Problem 58 - Calculus Early Transcendentals

Question

You are shopping for a 150,000,30 dollars -year (360 -month) loan to buy a house. The monthly payment is $$m(r)=\frac{150,000(r / 12)}{1-(1+r / 12)^{-360}},$$ where \(r\) is the annual interest rate. Suppose banks are currently offering interest rates between \(6 \%\) and \(8 \%\). a. Use the Intermediate Value Theorem to show there is a value of \(r\) in \((0.06,0.08)-\) an interest rate between \(6 \%\) and \(8 \%-\) that allows you to make monthly payments of 1000 dollars per month. b. Use a graph to illustrate your explanation to part (a). Then determine the interest rate you need for monthly payments of 1000 dollars.

Step-by-Step Solution

Answer

If so, approximately what would that interest rate be? Yes, there exists an interest rate between 0.06 and 0.08 that would allow for a monthly payment of $1000 for a loan amount of $150,000 over 360 months. To find the approximate value of the interest rate, we would graph the function and determine where it intersects with the monthly payment of $1000, which could be found using trial and error or a numerical method.

Step 1: Set up the equation to solve using the Intermediate Value Theorem

We need to find if there exists an interest rate r with a monthly payment of $1000. Given the function: $$m(r)=\frac{150,000(r / 12)}{1-(1+r / 12)^{-360}}$$ We should find a value of r which makes this equation equal to the desired $1000 monthly payment.

Step 2: Check if the function is continuous

The function $$m(r)=\frac{150,000(r / 12)}{1-(1+r / 12)^{-360}}$$ is continuous for all r. Since the domain of r is between 0.06 and 0.08, it is continuous on this interval as well.

Step 3: Find m(0.06) and m(0.08)

We need to find the values of the function at r=0.06 and r=0.08. $$m(0.06)=\frac{150,000(0.06 / 12)}{1-(1+0.06 / 12)^{-360}}$$ $$m(0.08)=\frac{150,000(0.08 / 12)}{1-(1+0.08 / 12)^{-360}}$$ Calculate these values to check if they are less or greater than $1000.

Step 4: Apply the Intermediate Value Theorem

If we find that one of the above values is less than \(1000 and the other greater than \)1000, the Intermediate Value Theorem would indicate that there exists an interest rate r between 0.06 and 0.08 such that the monthly payment is $1000.

Step 5: Graph the function and find r

Plot the function $$m(r)=\frac{150,000(r / 12)}{1-(1+r / 12)^{-360}}$$ for r in the range [0.06, 0.08] and observe where the curve crosses the $1000 payment line. This point will give us the desired interest rate r. Use trial and error or a numerical method to find the precise value of r at this intersection point.

Key Concepts

Continuity of Functions

Continuity in functions is an essential concept in calculus that can be understood by looking at how a function behaves graphically and numerically. A function is said to be continuous if you can draw its graph without lifting your pen from the paper. This implies that small changes in the input (or the variable) of the function result in small changes in the output (or the function's value).
In the context of the given problem, the function for the monthly payment, \[m(r)=\frac{150,000(r / 12)}{1-(1+r / 12)^{-360}}\]is continuous for values of the interest rate \(r\), meaning there are no gaps or jumps in the values of monthly payments as \(r\) varies between 0.06 and 0.08. This smooth and unbroken behavior allows us to use the Intermediate Value Theorem, which states that if a function is continuous on a closed interval \([a, b]\) and takes different values at \(a\) and \(b\), then it must take every value between them at some point within \((a, b)\).
For our case, because the function is continuous between 6% and 8% interest rates, and results in monthly payments that vary within a range including \(1000, the Intermediate Value Theorem ensures that there must be at least one interest rate in this range where the monthly payment is exactly \)1000.

Interest Rate Calculation

Calculating interest rates is a crucial step when trying to manage and evaluate loans effectively. An interest rate determines the additional amount paid on top of the borrowed money, affecting the overall financial cost of the loan.
For the loan in the exercise, finding the interest rate that results in the desired monthly payment of \(1000 requires us to tweak the rate \( r \) in our continuous payment function until the result meets this value. Initially, we calculated the payments at the boundaries of the interest rate interval, \( r = 0.06 \) and \( r = 0.08 \), by substituting these values into the function:- For \( m(0.06) \), using the formula gives a specific monthly payment.- For \( m(0.08) \), using the formula gives another specific monthly payment.If one of these calculated payments is less than \)1000 and the other is more, it confirms, through the Intermediate Value Theorem, that a suitable interest rate exists between these two values, ensuring a payment of exactly $1000. By iterating with various interest rate values between 6% and 8%, we can systematically narrow down and identify the precise rate required. This rate will provide a convenient balance between affordability and loan repayment duration.

Graphical Analysis in Calculus

Graphical analysis involves using graphs to visually interpret and analyze functions. It is a powerful tool in calculus that helps to understand the behavior of functions over specific intervals.
In this exercise, the graphical method is used to find the interest rate that corresponds to a \(1000 monthly payment by plotting the payment function over the range of interest rates from 6% to 8%:- The curve of this graph represents how the monthly payments change as the interest rate \( r \) varies.- To find the desired interest rate, the crucial step is identifying where this graph intersects the horizontal line at \( y = 1000 \). This intersection indicates the exact interest rate where the payments equal \)1000.
The graph provides a visual means of applying the Intermediate Value Theorem by indicating the exact interval location where the payment hits $1000. Through either manual inspection or using a graphing calculator, you can determine this cross point accurately. Calculators or software can also assist with zooming in and using numerical methods to pinpoint the intersection point with high precision. Overall, graphical analysis not only makes the function behavior tangible but also simplifies identifying solutions in complex situations.