Answer
$f(x) = 1 - \frac{x}{2} + \frac{1}{4 \times 2!} x^{2} - \frac{1}{8 \times 3!} x^{3} + \frac{1}{16 \times 4!}x^{4}$
Work Step by Step
$f(x) = e^{\frac{-x}{2}}$
$f(x) = f(0) + f'(0)x + \frac{f^{''}(0)}{2!} x^{2} + \frac{f^{'''}(0)}{3!} x^{3} + ... + \frac{f^{n}(0)}{n!}x^{n}$
$f(0) = e^{0} = 1$
$f^{'}(0) = \frac{-1}{2}e^{0} = \frac{-1}{2}$
$f^{''}(0) = \frac{1}{4}e^{0} = \frac{1}{4}$
$f^{'''}(0) = \frac{-1}{8}e^{0} = \frac{-1}{8}$
$f^{''''}(0) = \frac{1}{16}e^{0} = \frac{1}{16}$
Maclaurin polynomial for $n=4$ is
$f(x) = 1 - \frac{x}{2} + \frac{1}{4 \times 2!} x^{2} - \frac{1}{8 \times 3!} x^{3} + \frac{1}{16 \times 4!}x^{4}$
