Question

How do you do these problems?

Answer 1

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