To compare the estimated average rate of change of the two functions \(y = 9(13)x\) (exponential) and \(y = 9x^2 + 13x\) (quadratic) over the interval \([0.1, 0.6]\), we first need to calculate the values of each function at the endpoints of the interval.
Step 1: Calculate the function values at the endpoints
For the exponential function: \[ y = 9(13)x \]
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At \(x = 0.1\): \[ y(0.1) = 9(13)(0.1) = 9(1.3) = 11.7 \]
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At \(x = 0.6\): \[ y(0.6) = 9(13)(0.6) = 9(7.8) = 70.2 \]
For the quadratic function: \[ y = 9x^2 + 13x \]
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At \(x = 0.1\): \[ y(0.1) = 9(0.1^2) + 13(0.1) = 9(0.01) + 1.3 = 0.09 + 1.3 = 1.39 \]
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At \(x = 0.6\): \[ y(0.6) = 9(0.6^2) + 13(0.6) = 9(0.36) + 7.8 = 3.24 + 7.8 = 11.04 \]
Step 2: Calculate the average rate of change
The average rate of change over the interval \([a, b]\) is given by the formula: \[ \frac{f(b) - f(a)}{b - a} \]
For the exponential function: \[ \text{Average Rate} = \frac{y(0.6) - y(0.1)}{0.6 - 0.1} = \frac{70.2 - 11.7}{0.5} = \frac{58.5}{0.5} = 117 \]
For the quadratic function: \[ \text{Average Rate} = \frac{y(0.6) - y(0.1)}{0.6 - 0.1} = \frac{11.04 - 1.39}{0.5} = \frac{9.65}{0.5} = 19.3 \]
Step 3: Analyze the rates
- The average rate of change for the exponential function over the interval \([0.1, 0.6]\) is 117.
- The average rate of change for the quadratic function over the interval \([0.1, 0.6]\) is 19.3.
Conclusion
Both average rates of change are positive. Thus, neither function has a negative estimated average rate of change over the interval \([0.1, 0.6]\).
Final Response: neither function
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