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I find it really hard to find the range. I usually substitute the x's with y and then solve for y, but it does not always work for me. Do you have any advice?

Function in question:

$$f(x) = \frac{e^{-2x}}{x}$$

Vizag
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tyui
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  • The function I'm working on right now is f(x) = (e^(-2x)) / x. – tyui Jun 09 '19 at 19:16
  • What are the rules of the game? Is graphing allowed? Can you use a derivative? (that would make it lots easier).Any other pertinent info? – imranfat Jun 09 '19 at 19:21
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    There's no easy trick. Graph it - either by computer/calculator or by hand. You can use calculus to find minima/maxima and where it is increasing/decreasing. – Jair Taylor Jun 09 '19 at 19:24
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    Think also about the limits as $x$ approaches $\pm \infty$ or any singularities. – Jair Taylor Jun 09 '19 at 19:26
  • The only rule is that we cannot use any aid such as calculators or computers. Everything else should be okay! – tyui Jun 09 '19 at 19:28

2 Answers2

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Since $\lim_{x\to\infty}f(x)=0$, since $\lim_{x\to0^+}f(x)=-\infty$, and since $x>0\implies f(x)>0$, it follows from the intermediate value theorem that $f\bigl((0,\infty)\bigr)=(0,\infty)$.

On the other hand, $f|_{(-\infty,0)}$ attains its maximum at $-\frac12$ and it takes the value $-2e$ there. Furtermore,$$\lim_{x\to-\infty}f(x)=\lim_{x\to0^-}f(x)=-\infty.$$So, $f\bigl((0,\infty)\bigr)=(-\infty,-2e)$. So, the range of $f$ is $(-\infty,-2e)\cup(0,\infty)$.

José Carlos Santos
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1

Hint:

The domain of the function is $D_f=\mathbf R^*$. Its derivative is $f'(x)=-\dfrac{(2x+1)\mathrm e^{-2x}}{x^2}$. It has the opposite sign of the sign of $2x+1$, hence the function is

  • increasing on the interval $\bigl(-\infty,-\frac12\bigr]$,
  • decreasing on $\bigl[-\frac12, 0\bigr)$ and on $(0,+\infty]$.

Thus $f$ has a local maximum, $-2\mathrm e$, at $x=-\frac12$.

There remains to calculate the limits at $0, -\infty$ and $+\infty$; Can you finish the computations?

Bernard
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