If sin(1/x^2+1) is an anti derivative for f(x), then what is the limit of f(x)dx from 1 to 2?
1 answer:
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Answer:
1, 5, 17, 53, 161
Step-by-step explanation:
Given:
First five terms of the sequence:
Since we know that the graph of our quadratic function has x-intercepts at (6,0) and (18,0), 
and 
are the zeroes of our quadratic. To find our quadratic we are going to factor each zero backwards and multiply them:
Now that we have our quadratic function, we are going to use the average formula:
is the function evaluated at the <span>9th month
</span>
is the function evaluated at the 3rd month
![m= \frac{[9^{2}-24(9)+108]-[3^{2}-24(3)+108]}{9-3}](https://tex.z-dn.net/?f=m%3D%20%5Cfrac%7B%5B9%5E%7B2%7D-24%289%29%2B108%5D-%5B3%5E%7B2%7D-24%283%29%2B108%5D%7D%7B9-3%7D%20)
We can conclude that the closest approximate average rate of change for Edna's profits from the 3rd month to the 9th month is: <span>B. −11.63 dollars per month.</span>
Now that we have our quadratic function, we are going to use the average formula:
</span>
We can conclude that the closest approximate average rate of change for Edna's profits from the 3rd month to the 9th month is: <span>B. −11.63 dollars per month.</span>