Question

f

Answer 1

@Hero @UnkleRhaukus @TuringTest @mukushla

Answer 2

what have you got so far

Answer 3

i did part a and b but i dont get how to prove this one

Answer 4

what have you got as the 6th degree Taylor polynomial ?

Answer 5

45(x-5)^5/28

Answer 6

so what is this evaluated at f(6)

Answer 7

45/28

Answer 8

what was f before it got converted to Taylor

Answer 9

i dont think it gives it to us

Answer 10

yeah then im confused too,
im not sure what to compare 45/28 to.

Answer 11

i think i might have messed up the taylor let me try it again

Answer 12

actually i got 45/32

Answer 13

@nphuongsun93

Answer 14

What did you get for parts A, B

Answer 15

Part b i got 0 and a 1/2-1/6(x-5) + 1/8(x-5)^2-3/20(x-5)^3

Answer 16

D:
i don't know how to do these.. but i found a solution sheet on google.
< http://www.math.ksu.edu/~dbski/calculus/chapter9_testbank_solutions.pdf >
problem 2 part II

Answer 17

oh my i got those two wrong

Answer 18

how did you find that?

Answer 19

actually i dont think that is right.

Answer 20

in the pdf i post, for A, i think there's a typo f(5) = 1/2 not 1/5

Answer 21

yeah and the 3rd term

Answer 22

it says (x-2) not (x-5)

Answer 23

yea -.-
< http://blogs.friendscentral.org/jpcalc/files/2011/02/solutions-to-ap-series-87889095969800.pdf > new one

Answer 24

Haha yeah I just saw that one too. It makes sense. thank you

Answer 25

\[P_3(x)=f(5)+f'(5)(x-5)+\frac{f'''(5)}{2}(x-5)^2+\frac{f'''(5)}{3!}(x-5^3)\]

Answer 26

\[f^n(5)=\frac{{(-1)^nn!}}{2^n(n+2)}\]
\[f(5)=\frac{{(-1)^00!}}{2^0(0+2)}=\frac1{1\times2}=\frac12\]
\[f'(5)=\frac{{(-1)^11!}}{2^1(1+2)}=\frac{-1}{2\times3}=-\frac {1}{6}\]
\[f''(5)=\frac{{(-1)^22!}}{2^2(2+2)}=\frac{{2}}{4\times4}=\frac1{16}\]
\[f'''(5)=\frac{{(-1)^33!}}{2^3(3+2)}=\frac{{-6}}{8\times5}=-\frac6{40}\]

Answer 27

\[-6/40=-3/20\]

Answer 28

oh yeah
\[f''(5)=2/16=1/8\]

Answer 29

on the answer sheet it says the last term is 1/40

Answer 30

thats wrong im assuming

Answer 31

oh no you have to multiply by the factorial

Answer 32

@UnkleRhaukus I got it. You dont need to further explain

Answer 33

\[P_3(x)=f(5)+f'(5)(x-5)+\frac{f''(5)}{2}(x-5)^2+\frac{f'''(5)}{3!}(x-5^3)\]\[\qquad\downarrow\]\[\quad=\frac 12-\frac{1}6(x-5)+\frac 1{8}(x-5)^2-\frac3{20}(x-5)^3\]\[...\]