Approximately how much greater 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?(1 point)
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]
.

1 answer

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 will use the formula for the average rate of change of a function \( f \) over the interval \([a, b]\):

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

Step 1: Calculate the average rate of change for \( y = 16 \cdot 4^x \)

Let \( f(x) = 16 \cdot 4^x \).

  1. Calculate \( f(1.1) \): \[ f(1.1) = 16 \cdot 4^{1.1} \approx 16 \cdot 4^{1.1} \approx 16 \cdot 4.6416 \approx 74.2656 \]

  2. Calculate \( f(1.6) \): \[ f(1.6) = 16 \cdot 4^{1.6} \approx 16 \cdot 4^{1.6} \approx 16 \cdot 10.0794 \approx 161.271 \]

  3. Compute the average rate of change: \[ \text{Average Rate of Change} = \frac{f(1.6) - f(1.1)}{1.6 - 1.1} = \frac{161.271 - 74.2656}{0.5} = \frac{87.0054}{0.5} \approx 174.0108 \]

Step 2: Calculate the average rate of change for \( y = 16x^2 + 4x \)

Let \( g(x) = 16x^2 + 4x \).

  1. Calculate \( g(1.1) \): \[ g(1.1) = 16 \cdot (1.1)^2 + 4 \cdot (1.1) = 16 \cdot 1.21 + 4.4 = 19.36 + 4.4 = 23.76 \]

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

  3. Compute the average rate of change: \[ \text{Average Rate of Change} = \frac{g(1.6) - g(1.1)}{1.6 - 1.1} = \frac{47.36 - 23.76}{0.5} = \frac{23.6}{0.5} = 47.2 \]

Step 3: Calculate the difference

Now we can find how much greater the estimated average rate of change of the first function is than that of the second function:

\[ \text{Difference} = 174.0108 - 47.2 \approx 126.8108 \]

Conclusion

The difference is approximately 127, which suggests that it is closer to 100 than to 10 or 1. Thus, the estimated average rate of change of the function \( y=16 \cdot 4^x \) is approximately 100 units greater than that of \( y=16x^2 + 4x \) over the interval \([1.1, 1.6]\).