Question

Suppose `T_4(x) = 7-3(x-2)+7(x-2)^2-6(x-2)^3+8(x-2)^4` is the 4th-degree Taylor polynomial centered at `x=2` for some function f, what is the value of `f^((3))(2)`?

Answer 1

`-36`

Explanation:

`T_4(x)=(f^((0))(2))/(0!)(x-2)^0+(f^((1))(2))/(1!)(x-2)^1+(f^((2))(2))/(2!)(x-2)^2+(f^((3))(2))/(3!)(x-2)^3+(f^((4))(2))/(4!)(x-2)^4`

By comparing the coefficient of the fourth term,

`Rightarrow (f^((3))(2))/(3!)=-6`

By multiplying both sides by `3!`,

`Rightarrow f^((3))(2)=-6cdot3! =-36`

I hope that this was clear.