Okay so here's a link http://apcentral.collegeboard.com/apc/public/repository/ap10_calculus_bc_scoring_guidelines.pdf I get the first part of part a, but not the second part where it asks to write terms for the "f" function... can someone explain?

Question

ganeshie8 · Answer

i see many questions in that link
which question u talking about ha ?

anonymous · Answer

Second part of part A

anonymous · Answer

Am I taking derivatives of cosx-1/x^2 or of -1/2? but then the latter is just zero...

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=taylor+series+%28cosx-1%29%2Fx%5E2

ganeshie8 · Answer

you're taking taylor series about 0, or around 0.

$\lim \limits_{x \to 0} \frac{\cos x - 1}{x^2} = \frac{-1}{2}$

ganeshie8 · Answer

the given piecewise function is just to make the domain valid for all real numbers

ganeshie8 · Answer

$\large \cos x =  1 - \frac{x^2}{2} + \frac{x^4}{4!} - ... $

ganeshie8 · Answer

$\large f(x) =  \frac{\cos x - 1}{x^2} =   \frac{1 - \frac{x^2}{2} + \frac{x^4}{4!} - ...  -1}{x^2}$

ganeshie8 · Answer

$\large=   \frac{- \frac{x^2}{2} + \frac{x^4}{4!} - ...  }{x^2}$

ganeshie8 · Answer

cancel x^2

ganeshie8 · Answer

see if that makes some sense :)

anonymous · Answer

wait what did you cancel x^2 with?

ganeshie8 · Answer

$\large=   \frac{- \frac{x^2}{2} + \frac{x^4}{4!} - ...  }{x^2} $

ganeshie8 · Answer

notice that u can factor x^2 in the numerator

ganeshie8 · Answer

$\large=   \frac{- \frac{x^2}{2} + \frac{x^4}{4!} - ...  }{x^2} $


$\large=   \frac{x^2(- \frac{1}{2} + \frac{x^2}{4!} - ...)  }{x^2} $

ganeshie8 · Answer

x^2 cancels out in numerator and denominator

ganeshie8 · Answer

leaving u wid  :

$\large=   - \frac{1}{2} + \frac{x^2}{4!} - ... $

anonymous · Answer

ok thank you! and also, why does -1 get tacked on at the end and not a part of every term?

ganeshie8 · Answer

good question :)

ganeshie8 · Answer

$\large f(x) =  \frac{\color{red}{\cos x} - 1}{x^2} $

ganeshie8 · Answer

right ?

ganeshie8 · Answer

u just replace $\color{red}{\cos x}$ with its taylor series equivalent

ganeshie8 · Answer

$\large f(x) =  \frac{\color{red}{\cos x} - 1}{x^2} $

$\large f(x) =  \frac{\color{red}{1 - \frac{x^2}{2} + \frac{x^4}{4!} - ...} - 1}{x^2} $

ganeshie8 · Answer

the 1 in taylor series of cosx cancels with the previously existing -1 on numerator

ganeshie8 · Answer

right ?

anonymous · Answer

oh yeah! thank you so much

ganeshie8 · Answer

np.. . u wlc :)

anonymous · Answer

can you or anyone else help me with part c? sorryyyy haha