The Taylor polynomial of order 3 for xe at 0x = is

It seems like there's a bit of confusion in your question. If you're asking for the Taylor polynomial of order 3 for the function f(x) = xe^x at x = 0, here's how you can calculate it:

The Taylor series of a function f(x) about x = a is given by:

P(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

The derivatives of f(x) = xe^x are:

f'(x) = e^x + xe^x, f''(x) = 2e^x + xe^x, f'''(x) = 3e^x + xe^x.

At x = 0, these become:

f(0) = 0, f'(0) = 1, f''(0) = 2, f'''(0) = 3.

So, the Taylor polynomial of order 3 for f(x) = xe^x at x = 0 is:

P(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! = 0 + 1x + 2x^2/2 + 3*x^3/6 = x + x^2 + x^3/2.