Answer
$f(x) = 0 + x + \frac{2}{2!}x^{2} + \frac{3}{3!}x^{3} + \frac{4}{4!}x^{4}$
Work Step by Step
$f(x) = xe^{x}$
$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} + ... + \frac{f^{n}(0)}{n!}x^{n}$
$f(0) = 0e^{0} = 0$
$f'(0) = (0+1)e^{0} =1$
$f''(0) = (0+2)e^{0} =2$
$f'''(0) = (0+3)e^{0} =3$
$f''''(0) = (0+4)e^{0} =4$
Maclaurin polynomial for $n=4$ is
$f(x) = 0 + x + \frac{2}{2!}x^{2} + \frac{3}{3!}x^{3} + \frac{4}{4!}x^{4}$
