Answer
Taylor`s Series: $e^{3x}=\dfrac{1}{e^3}\Sigma_{n = 0}^{\infty} \dfrac{3^n(x+1)^n}{n!}$ center at $c=-1$ with convergence interval $(-\infty,\infty)$.
Work Step by Step
Taylor`s Series: $T(x)=\Sigma_{n = 0}^{\infty} \dfrac{f^n (c)}{n!}(x-c)^n$
If we consider $c=0$, then $T(x)$ is also known as Maclaurin Series.
We have the Maclaurin series for $f(x)$ as follows:
$f(x)=f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0) x^2}{2!}+\dfrac{f^{\prime \prime \prime}(0) x^3}{3!}+\dfrac{f^{\prime \prime \prime \prime}(0) x^4}{4!}+...$
From Table $2$, we notice $f(x)=e^x=\Sigma_{n = 0}^{\infty}(-1)^n\dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+.....$; converges for $x$
Now, Taylor`s Series becomes: $f(x)=e^{-3}e^{3(x+1)}=\dfrac{1}{e^3}\Sigma_{n = 0}^{\infty} \dfrac{(3(x+1))^n}{n!}$; converges for $x$
So, we get Taylor`s Series: $e^{3x}=\dfrac{1}{e^3}\Sigma_{n = 0}^{\infty} \dfrac{3^n(x+1)^n}{n!}$ center at $c=-1$ with convergence interval $(-\infty,\infty)$.
