Answer
$1, 2, 2, \frac{4}{3}$
Work Step by Step
We are given:
$f(x) = e^{2x}$, $f(0) = e^{0} = 1$
Calculate the first three derivatives and their value at $x=0$:
$f'(x) = 2e^{2x}$,
$f'(0) = 2e^{0} = 2$
$f''(x) = 2(2e^{2x})=4e^{2x}$,
$f'(0) = 4e^{0} = 4$
$f''''(x) = 4(2e^{2x})=8e^{2x}$,
$f'''(0) = 8e^{0} = 8$
Write the third Maclaurin polynomial:
$p_n(x)=\displaystyle\sum_{n=0}^N\frac{f^{(n)}(0)}{n!}x^n$
$p_3(x)=f(0)+\frac{f''(0)}{1!}+\frac{f'''(0)}{2!}+\frac{f''''(0)}{3!}$
$=1+\frac{2}{1}x+\frac{4}{2}x^2+\frac{8}{6}x^3$
$=1+2x+2x^2+\frac{4}{3}x^3$
