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I am having trouble in finding the range of this function:-
$y = \frac{e^{x}}{1 + [x]}$
Additional details are:-
$x \geq0$
and [x] is the Greatest Integer function of x.

xyres
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2 Answers2

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Hint Restrict the range in $\{0\}\bigcup (n,n+1]$ where you know the values of $[x]$

In that way you find the range in each interval which it is

$$\{e^{0}\}\bigcup (\frac{e^n}{n+2},\frac{e^{n+1}}{n+2}]$$ Which is not a very good from, now observe that $$(\frac{e^n}{n+2},\frac{e^{n+1}}{n+2}]\bigcup (\frac{e^{n+1}}{n+3},\frac{e^{n+2}}{n+3}]= (\frac{e^n}{n+2},\frac{e^{n+2}}{n+3}]$$ Indeed you only need to prove that \begin{align} \frac{e^{n+1}}{n+3}<\frac{e^{n+1}}{n+2}\\ \frac{e^{n+1}}{n+3}>\frac{e^{n}}{n+2}\\\frac{e^{n+1}}{n+3}<\frac{e^{n+2}}{n+3} \end{align} The first and the third are obvious, and the second follows from the monotonicity of $$f(x)=\frac{e^x}{x+2}, \mathrm{\, since \,}f'>0$$. What that implies for the union of all intervals?

clark
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Hint : Find $y'$ and prove $y'>0 \ ; \ \forall x\ge0$

Proving this proves that it's value keeps on increasing for all $x\ge0$ and so range is $[1,\infty)$

ABC
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  • But the function has discontinuities? So $y'$ doesn't exist in all points. – Jeppe Stig Nielsen Mar 31 '13 at 16:28
  • @JeppeStigNielsen I had read it as x itself. Now I have edited answer. So, range$\in[1,\infty)$ . – ABC Mar 31 '13 at 16:46
  • @exploringnet To be sure I understand you argument if $f(x)>g(x)$ that implies $g([0,\infty)) \subset f([0,\infty))$ – clark Mar 31 '13 at 16:49
  • @clark Hmm... Yes i made this assertion that if $f(x)>g(x) \forall x\ge 0; \ Range(g[0,\infty))\in \ Range(f[0,\infty))$ . Please tell if it is wrong, I'm also a student though. – ABC Mar 31 '13 at 16:54
  • @exploringnet for example take $f(x)=[x]+5$ and $g(x)=x$,or take the function $f(x)=1, g(x)=0$ – clark Mar 31 '13 at 16:58
  • @clark So, do we have to consider that the functions must have common origin so that at least some minimum values are not lost? Or this is total FAIL? – ABC Mar 31 '13 at 17:03
  • Though , It would work for this question. I hope. – ABC Mar 31 '13 at 17:12
  • @exploringnet the biggest problem is that $\frac{e^x}{1+[x]}$ is not continuous so you cannot know if its range is an interval. What you say would have worked greatly if $f(x)>g(x), f(0)=g(0)$ and both $f,g$ are continuous. if $f$ is not continuous it can very few values and still manage to be above other continuous functions. – clark Mar 31 '13 at 17:12
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    done! understood the mistake.I am now not going to delete this answer because of this useful discussion which would help future visitors – ABC Mar 31 '13 at 17:16
  • I think your answer before the edit was better. Your hint was to differentiate the function. After differentiating it I found that it is an increasing function. Therefore its range can be obtained at the extreme points of the interval. – xyres Mar 31 '13 at 17:24
  • @weasel I understood your way. At least my 1st solution provided a standard way or the first way to tackle these problems.Which seems to be increasing till $\infty$ – ABC Mar 31 '13 at 17:28