#What value to use for c
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so shouldnt the error be (f'''(61)(61-64)^3)/6
the t greater than 49 ?
well yes, thats what they eventually do
t greater than or equal to 4 because we got approx 7.810 rounding it down to 7, and we were in square root, so now we square to 49
so you would get the correct answer if you put 61 instead
I think we would get a more accurate answer, but calculation would get worse
64 and 49 are perfect squares so good to approximate within
so it doesn't matter which number you choose from the interval, is it best to pick the biggest number to minimize the error range, or the number that is best to calculate with?
depends, while approximating you probably won't have calculator, if you do, approximating with 61 is a good idea, otherwise I'd say 49