Question
Approximately how much grater is the estimated average rate of change of the function y = 16 • 4^x over the interval [1.1, 1.6] than the estimated average rate of change of the function y = 16x^2 + 4x over the same interval: 1, 10, or 100?
The estimated average rate of change of the function y = 16 • 4^x is approximately _ greater than the estimated average rate of change of the function y = 16x^2 + 4x over the interval [1.1, 1.6].
The estimated average rate of change of the function y = 16 • 4^x is approximately _ greater than the estimated average rate of change of the function y = 16x^2 + 4x over the interval [1.1, 1.6].
Answers
Answered by
GPT-4o mini
To find the estimated average rate of change of the functions \(y = 16 \cdot 4^x\) and \(y = 16x^2 + 4x\) over the interval \([1.1, 1.6]\), we use the formula for the average rate of change over an interval \([a, b]\), given by:
\[
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
\]
### Step 1: Calculate the average rate of change for \(y = 16 \cdot 4^x\)
Let’s calculate \(f(1.1)\) and \(f(1.6)\) for \(f(x) = 16 \cdot 4^x\):
\[
f(1.1) = 16 \cdot 4^{1.1}\approx 16 \cdot 4^{1.1} \approx 16 \cdot 4 \cdot 4^{0.1} \approx 16 \cdot 4 \cdot 1.1487 \approx 73.517
\]
\[
f(1.6) = 16 \cdot 4^{1.6} \approx 16 \cdot 4^{1.6} \approx 16 \cdot 4 \cdot 4^{0.6} \approx 16 \cdot 4 \cdot 2.297 \approx 146.253
\]
Now calculate the average rate of change over \([1.1, 1.6]\):
\[
\text{Average Rate of Change} = \frac{f(1.6) - f(1.1)}{1.6 - 1.1} = \frac{146.253 - 73.517}{0.5} \approx \frac{72.736}{0.5} \approx 145.472
\]
### Step 2: Calculate the average rate of change for \(y = 16x^2 + 4x\)
Let’s calculate \(g(1.1)\) and \(g(1.6)\) for \(g(x) = 16x^2 + 4x\):
\[
g(1.1) = 16(1.1^2) + 4(1.1) = 16(1.21) + 4(1.1) = 19.36 + 4.4 = 23.76
\]
\[
g(1.6) = 16(1.6^2) + 4(1.6) = 16(2.56) + 4(1.6) = 40.96 + 6.4 = 47.36
\]
Now calculate the average rate of change over \([1.1, 1.6]\):
\[
\text{Average Rate of Change} = \frac{g(1.6) - g(1.1)}{1.6 - 1.1} = \frac{47.36 - 23.76}{0.5} = \frac{23.6}{0.5} = 47.2
\]
### Step 3: Finding the difference in average rates of change
Now we subtract the two average rates of change:
\[
\text{Difference} = 145.472 - 47.2 \approx 98.272
\]
### Conclusion
The estimated average rate of change of the function \(y = 16 \cdot 4^x\) is approximately 98.272 greater than the estimated average rate of change of the function \(y = 16x^2 + 4x\) over the interval \([1.1, 1.6]\).
Thus, the closest answer to the difference is **100**.
\[
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
\]
### Step 1: Calculate the average rate of change for \(y = 16 \cdot 4^x\)
Let’s calculate \(f(1.1)\) and \(f(1.6)\) for \(f(x) = 16 \cdot 4^x\):
\[
f(1.1) = 16 \cdot 4^{1.1}\approx 16 \cdot 4^{1.1} \approx 16 \cdot 4 \cdot 4^{0.1} \approx 16 \cdot 4 \cdot 1.1487 \approx 73.517
\]
\[
f(1.6) = 16 \cdot 4^{1.6} \approx 16 \cdot 4^{1.6} \approx 16 \cdot 4 \cdot 4^{0.6} \approx 16 \cdot 4 \cdot 2.297 \approx 146.253
\]
Now calculate the average rate of change over \([1.1, 1.6]\):
\[
\text{Average Rate of Change} = \frac{f(1.6) - f(1.1)}{1.6 - 1.1} = \frac{146.253 - 73.517}{0.5} \approx \frac{72.736}{0.5} \approx 145.472
\]
### Step 2: Calculate the average rate of change for \(y = 16x^2 + 4x\)
Let’s calculate \(g(1.1)\) and \(g(1.6)\) for \(g(x) = 16x^2 + 4x\):
\[
g(1.1) = 16(1.1^2) + 4(1.1) = 16(1.21) + 4(1.1) = 19.36 + 4.4 = 23.76
\]
\[
g(1.6) = 16(1.6^2) + 4(1.6) = 16(2.56) + 4(1.6) = 40.96 + 6.4 = 47.36
\]
Now calculate the average rate of change over \([1.1, 1.6]\):
\[
\text{Average Rate of Change} = \frac{g(1.6) - g(1.1)}{1.6 - 1.1} = \frac{47.36 - 23.76}{0.5} = \frac{23.6}{0.5} = 47.2
\]
### Step 3: Finding the difference in average rates of change
Now we subtract the two average rates of change:
\[
\text{Difference} = 145.472 - 47.2 \approx 98.272
\]
### Conclusion
The estimated average rate of change of the function \(y = 16 \cdot 4^x\) is approximately 98.272 greater than the estimated average rate of change of the function \(y = 16x^2 + 4x\) over the interval \([1.1, 1.6]\).
Thus, the closest answer to the difference is **100**.