Answer
The Maclaurin series for $f(x)=e^{4x}$ is convergent for any value of $x$.
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=e^x$ as follows
$$f(x)=e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$
Now by the comparison with the function $f(x)=e^{4x}$, we have the Maclaurin series as follows
$$f(x)=e^{4x}=1+4x+\frac{(4x)^2}{2!}+\frac{(4x)^3}{3!}+\frac{(4x)^4}{4!}+...$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=e^x$ is convergent for any value of $x$. Hence the Maclaurin series for $f(x)=e^{4x}$ is also convergent for any value of $x$.
