To compare the rates of change between the two investment options, we need to calculate the functions at years 9 and 10 and then determine the rate of change for each option.
Option #1: Exponential Growth
The function for Option #1 is: \[ p(x) = 5,000(1.06)^x \]
Calculating for years 9 and 10:
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For year 9: \[ p(9) = 5,000(1.06)^9 \]
Approximating \( (1.06)^9 \): \[ (1.06)^9 \approx 1.715 \] Thus, \[ p(9) \approx 5,000 \times 1.715 \approx 8,575 \]
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For year 10: \[ p(10) = 5,000(1.06)^{10} \]
Approximating \( (1.06)^{10} \): \[ (1.06)^{10} \approx 1.822 \] Thus, \[ p(10) \approx 5,000 \times 1.822 \approx 9,110 \]
Now, calculate the rate of change between years 9 and 10: \[ \text{Rate of Change for Option #1} = p(10) - p(9) = 9,110 - 8,575 = 535 \]
Option #2: Linear Growth
The function for Option #2 is: \[ q(x) = 500x + 5,000 \]
Calculating for years 9 and 10:
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For year 9: \[ q(9) = 500(9) + 5,000 = 4,500 + 5,000 = 9,500 \]
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For year 10: \[ q(10) = 500(10) + 5,000 = 5,000 + 5,000 = 10,000 \]
Now, calculate the rate of change between years 9 and 10: \[ \text{Rate of Change for Option #2} = q(10) - q(9) = 10,000 - 9,500 = 500 \]
Comparison of Rates of Change
- Rate of Change for Option #1: 535
- Rate of Change for Option #2: 500
Conclusion
Since Option #1 has a greater rate of change between years 9 and 10, the answer is:
Option # 1 has the greater rate of change between years 9 and 10. If Miguel plans to invest his money for only 10 years, he should choose Option # 1.
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