In the Maclaurin series expansion of $f(x) = e^{3x}$, the coefficient of $x^3$ is equal to ________ (round off to 1 decimal place).
Answer Key:
4.5
Step-by-Step Solution & Explanation
The Maclaurin series for $e^u$ is:
$e^u = \sum_{n=0}^\infty \frac{u^n}{n!} = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$
Substituting $u = 3x$:
$e^{3x} = 1 + (3x) + \frac{(3x)^2}{2} + \frac{(3x)^3}{6} + \dots$
The term containing $x^3$ is:
$\frac{27x^3}{6} = \frac{9}{2}x^3 = 4.5 x^3$
Hence, the coefficient of $x^3$ is $4.5$.