Approximately how much greater is the estimated average rate of change of the function y=16⋅4x

Question

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].