Question

Approximately how much greater is the estimated average rate of change of the function y=16⋅4x
over the interval [1.1,1.6]
than the estimated average rate of change of the function y=16x2+4x
over the same interval: 1, 10, or 100?(1 point)
The estimated average rate of change of the function y=16⋅4x
is approximately
greater than the estimated average rate of change of the function y=16x2+4x
over the interval [1.1,1.6]
.
Remaining Attempts : 3
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Answer 1

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 can use the formula for average rate of change, which is:

\[
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
\]

where \( a \) and \( b \) are the endpoints of the interval.

### For \( y = 16 \cdot 4^x \):
1. Calculate \( f(1.1) \):
\[
f(1.1) = 16 \cdot 4^{1.1} \approx 16 \cdot 4^{1.1} \approx 16 \cdot 4.57 \approx 73.12
\]

2. Calculate \( f(1.6) \):
\[
f(1.6) = 16 \cdot 4^{1.6} \approx 16 \cdot 4^{1.6} \approx 16 \cdot 10.08 \approx 161.28
\]

3. Now, apply the average rate of change formula:
\[
\text{Average Rate of Change} = \frac{f(1.6) - f(1.1)}{1.6 - 1.1} = \frac{161.28 - 73.12}{0.5} = \frac{88.16}{0.5} \approx 176.32
\]

### For \( y = 16x^2 + 4x \):
1. Calculate \( f(1.1) \):
\[
f(1.1) = 16 \cdot (1.1)^2 + 4 \cdot (1.1) = 16 \cdot 1.21 + 4 \cdot 1.1 = 19.36 + 4.4 = 23.76
\]

2. Calculate \( f(1.6) \):
\[
f(1.6) = 16 \cdot (1.6)^2 + 4 \cdot (1.6) = 16 \cdot 2.56 + 4 \cdot 1.6 = 40.96 + 6.4 = 47.36
\]

3. Now, apply the average rate of change formula:
\[
\text{Average Rate of Change} = \frac{f(1.6) - f(1.1)}{1.6 - 1.1} = \frac{47.36 - 23.76}{0.5} = \frac{23.6}{0.5} \approx 47.2
\]

### Compare the Average Rates of Change:
1. Average rate of change for \( y = 16 \cdot 4^x \): \( \approx 176.32 \)
2. Average rate of change for \( y = 16x^2 + 4x \): \( \approx 47.2 \)

### Finding the Difference:
\[
\text{Difference} = 176.32 - 47.2 \approx 129.12
\]

### Conclusion:
The estimated average rate of change of the function \( y = 16 \cdot 4^x \) is approximately \( 129.12 \) greater than that of \( y = 16x^2 + 4x \).

Since we need to determine if this difference is approximately 1, 10, or 100:
- Clearly, \( 129.12 \) is **greater than** 100.

Thus, the answer is **100** (approximately).