Taylor polynomial??

Question

Determine the third taylor polynomial of 1/(5-x) for x=4.

Answer 1

f(x) = 1/(5-x) = (5-x)^(-1), so f(4) = 1,
f'(x) = (5-x)^(-2) [chain rule is easier than quotient rule here], so f'(4) = 1,
f''(x) = 2(5-x)^(-3), so f''(4) = 2,
f'''(x) = 6(5-x)^(-4), so f'''(4) = 6.

Taylor sez;

f(4) + f'(4)*(x-4) + f''(4)/2!*(x-4)^2 + f'''(4)/3!*(x-4)^3 +...

For this one, that's

1 + (1/1)(x-4) + (2/2)(x-4)^2 + (6/6)(x-4)^3, and you stop at 3 for the third degree.

Answer 2

f(x)= (5-x)^-1 f(4)=1

f´(x) =(5-x)^-2 f´(4)=1
f´´(x) =2(5-x)^-3 f´´(4) = 2
f´´´(x) = 6(5-x)^-4 f´´´(4)=6

f(x) =1 +(x-4) +(x-4)^2 +(x-4)^3

Answer 3

i think you mnean taylor polynome until degree is 3 ?

or do a long polynomial division
or use differentiation ( more work i guess )

5-x / 1 \ x^-1
.........1 + 5^x-1
------------------
...........1-5^x_1

etc