4. try again get help practice similar attempt 3: 1 out of 2 parts have been answered correctly. 7 attempts…

Explanation:

Step1: Recall Maclaurin series formula

The Maclaurin series of a function $f(x)$ is given by $\sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}$, where $f^{(n)}(x)$ is the $n$-th derivative of $f(x)$.

Step2: Find derivatives of $f(x)=e^{3x}$

Let $f(x)=e^{3x}$. Then $f^{(n)}(x)=3^{n}e^{3x}$ for $n = 0,1,2,\cdots$. Evaluating at $x = 0$, we have $f^{(n)}(0)=3^{n}e^{0}=3^{n}$.

Step3: Write out the Maclaurin series

Substitute $f^{(n)}(0)=3^{n}$ into the Maclaurin - series formula: $\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}=\sum_{n = 0}^{\infty}\frac{3^{n}}{n!}x^{n}$.

Step4: Find the interval of convergence

Use the ratio - test. Let $a_{n}=\frac{3^{n}x^{n}}{n!}$. Then $\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\lim_{n\rightarrow\infty}\left|\frac{\frac{3^{n+1}x^{n + 1}}{(n+1)!}}{\frac{3^{n}x^{n}}{n!}}\right|=\lim_{n\rightarrow\infty}\left|\frac{3x}{n + 1}\right| = 0$ for all $x\in(-\infty,\infty)$. So the interval of convergence is $(-\infty,\infty)$.

Answer:

The Maclaurin series for $f(x)=e^{3x}$ is $\sum_{n = 0}^{\infty}\frac{3^{n}x^{n}}{n!}$ and the interval of convergence is $(-\infty,\infty)$