Wednesday October 9th
Last class we started with a POD as usual. We then went over check 2.45 because many people had questions on the last problem. I don't have this to go over but there was a similar problem on the limits worksheet, so I will explain that. Exercise 5 on that worksheet said:
lim (1+h)^4 -1 0 1
h--> 0 h 1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
1(1^4)(h^0) + 4(1^3)(h^1) + 6(1^2)(h^2) + 4(1^1)(h^3) + 1(1^1)(h^4) substitute into function
(1 + 4h + 6h^2 + 4h^3 + h^4) -1 ones cancel, factor h out of polynomial
h
h(4 + 6h + 4h^2 + h^3) h cancels, substitute to find answer
h
4 + 0 + 0 + 0 = 4 final answer
Basically you need to expand the polynomial in the numerator. when expanding anything more than the power of 2 or 3 it is easier and faster to use pascal's triangle. This is review from pre-calc but you use the triangle by looking at the row of the triangle that corresponds with the degree you are raising the polynomial to. Then for each number in the series you multiply the first value in the original polynomial to the highest degree (and in descending order as you move onto the next term) and the second value to the 0th degree which is always one (and in increasing order as you move to the next term). It may be easier to understand looking at the example above.
We also went over part two of the FRQ 2003 #6 which was a helpful to show how the AP wants the answer found when there is no calculator. Then we did the two part exploration. Finally we received our homework...
HOMEWORK DUE FRIDAY: the new FRQ 2008 #2 we got last class as well as check 2.5. No bookwork
Josh Gervais
sorry about the confusion in the limit problem, everyone. It looked normal when I typed it, just try to bear with it
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