Answer:
18e⁶
⁵/₁₂₈
Step-by-step explanation:
Rₙ(x) = f⁽ⁿ⁺¹⁾(c) / (n+1)! (x − a)ⁿ⁺¹, and a < c < x.
f(x) = eˣ, a = 0, and n = 1. Thus R₁ is:
R₁(x) = f"(z)/2! x²
R₁(x) = eᶻ/2 x²
|R₁| is a maximum when |f"(z)| is a maximum. On the domain 0 < z < 6, that maximum is e⁶. At x = 6, the upper bound of |R₁| is:
|R₁| = 18e⁶
This time, f(x) = 1 / √(1 + x) = (1 + x)^-½. a = 0, and n = 2.
R₂(x) = f⁽³⁾(z)/3! x³
Find f⁽³⁾(x):
f'(x) = -½ (1 + x)^-³/₂
f"(x) = ¾ (1 + x)^-⁵/₂
f⁽³⁾(x) = -¹⁵/₈ (1 + x)^-⁷/₂
On the domain -½ < z < 0, |f⁽³⁾(z)| is a maximum at z = 0.
|f⁽³⁾(z)| = ¹⁵/₈
Therefore, at x = -½, the upper bound of R₂ is:
|R₂| = (¹⁵/₈)/6 |(-½)³|
|R₂| = ⁵/₁₂₈
