function like the one in our savings problem provides assurance that there is a solution, an interest rate that allows reaching a $7000 balance after 10 years.