Question

NA

Answer 1

is there a question?

Answer 2

you want a polynomial with those conditions?

Answer 3

yes

Answer 4

oh they have gone

Answer 5

damn

Answer 6

i think we can do it

Answer 7

f(2)=-3, f'(2)=5, f''(2)=3, f'''(2)=-8
taylor polynomial for f about x = 2 and use it to approximate f(1.5)

Answer 8

ok we can do this
first off it is going to be degree 3 right?

Answer 9

yes

Answer 10

f(x) = -3 + 5(x-2) + 3(x-2)^2/2! - 8(x-2)^3/3!
this is what im getting.

Answer 11

and we are expanding about \(2\) so it is going to look like
\[a_3(x-2)^2+a_2(x-2)^2+a_1(x-2)+a_0\]

Answer 12

ok

Answer 13

\(a_0=-3\) right?

Answer 14

f(2) = -3 yes

Answer 15

ok so \(f'(x)=3a_3(x-2)^2+2a_2(x-2)+a_1\)

Answer 16

ok

Answer 17

and since \(f'(2)=5\) you get \(a_1=5\)

Answer 18

ok

Answer 19

oh i didn't see
you may have the answer above right?

Answer 20

yes i posted my answer, im wondering what is wrong with that specific answer

Answer 21

\[f''(x)=6a_2(x-2)+2a_2\]

Answer 22

\[f''(2)=2a_2=3\] making \(a_2=\frac{3}{2}\)

Answer 23

ok

Answer 24

and finally \[f'''(x)=6a_3\] and \[f'''(2)=8\] so
\[6a_3=8\] and
\[a_3=\frac{4}{3}\]

Answer 25

oh \(-8\) sorry i didn't read it correctly
so \(a_3=-\frac{4}{3}\)

Answer 26

looks like you got it

Answer 27

ok thanks

Answer 28

yw