The cash price of a car is 970,000 Php. The dealer offers you a loan arrangement of 22,339.50 Php per month payable for 5 years. How much interest is being imposed on this loan? Make a table of f(x) for intervals of 2% for x to determine the initial range of the root. Then, using this initial range, apply false-position method for 3 iterations to approach the root.

The problem involves determining the interest rate on a car loan given monthly payments. We'll treat this as a problem of solving for the interest rate \(x\) in the context of a loan. The monthly payment formula for an amortized loan is:

\[
P = \frac{A \cdot \left(\frac{x}{12}\right)}{1 - \left(1 + \frac{x}{12}\right)^{-n}}
\]

Where:
- \(P = 22,339.50\) Php (monthly payment)
- \(A = 970,000\) Php (loan amount or principal)
- \(n = 5 \times 12 = 60\) (total number of payments, as 5 years = 60 months)
- \(x\) is the annual interest rate (what we are solving for).

We will:
1. Create a function \(f(x)\) that represents the difference between the monthly payment formula and the actual payment.
2. Create a table of \(f(x)\) for different interest rates in 2% intervals to determine the initial range of the root.
3. Use the false-position method to refine the solution for \(x\), performing 3 iterations.

The function \(f(x)\) is:

\[
f(x) = \frac{A \cdot \left(\frac{x}{12}\right)}{1 - \left(1 + \frac{x}{12}\right)^{-60}} - 22,339.50
\]

I'll compute the values for different \(x\) intervals and apply the false-position method for 3 iterations. Let's begin by calculating the values of \(f(x)\) for the interest rate intervals.

Here is the table of \( f(x) \) values for interest rates at 2% intervals:

\[
\begin{array}{|c|c|}
\hline
\text{Interest Rate (x)} & f(x) \\
\hline
0.02 & -5337.57 \\
0.04 & -4475.47 \\
0.06 & -3586.68 \\
0.08 & -2671.40 \\
0.10 & -1729.87 \\
0.12 & -762.39 \\
0.14 & 230.70 \\
0.16 & 1249.02 \\
0.18 & 2292.12 \\
\hline
\end{array}
\]

From this table, the root lies between 12% and 14%, as the function changes sign in this interval. Now, we will apply the **false-position method** using this initial range.

### Steps of False-Position Method:
1. Identify \( x_0 = 0.12 \) and \( x_1 = 0.14 \) as the initial range where the root lies.
2. Use the formula to compute the next approximation:

\[
x_{\text{new}} = x_1 - \frac{f(x_1) \cdot (x_1 - x_0)}{f(x_1) - f(x_0)}
\]

Let's proceed with 3 iterations of the false-position method.The results of applying the false-position method for 3 iterations are as follows:

\[
\begin{array}{|c|c|c|c|c|}
\hline
\text{Iteration} & x_0 & x_1 & x_{\text{new}} & f(x_{\text{new}}) \\
\hline
1 & 0.12 & 0.14 & 0.13535 & -2.264 \\
2 & 0.13535 & 0.14 & 0.13540 & -0.00657 \\
3 & 0.13540 & 0.14 & 0.13540 & -0.00002 \\
\hline
\end{array}
\]

After 3 iterations, the estimated interest rate is approximately **13.54%**.

### Conclusion:
The interest being imposed on this loan is about **13.54%**.

Would you like more details, or do you have any questions?

Here are some related questions:
1. How does the false-position method compare to other root-finding methods?
2. What factors affect the accuracy of the false-position method?
3. How does compound interest influence loan payments?
4. Can interest rates vary during the loan period?
5. How can overpayment impact the total interest paid on a loan?

**Tip:** For faster convergence in the false-position method, choose initial points that are close to the root.

Learn how to calculate the interest rate on a car loan of 970,000 Php with monthly payments of 22,339.50 Php for 5 years. This guide walks you through using the amortization formula and the false-position method to approximate the interest rate to 13.54% over 3 iterations. Suitable for advanced high school and college students.

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