(b) the taylor series is not alternating when x < 32, so we cant use the alternating series estimation…

(b) the taylor series is not alternating when x < 32, so we cant use the alternating series estimation theorem in this example. but we can use taylors |r₂(x)| ≤ m/3! |x - 32|³ where |f(x)| ≤ m. because x ≥ 31, we have x¹⁴/⁵ ≥ 31¹⁴/⁵, and so f(x) = 1/x¹⁴/⁵ ≤ 1/31¹⁴/⁵ < 0.0000192 (rounded to seven decimal places). therefore we can take m = 0.0000192. also 31 ≤ x ≤ 33, so -1 ≤ x - 32 ≤ 1 and |x - 32| ≤ 1. then taylors inequality gives |r₂(x)| ≤ 0.0000192/3! · 1³ = 0.0000192/6 < 0.0000032 (rounded to seven decimal places). thus, if 31 ≤ x ≤ 33, the approximation in part (a) is accurate to within 0.0000032 (rounded to seven decimal places). need help? read it

(b) the taylor series is not alternating when x < 32, so we cant use the alternating series estimation theorem in this example. but we can use taylors |r₂(x)| ≤ m/3! |x - 32|³ where |f(x)| ≤ m. because x ≥ 31, we have x¹⁴/⁵ ≥ 31¹⁴/⁵, and so f(x) = 1/x¹⁴/⁵ ≤ 1/31¹⁴/⁵ < 0.0000192 (rounded to seven decimal places). therefore we can take m = 0.0000192. also 31 ≤ x ≤ 33, so -1 ≤ x - 32 ≤ 1 and |x - 32| ≤ 1. then taylors inequality gives |r₂(x)| ≤ 0.0000192/3! · 1³ = 0.0000192/6 < 0.0000032 (rounded to seven decimal places). thus, if 31 ≤ x ≤ 33, the approximation in part (a) is accurate to within 0.0000032 (rounded to seven decimal places). need help? read it

Answer

Explanation:

Step1: Identify the Taylor - remainder formula

We use the Taylor's Inequality (|R_{n}(x)|\leq\frac{M}{n!}|x - a|^{n + 1}), here (n = 2), (a=32), so (|R_{2}(x)|\leq\frac{M}{3!}|x - 32|^{3}).

Step2: Find an upper - bound for (|f'''(x)|)

Given (x\geq31), for (f'''(x)=\frac{1}{x^{14/5}}), we know that (x^{14/5}\geq31^{14/5}), so (|f'''(x)|\leq\frac{1}{31^{14/5}}\approx0.0000192), and we take (M = 0.0000192).

Step3: Determine the range of (|x - 32|)

Since (31\leq x\leq33), then (- 1\leq x - 32\leq1), so (|x - 32|\leq1).

Step4: Calculate the upper - bound of (|R_{2}(x)|)

Substitute (M = 0.0000192), (n = 2), and (|x - 32|\leq1) into the Taylor's Inequality: (|R_{2}(x)|\leq\frac{0.0000192}{3!}\times1^{3}=\frac{0.0000192}{6}\approx0.0000032).

Answer:

The approximation is accurate to within (0.0000032).