Final answer:
The average rate of change of the skateboarder's distance when riding along the ramp modeled by \(f(x)=\frac{8}{(x-4)^2}+5\) from x=1 foot to x=2 feet is \(\frac{10}{9}\), which is approximately 1.11.
Step-by-step explanation:
To calculate the average rate of change of a function, we use the formula \(\frac{f(x_2) - f(x_1)}{x_2 - x_1}\), where \(x_1\) and \(x_2\) are the points between which we are finding the average rate of change. Here, we are calculating the average rate of change from \(x=1\) foot to \(x=2\) feet for the function \(f(x)=\frac{8}{(x-4)^2}+5\).
We first find the values of the function at these points:
- \(f(1) = \frac{8}{(1-4)^2}+5 = \frac{8}{9}+5\)
- \(f(2) = \frac{8}{(2-4)^2}+5 = \frac{8}{4}+5 = 2+5\)
Now we can calculate the average rate of change:
\(\frac{f(2)-f(1)}{2-1} = \frac{(7 - (\frac{8}{9}+5))}{1} = \frac{7 - \frac{8}{9} - 5}{1} = \frac{7 - \frac{53}{9}}{1} = \frac{7\times9 - 53}{9} = \frac{63 - 53}{9} = \frac{10}{9}\)
Thus, the average rate of change of the skateboarder's distance as they ride along the ramp from \(x=1\) foot to \(x=2\) feet is \(\frac{10}{9}\) or approximately 1.11 (to two decimal places).