Extremal values/asymptotes

We have 
\[f=e^{-2x}(2x)^{-1}.\]
Note that $f(x) \neq0$. By product rule 
\[f'=-2 e^{-2x}(2x)^{-1}- 2e^{-2x}(2x)^{-2}=-2 e^{-2x}(2x)^{-1}(1+(2x)^{-1})\]
\[=-2f(1+(2x)^{-1})=0\]
\[ \Rightarrow (1+(2x)^{-1})=0    \Rightarrow (2x)^{-1}=-1  \Rightarrow 2x=-1\]
\[\Rightarrow x=-\frac{1}{2}   \text{is the only extremal point}.\]

To compute $f''$, note that 
\[f'=-2f(x)(1+(2x)^{-1})  \Rightarrow f'' =-2 f'(x)(1+(2x)^{-1}) -2f(x)[-2 (2x)^{-2}].\]
Hence
\[f''(-\frac{1}{2})=-2 f'(-\frac{1}{2})(1+(2(-\frac{1}{2}))^{-1}) -2f(-\frac{1}{2})[-2 (2(-\frac{1}{2}))^{-2}]\]
\[=0+8f(-\frac{1}{2})(-\frac{1}{2})^2=2f(-\frac{1}{2})=2 e^{1}(-1)=-2e<0,\]
since $f'(-\frac{1}{2})=0.$ Thus $x=-\frac{1}{2}$ is a local maximum point. 

Note that 
\[\lim_{x\rightarrow \infty} \frac{e^{-2x}}{2x}=0,\]
hence $x=0$ is a horizental asymptote. 

Also 
\[\lim_{x\rightarrow 0^{\pm}} \frac{e^{-2x}}{2x}=\pm \infty,\] hence $y=0$ is a veritical asymptote.