To compare the estimated average rate of change of the exponential function \( y = 9(13)^x \) and the quadratic function \( y = 9x^2 + 13x \) over the interval \([0.1, 0.6]\), we can calculate the average rate of change for each function over this interval.
The average rate of change of a function \( f \) over an interval \([a, b]\) is given by:
\[ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \]
For the Exponential Function \( y = 9(13)^x \):
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Compute \( f(0.1) \): \[ f(0.1) = 9(13)^{0.1} \approx 9(1.34986) \approx 12.14874 \]
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Compute \( f(0.6) \): \[ f(0.6) = 9(13)^{0.6} \approx 9(5.2525) \approx 47.2725 \]
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Calculate the average rate of change: \[ \text{Average Rate of Change} = \frac{47.2725 - 12.14874}{0.6 - 0.1} = \frac{35.12376}{0.5} \approx 70.24752 \]
For the Quadratic Function \( y = 9x^2 + 13x \):
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Compute \( f(0.1) \): \[ f(0.1) = 9(0.1)^2 + 13(0.1) = 9(0.01) + 1.3 = 0.09 + 1.3 = 1.39 \]
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Compute \( f(0.6) \): \[ f(0.6) = 9(0.6)^2 + 13(0.6) = 9(0.36) + 7.8 = 3.24 + 7.8 = 11.04 \]
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Calculate the average rate of change: \[ \text{Average Rate of Change} = \frac{11.04 - 1.39}{0.6 - 0.1} = \frac{9.65}{0.5} = 19.3 \]
Summary of Results:
- Average Rate of Change for the Exponential Function: \( \approx 70.25 \) (positive)
- Average Rate of Change for the Quadratic Function: \( 19.3 \) (positive)
Conclusion:
Both functions have positive estimated average rates of change over the interval \([0.1, 0.6]\). Therefore, the answer to the question is:
Neither function has a negative estimated average rate of change over the given interval.