Answer:
See the explanation for the answer.
Step-by-step explanation:
Given function:
![f(x) = x^{1/4}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%5E%7B1%2F4%7D)
The n-th order Taylor polynomial for function f with its center at a is:
![p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}](https://tex.z-dn.net/?f=p_%7Bn%7D%28x%29%20%3D%20f%28a%29%20%2B%20f%27%28a%29%20%28x-a%29%2B%5Cfrac%7Bf%27%27%28a%29%7D%7B2%21%7D%20%28x-a%29%5E%7B2%7D%20%2B...%2B%5Cfrac%7Bf%5E%7B%28n%29%7Da%7D%7Bn%21%7D%20%28x-a%29%5E%7Bn%7D)
As n = 3 So,
![p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}](https://tex.z-dn.net/?f=p_%7B3%7D%28x%29%20%3D%20f%28a%29%20%2B%20f%27%28a%29%20%28x-a%29%2B%5Cfrac%7Bf%27%27%28a%29%7D%7B2%21%7D%20%28x-a%29%5E%7B2%7D%20%2B...%2B%5Cfrac%7Bf%5E%7B%283%29%7Da%7D%7B3%21%7D%20%28x-a%29%5E%7B3%7D)
![p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}](https://tex.z-dn.net/?f=p_%7B3%7D%28x%29%20%3D%20f%28a%29%20%2B%20f%27%28a%29%20%28x-a%29%2B%5Cfrac%7Bf%27%27%28a%29%7D%7B2%21%7D%20%28x-a%29%5E%7B2%7D%20%2B...%2B%5Cfrac%7Bf%5E%7B%283%29%7Da%7D%7B6%7D%20%28x-a%29%5E%7B3%7D)
![p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} } (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} + (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}](https://tex.z-dn.net/?f=p_%7B3%7D%28x%29%20%3D%20a%5E%7B1%2F4%7D%20%2B%20%5Cfrac%7B1%7D%7B4a%5E%7B%203%2F4%7D%20%7D%20%20%28x-a%29%2B%20%28%5Cfrac%7B1%7D%7B2%7D%29%28-%5Cfrac%7B3%7D%7B16a%5E%7B7%2F4%7D%20%7D%20%29%20%28x-a%29%5E%7B2%7D%20%2B%20%20%28%5Cfrac%7B1%7D%7B6%7D%29%28%5Cfrac%7B21%7D%7B64a%5E%7B11%2F4%7D%20%7D%20%29%20%28x-a%29%5E%7B3%7D)
![p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} } (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} + (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}](https://tex.z-dn.net/?f=p_%7B3%7D%28x%29%20%3D%2081%5E%7B1%2F4%7D%20%2B%20%5Cfrac%7B1%7D%7B4%2881%29%5E%7B%203%2F4%7D%20%7D%20%20%28x-81%29%2B%20%28%5Cfrac%7B1%7D%7B2%7D%29%28-%5Cfrac%7B3%7D%7B16%2881%29%5E%7B7%2F4%7D%20%7D%20%29%20%28x-81%29%5E%7B2%7D%20%2B%20%20%28%5Cfrac%7B1%7D%7B6%7D%29%28%5Cfrac%7B21%7D%7B64%2881%29%5E%7B11%2F4%7D%20%7D%20%29%20%28x-81%29%5E%7B3%7D)
= 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6
(0.0000018522752) (x-81)³
= 0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254
(x-81)³ + 2.25
Hence approximation at given quantity i.e.
x = 94
Putting x = 94
= 0.0092592593 (94) - 0.000042866941 (94 - 81)² +
0.00000030871254 (94-81)³ + 2.25
= 0.87037
03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +
2.25
= 0.87037
03742 - 0.000042866941 (169) +
0.00000030871254(2197) + 2.25
= 0.87037
03742 - 0.007244513029 + 0.0006782414503 + 2.25
= 3.113804102621
Compute the absolute error in the approximation assuming the exact value is given by a calculator.
Compute
as
using calculator
Exact value:
(94) = 3.113737258478
Compute absolute error:
Err = | 3.113804102621 - 3.113737258478 |
Err (94) = 0.000066844143
If you round off the values then you get error as:
|3.11380 - 3.113737| = 0.000063
Err (94) = 0.000063
If you round off the values up to 4 decimal places then you get error as:
|3.1138 - 3.1137| = 0.0001
Err (94) = 0.0001