if f(x) = (x-2)^(1/3)
use Tylor's polynomial of degree two about c=3 to estimate the value of (1.36)^(1/3)
thanks
To estimate the value of (1.36)^(1/3) using Taylor's polynomial of degree two about c=3, we need to calculate the polynomial approximation and evaluate it at x=1.36.
The Taylor's polynomial of degree n for a function f(x) centered at c is given by:
P(x) = f(c) + f'(c)(x - c) + f''(c)(x - c)^2/2! + ... + f^n(c)(x - c)^n/n!
In this case, we need to find the Taylor's polynomial of degree two for f(x) = (x-2)^(1/3) centered at c=3.
First, let's find the derivatives of f(x):
f'(x) = (1/3)(x-2)^(-2/3)
f''(x) = (-2/3)(-1/3)(x-2)^(-5/3)
Next, we evaluate each derivative at c=3:
f(3) = (3-2)^(1/3) = 1^(1/3) = 1
f'(3) = (1/3)(3-2)^(-2/3) = (1/3)(1)^(-2/3) = 1/3
f''(3) = (-2/3)(-1/3)(3-2)^(-5/3) = (-2/3)(-1/3)(1)^(-5/3) = 2/27
Now we can substitute these values into the Taylor's polynomial equation:
P(x) = f(3) + f'(3)(x - 3) + f''(3)(x - 3)^2/2!
Simplifying:
P(x) = 1 + (1/3)(x - 3) + (2/27)(x - 3)^2
To estimate the value of (1.36)^(1/3), we substitute x=1.36 into the polynomial:
P(1.36) = 1 + (1/3)(1.36 - 3) + (2/27)(1.36 - 3)^2
Now we can calculate the value of P(1.36) to obtain our estimate.