
Andy C. answered 04/22/18
Tutor
4.9
(27)
Math/Physics Tutor
Taylor expansion :
f(8)(x-8)^0/0! +
f'(8) (x-8)^1/1! +
f''(8)(x-8)^2 / 2! +
f'''(8)(x-8)^3/3!
= 4 + 5*(x-8) + -4(x-8)^2/2 + (x-8)^3/6
= 4 + 5(x-8) - 2(x-8)^2 + (1/6)(x-8)^3
The approximation is
4 + 5(0.1) - 2(0.1)^2 + (1/6)(0.1)^3 =
4 + 0.5 - 0.02 + (1/6)(0.001) = 4.48016
Bobby T.
04/22/18