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 : 2

Answer 1

To find the average rate of change of each function over the interval [1.1, 1.6], we can use the formula:

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

where \( f(x) \) is the function, \( a = 1.1 \), and \( b = 1.6 \).

1. **Calculating for \( y = 16 \cdot 4^x \):**

First, we need to evaluate \( y \) at \( x = 1.1 \) and \( x = 1.6 \).

\[
f(1.1) = 16 \cdot 4^{1.1}
\]

\[
f(1.6) = 16 \cdot 4^{1.6}
\]

Now calculating these values:

- \( 4^{1.1} \approx 4.64 \) (approximately)
- \( 4^{1.6} \approx 9.24 \) (approximately)

Then,

\[
f(1.1) \approx 16 \cdot 4.64 \approx 74.24
\]

\[
f(1.6) \approx 16 \cdot 9.24 \approx 147.84
\]

Now we compute the average rate of change:

\[
\text{Average Rate of Change} = \frac{147.84 - 74.24}{1.6 - 1.1} = \frac{73.6}{0.5} = 147.2
\]

2. **Calculating for \( y = 16x^2 + 4x \):**

We do the same for this function:

\[
g(1.1) = 16(1.1^2) + 4(1.1) = 16(1.21) + 4.4 \approx 19.36 + 4.4 \approx 23.76
\]

\[
g(1.6) = 16(1.6^2) + 4(1.6) = 16(2.56) + 6.4 \approx 40.96 + 6.4 \approx 47.36
\]

Now, compute the average rate of change:

\[
\text{Average Rate of Change} = \frac{47.36 - 23.76}{1.6 - 1.1} = \frac{23.6}{0.5} = 47.2
\]

3. **Finding the difference:**

Now we find the difference between the two average rates of change:

\[
\text{Difference} = 147.2 - 47.2 = 100
\]

Thus, the estimated average rate of change of the function \( y = 16 \cdot 4^x \) is approximately **100** greater than that of the function \( y = 16x^2 + 4x \) over the interval [1.1, 1.6].