$Extremal values/asymptotes$

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6. Consider the function $f(x) = (e^{?2x})/(2x)$ defined in all of R. Find and classify the extremal values of f. Does f have any asymptotes? If so, find them.

Calculus

Stickix02

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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.

Nirenberg

Nirenberg

Please leave a comment if you need any clarifications.
Stickix02

When evaluating the dobble derivative it seem like you are leaving the (-) out before 2 which i believe gives 2e greater than 0 instead of -2e less than zero. If so, wouldn't it be a global max then?
Nirenberg

No, I doubled checked my computations and there seems to be no mistake. I also graphed the function, and the critical point is indeed a maximum point.
Stickix02

Perfect, thank you so much. I believe i forgot that (2x)^-2 equals 1/4x^2

$Extremal values/asymptotes$

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