Past exam of the mathematical methods II course of the University of Oxford / 2019 / 1 / Solution

The successive derivatives are $f^{(n)}(x)=(-6{)}^n{e}^{-6x}$, so $f^{(n)}(-4)=(-6{)}^n{e}^{24}$. Substitution in the Taylor series formula about $a=-4$ givesThe ratio test shows convergence for every finite real or complex $x$: the absolute ratio of successive terms is $6|x+4|/(n+1)\to 0$. This is also the usual exponential function series applied to $e^{24}{e}^{-6(x+4)}$, proving that the series equals the stated function everywhere.