Find the first four terms in the Maclaurin series for $f ( x ) = x e ^ { - x }$ .
A) $x - x ^ { 2 } + x ^ { 3 } - x ^ { 4 }$
B) $x - \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }$
C) $x - x ^ { 2 } + \frac { 1 } { 2 } x ^ { 3 } - \frac { 1 } { 6 } x ^ { 4 }$
D) $x - 2 x ^ { 2 } + 3 x ^ { 3 } - 4 x ^ { 4 }$
E) $x + \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 6 } x ^ { 3 } + \frac { 1 } { 24 } x ^ { 4 }$
F) $x + x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } + \frac { 1 } { 8 } x ^ { 4 }$
G) $x + x ^ { 2 } + \frac { 1 } { 2 } x ^ { 3 } + \frac { 1 } { 6 } x ^ { 4 }$
H)$\frac { 1 } { 2 } x - \frac { 1 } { 6 } x ^ { 2 } + \frac { 1 } { 24 } x ^ { 3 } - \frac { 1 } { 120 } x ^ { 4 }$

Question

Find the first four terms in the Maclaurin series for $f ( x ) = x e ^ { - x }$ .
A) $x - x ^ { 2 } + x ^ { 3 } - x ^ { 4 }$ 
B) $x - \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }$ 
C) $x - x ^ { 2 } + \frac { 1 } { 2 } x ^ { 3 } - \frac { 1 } { 6 } x ^ { 4 }$ 
D) $x - 2 x ^ { 2 } + 3 x ^ { 3 } - 4 x ^ { 4 }$ 
E) $x + \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 6 } x ^ { 3 } + \frac { 1 } { 24 } x ^ { 4 }$ 
F) $x + x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } + \frac { 1 } { 8 } x ^ { 4 }$ 
G) $x + x ^ { 2 } + \frac { 1 } { 2 } x ^ { 3 } + \frac { 1 } { 6 } x ^ { 4 }$ 
H)$\frac { 1 } { 2 } x - \frac { 1 } { 6 } x ^ { 2 } + \frac { 1 } { 24 } x ^ { 3 } - \frac { 1 } { 120 } x ^ { 4 }$

Quizplus · Accepted Answer

The Maclaurin series for $f(x) = xe^{-x}$ can be found by taking derivatives of $f(x)$ and evaluating them at $x=0$, then applying the formula for the Maclaurin series. The correct series expansion up to the $x^4$ term is $x - x^2 + \frac{1}{2}x^3 - \frac{1}{6}x^4$, which matches option C.

Find the First Four Terms in the Maclaurin Series For f