Step-by-step explanation:
Find the next term of the Taylor series.
f⁽⁰⁾(x) = x^⅓
f⁽¹⁾(x) = ⅓ x^-⅔
f⁽²⁾(x) = -²/₉ x^-⁵/₃
f⁽³⁾(x) = ¹⁰/₂₇ x^-⁸/₃
f⁽⁴⁾(x) = -⁸⁰/₈₁ x^-¹¹/₃
So the fourth term would be:
(-⁸⁰/₈₁ z^-¹¹/₃) (x − 1)⁴ / 4!
For 0.8 ≤ z ≤ 1.2, │f⁽⁴⁾(z)│is a maximum at z = 0.8. Therefore:
│R₃(x)│≤ │(-⁸⁰/₈₁ (0.8)^-¹¹/₃) (0.8 − 1)⁴ / 4!│
│R₃(x)│≤ 0.00014923
Looks like you accidentally wrote an extra zero.
