Final answer:
The 4th degree Taylor polynomial for f(x) = xe^x at x_0 = 0 is T_4(x) = x + x^2 + 1/2x^3 + 1/6x^4.
Step-by-step explanation:
To find the 4th degree Taylor polynomial for the function f(x) = xex at x0 = 0, we need to calculate the derivatives of the function up to the fourth derivative and evaluate them at x0. The general form of a Taylor polynomial of degree n at x0 is given by:
Tn(x) = f(x0) + f'(x0)(x - x0) + \frac{f''(x0)}{2!}(x - x0)2 + \dots + \frac{f(n)(x0)}{n!}(x - x0)n
For f(x) = xex, we have:
Evaluating each derivative at x0 = 0, we get:
Therefore, the 4th degree Taylor polynomial at x0 = 0 is:
T4(x) = 0 + 1(x - 0) + \frac{2}{2!}x2 + \frac{3}{3!}x3 + \frac{4}{4!}x4
Simplified, we get:
T4(x) = x + x2 + \frac{1}{2}x3 + \frac{1}{6}x4